The North Alpine foreland basin (NAFB) in Germany is characterized by various types of sedimentologic features that make it an excellent terrestrial analog of regions affected by high-energy asteroid impact-quakes on Mars. Impact events have shaped all planetary bodies in the inner Solar System over the past >4 Gyr. The well-preserved Ries impact crater (Baden-Württemberg and Bavaria), formed around 14.8 Ma, has recently been linked to an earthquake-produced seismite horizon in Mid-Miocene NAFB sediments that exhibits typical dewatering structures and is associated with sand spikes, seismically produced pin-shaped pseudo-concretions. In this terrestrial setting, the sand spike tails systematically point away from the Ries crater. On its path across Gale Crater, the Mars rover Curiosity seems to have observed a similar seismite horizon in early Hesperian lacustrine deposits including clastic dikes, convolute bedding, and, likely, sand spikes. Their orientation suggests that the nearby Slagnos impact crater might be the seismic source for the formation of those seismites. The Ries impact–seismite deposits can be traced over a distance of more than 200 km from the source crater (northern Switzerland), which makes the NAFB an excellent terrestrial analog for similar deposits and their sedimentologic inventory within Gale Crater's lake deposits on Mars.
Citation: Elmar Buchner, Volker J Sach, Martin Schmieder. Ries impact deposits in the North Alpine Foreland Basin of Germany as a terrestrial analog site for impact-produced seismites and sand spikes on planet Mars[J]. AIMS Geosciences, 2025, 11(1): 68-90. doi: 10.3934/geosci.2025005
[1] |
Zhi-Wei Sun .
New series for powers of |
[2] | Harman Kaur, Meenakshi Rana . Congruences for sixth order mock theta functions λ(q) and ρ(q). Electronic Research Archive, 2021, 29(6): 4257-4268. doi: 10.3934/era.2021084 |
[3] |
Jorge Garcia Villeda .
A computable formula for the class number of the imaginary quadratic field |
[4] | Fedor Petrov, Zhi-Wei Sun . Proof of some conjectures involving quadratic residues. Electronic Research Archive, 2020, 28(2): 589-597. doi: 10.3934/era.2020031 |
[5] |
Jin-Yun Guo, Cong Xiao, Xiaojian Lu .
On |
[6] | Dušan D. Repovš, Mikhail V. Zaicev . On existence of PI-exponents of unital algebras. Electronic Research Archive, 2020, 28(2): 853-859. doi: 10.3934/era.2020044 |
[7] |
Victor J. W. Guo .
A family of |
[8] | Chen Wang . Two congruences concerning Apéry numbers conjectured by Z.-W. Sun. Electronic Research Archive, 2020, 28(2): 1063-1075. doi: 10.3934/era.2020058 |
[9] | Dmitry Krachun, Zhi-Wei Sun . On sums of four pentagonal numbers with coefficients. Electronic Research Archive, 2020, 28(1): 559-566. doi: 10.3934/era.2020029 |
[10] | Hai-Liang Wu, Zhi-Wei Sun . Some universal quadratic sums over the integers. Electronic Research Archive, 2019, 27(0): 69-87. doi: 10.3934/era.2019010 |
The North Alpine foreland basin (NAFB) in Germany is characterized by various types of sedimentologic features that make it an excellent terrestrial analog of regions affected by high-energy asteroid impact-quakes on Mars. Impact events have shaped all planetary bodies in the inner Solar System over the past >4 Gyr. The well-preserved Ries impact crater (Baden-Württemberg and Bavaria), formed around 14.8 Ma, has recently been linked to an earthquake-produced seismite horizon in Mid-Miocene NAFB sediments that exhibits typical dewatering structures and is associated with sand spikes, seismically produced pin-shaped pseudo-concretions. In this terrestrial setting, the sand spike tails systematically point away from the Ries crater. On its path across Gale Crater, the Mars rover Curiosity seems to have observed a similar seismite horizon in early Hesperian lacustrine deposits including clastic dikes, convolute bedding, and, likely, sand spikes. Their orientation suggests that the nearby Slagnos impact crater might be the seismic source for the formation of those seismites. The Ries impact–seismite deposits can be traced over a distance of more than 200 km from the source crater (northern Switzerland), which makes the NAFB an excellent terrestrial analog for similar deposits and their sedimentologic inventory within Gale Crater's lake deposits on Mars.
The classical rational Ramanujan-type series for
∞∑k=0bk+cmka(k)=λ√dπ,(∗) |
where
(2kk)3, (2kk)2(3kk), (2kk)2(4k2k), (2kk)(3kk)(6k3k). |
In 1997 Van Hamme [47] conjectured that such a series
p−1∑k=0bk+cmka(k)≡cp(εddp) (mod p3), |
where
∑pn−1k=0(21k+8)(2kk)3−p∑n−1k=0(21k+8)(2kk)3(pn)3(2nn)3∈Zp, |
where
During the period 2002–2010, some new Ramanujan-type series of the form
∞∑n=05n+164nDn=8√3π, |
where
p−1∑k=05k+164kDk≡p(p3) (mod p3)for any prime p>3. |
The author [45,Conjecture 77] conjectured further that
1(pn)3(pn−1∑k=05k+164kDk−(p3)pn−1∑k=05k+164rDk)∈Zp |
for each odd prime
Let
Tn(b,c)=⌊n/2⌋∑k=0(n2k)(2kk)bn−2kck=⌊n/2⌋∑k=0(nk)(n−kk)bn−2kck. |
Note also that
T0(b,c)=1, T1(b,c)=b, |
and
(n+1)Tn+1(b,c)=(2n+1)bTn(b,c)−n(b2−4c)Tn−1(b,c) |
for all
For
Pn(x):=n∑k=0(nk)(n+kk)(x−12)k. |
It is well-known that if
Tn(b,c)=(√b2−4c)nPn(b√b2−4c)for all n∈N. |
Via the Laplace-Heine asymptotic formula for Legendre polynomials, for any positive real numbers
Tn(b,c)∼(b+2√c)n+1/224√c√nπas n→+∞ |
(cf. [40]). For any real numbers
lim |
In 2011, the author posed over 60 conjectural series for
Type Ⅰ.
Type Ⅱ.
Type Ⅲ.
Type Ⅳ.
Type Ⅴ.
Type Ⅵ.
Type Ⅶ.
In general, the corresponding
\sum\limits_{k = 0}^\infty \frac{3990k+1147}{(-288)^{3k}}T_k(62,95^2)^3 = \frac{432}{95\pi}(94\sqrt2+195\sqrt{14}) |
as well as its
\sum\limits_{k = 0}^{p-1} \frac{3990k+1147}{(-288)^{3k}}T_k(62,95^2)^3{\equiv} \frac p{19} \left(4230 \left( \frac{-2}p \right)+17563 \left( \frac{-14}p \right) \right)\ ({\rm{mod}}\ {p^2}), |
where
In 1905, J. W. L. Glaisher [15] proved that
\sum\limits_{k = 0}^\infty \frac{(4k-1) \binom{2k}k^4}{(2k-1)^4256^k} = - \frac 8{\pi^2}. |
This actually follows from the following finite identity observed by the author [38]:
\sum\limits_{k = 0}^n \frac{(4k-1) \binom{2k}k^4}{(2k-1)^4256^k} = -(8n^2+4n+1) \frac{ \binom{2n}n^4}{256^n}\quad\ \text{for all}\ n\in{\Bbb N}. |
Motivated by Glaisher's identity and Ramanujan-type series for
Theorem 1.1. We have the following identities:
\begin{align} \sum\limits_{k = 0}^\infty \frac{k(4k-1) \binom{2k}k^3}{(2k-1)^2(-64)^k}& = - \frac1{\pi}, \end{align} | (1.1) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(4k-1) \binom{2k}k^3}{(2k-1)^3(-64)^k}& = \frac2{\pi}, \end{align} | (1.2) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(12k^2-1) \binom{2k}k^3}{(2k-1)^2 256^k}& = - \frac2{\pi}, \end{align} | (1.3) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{k(6k-1) \binom{2k}k^3}{(2k-1)^3256^k}& = \frac1{2\pi}, \end{align} | (1.4) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(28k^2-4k-1) \binom{2k}k^3}{(2k-1)^2(-512)^k}& = - \frac{3\sqrt2}{\pi}, \end{align} | (1.5) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(30k^2+3k-2) \binom{2k}k^3}{(2k-1)^3(-512)^k}& = \frac{27\sqrt2}{8\pi}, \end{align} | (1.6) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(28k^2-4k-1) \binom{2k}k^3}{(2k-1)^2 4096^k}& = - \frac3{\pi}, \end{align} | (1.7) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(42k^2-3k-1) \binom{2k}k^3}{(2k-1)^3 4096^k}& = \frac{27}{8\pi}, \end{align} | (1.8) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(34k^2-3k-1) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-192)^k}& = - \frac{10}{\sqrt3\,\pi}, \end{align} | (1.9) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(64k^2-11k-7) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-192)^k}& = - \frac{125\sqrt3}{9\pi}, \end{align} | (1.10) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(14k^2+k-1) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)216^k}& = - \frac{\sqrt3}{\pi}, \end{align} | (1.11) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(90k^2+7k+1) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)216^k}& = \frac{9\sqrt3}{2\pi}, \end{align} | (1.12) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(34k^2-3k-1) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-12)^{3k}}& = - \frac{2\sqrt3}{\pi}, \end{align} | (1.13) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(17k+5) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-12)^{3k}}& = \frac{9\sqrt3}{\pi}, \end{align} | (1.14) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(111k^2-7k-4) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)1458^k}& = - \frac{45}{4\pi}, \end{align} | (1.15) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(1524k^2+899k+263) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)1458^k}& = \frac{3375}{4\pi}, \end{align} | (1.16) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(522k^2-55k-13) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-8640)^k}& = - \frac{54\sqrt{15}}{5\pi}, \end{align} | (1.17) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(1836k^2+2725k+541) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-8640)^k}& = \frac{2187\sqrt{15}}{5\pi}, \end{align} | (1.18) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(529k^2-45k-16) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)15^{3k}}& = - \frac{55\sqrt3}{2\pi}, \end{align} | (1.19) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(77571k^2+68545k+16366) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)15^{3k}}& = \frac{59895\sqrt3}{2\pi}, \end{align} | (1.20) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(574k^2-73k-11) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-48)^{3k}}& = -20 \frac{\sqrt3}{\pi}, \end{align} | (1.21) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(8118k^2+9443k+1241) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-48)^{3k}}& = \frac{2250\sqrt3}{\pi}, \end{align} | (1.22) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(978k^2-131k-17) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-326592)^k}& = - \frac{990\sqrt7}{49\pi}, \end{align} | (1.23) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(592212k^2+671387k^2+77219) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-326592)^k}& = \frac{4492125\sqrt7}{49\pi}, \end{align} | (1.24) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(116234k^2-17695k-1461) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)(-300)^{3k}}& = -2650 \frac{\sqrt3}{\pi}, \end{align} | (1.25) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(223664832k^2+242140765k+18468097) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)(-300)^{3k}}& = 33497325 \frac{\sqrt3}{\pi}, \end{align} | (1.26) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(122k^2+3k-5) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)648^k}& = - \frac{21}{2\pi}, \end{align} | (1.27) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(1903k^2+114k+41) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)648^k}& = \frac{343}{2\pi}, \end{align} | (1.28) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(40k^2-2k-1) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-1024)^k}& = - \frac4{\pi}, \end{align} | (1.29) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(8k^2-2k-1) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-1024)^k}& = - \frac{16}{5\pi}, \end{align} | (1.30) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(176k^2-6k-5) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)48^{2k}}& = -8 \frac{\sqrt3}{\pi}, \end{align} | (1.31) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(208k^2+66k+23) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)48^{2k}}& = \frac{128}{\sqrt3\,\pi}, \end{align} | (1.32) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(6722k^2-411k-152) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-63^2)^k}& = -195 \frac{\sqrt7}{\pi}, \end{align} | (1.33) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(281591k^2-757041k-231992) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-63^2)^k}& = -274625 \frac{\sqrt7}{\pi}, \end{align} | (1.34) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(560k^2-42k-11) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)12^{4k}}& = -24 \frac{\sqrt2}{\pi}, \end{align} | (1.35) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(112k^2+114k+23) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)12^{4k}}& = \frac{256\sqrt2}{5\pi}, \end{align} | (1.36) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(248k^2-18k-5) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-3\times2^{12})^k}& = - \frac{28}{\sqrt3\,\pi}, \end{align} | (1.37) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(680k^2+1482k+337) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-3\times2^{12})^k}& = \frac{5488\sqrt3}{9\pi}, \end{align} | (1.38) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(1144k^2-102k-19) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-2^{10}3^4)^k}& = - \frac{60}{\pi}, \end{align} | (1.39) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(3224k^2+4026k+637) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-2^{10}3^4)^k}& = \frac{2000}{\pi}, \end{align} | (1.40) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(7408k^2-754k-103) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)28^{4k}}& = - \frac{560\sqrt3}{3\pi}, \end{align} | (1.41) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(3641424k^2+4114526k+493937) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)28^{4k}}& = 896000 \frac{\sqrt3}{\pi}, \end{align} | (1.42) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(4744k^2-534k-55) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-2^{14}3^45)^k}& = - \frac{1932\sqrt5}{25\pi}, \end{align} | (1.43) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(18446264k^2+20356230k+1901071) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-2^{14}3^45)^k}& = 66772496 \frac{\sqrt5}{25\pi}, \end{align} | (1.44) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(413512k^2-50826k-3877) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)(-2^{10}21^4)^k}& = - \frac{12180}{\pi}, \end{align} | (1.45) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(1424799848k^2+1533506502k+108685699) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)(-2^{10}21^4)^k}& = \frac{341446000}{\pi}, \end{align} | (1.46) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(71312k^2-7746k-887) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)1584^{2k}}& = -840 \frac{\sqrt{11}}{\pi}, \end{align} | (1.47) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(50678512k^2+56405238k+5793581) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)1584^{2k}}& = 5488000 \frac{\sqrt{11}}{\pi}, \end{align} | (1.48) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(7329808k^2-969294k-54073) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)396^{4k}}& = -120120 \frac{\sqrt2}{\pi}, \end{align} | (1.49) |
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^\infty \frac{(2140459883152k^2+2259867244398k+119407598201) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)396^{4k}} \\&\qquad\qquad\qquad = 44\times1820^3 \frac{\sqrt2}{\pi}, \end{aligned} \end{equation} | (1.50) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(164k^2-k-3) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)20^{3k}}& = - \frac{7\sqrt5}{2\pi}, \end{align} | (1.51) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(2696k^2+206k+93) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)20^{3k}} & = \frac{686}{\sqrt5\,\pi}, \end{align} | (1.52) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(220k^2-8k-3) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-2^{15})^k} & = - \frac{7\sqrt2}{\pi}, \end{align} | (1.53) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(836k^2-1048k-309) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-2^{15})^k} & = - \frac{686\sqrt2}{\pi}, \end{align} | (1.54) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(504k^2-11k-8) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-15)^{3k}} & = - \frac{9\sqrt{15}}{\pi}, \end{align} | (1.55) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(189k^2-11k-8) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-15)^{3k}}& = - \frac{243\sqrt{15}}{35\pi}, \end{align} | (1.56) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(516k^2-19k-7) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(2\times30^3)^k} & = - \frac{11\sqrt{15}}{2\pi}, \end{align} | (1.57) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(3237k^2+1922k+491) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(2\times30^3)^k} & = \frac{3993\sqrt{15}}{10\pi}, \end{align} | (1.58) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(684k^2-40k-7) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-96)^{3k}}& = - \frac{9\sqrt6}{\pi}, \end{align} | (1.59) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(2052k^2+2536k+379) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-96)^{3k}}& = \frac{486\sqrt6}{\pi}, \end{align} | (1.60) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(2556k^2-131k-29) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)66^{3k}}& = - \frac{63\sqrt{33}}{4\pi}, \end{align} | (1.61) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(203985k^2+212248k+38083) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)66^{3k}}& = \frac{83349\sqrt{33}}{4\pi}, \end{align} | (1.62) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(5812k^2-408k-49) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-3\times160^3)^k}& = - \frac{253\sqrt{30}}{9\pi}, \end{align} | (1.63) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(3471628k^2+3900088k+418289) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-3\times160^3)^k}& = \frac{32388554\sqrt{30}}{135\pi}, \end{align} | (1.64) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(35604k^2-2936k-233) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-960)^{3k}}& = -189 \frac{\sqrt{15}}{\pi}, \end{align} | (1.65) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(13983084k^2+15093304k+1109737) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-960)^{3k}}& = \frac{4500846\sqrt{15}}{5\pi}, \end{align} | (1.66) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(157752k^2-11243k-1304) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)255^{3k}}& = - \frac{513\sqrt{255}}{2\pi}, \end{align} | (1.67) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(28240947k^2+31448587k+3267736) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)255^{3k}}& = \frac{45001899\sqrt{255}}{70\pi}, \end{align} | (1.68) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{(2187684k^2-200056k-11293) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-5280)^{3k}} & = -1953 \frac{\sqrt{330}}{\pi}, \end{align} | (1.69) |
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^\infty \frac{(101740699836k^2+107483900696k+5743181813) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-5280)^{3k}} \\&\qquad\qquad\qquad = \frac{4966100118\sqrt{330}}{5\pi}, \end{aligned} \end{equation} | (1.70) |
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^\infty \frac{(16444841148k^2-1709536232k-53241371) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-640320)^{3k}} \\&\qquad\qquad\qquad = -1672209 \frac{\sqrt{10005}}{\pi}, \end{aligned} \end{equation} | (1.71) |
and
\begin{equation} \begin{aligned} \sum\limits_{k = 0}^\infty \frac{P(k) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-640320)^{3k}} = \frac{18\times557403^3\sqrt{10005}}{5\pi}, \end{aligned} \end{equation} | (1.72) |
where
\begin{align*} P(k): = &637379600041024803108 k^2 + 657229991696087780968 k \\&+ 19850391655004126179. \end{align*} |
Recall that the Catalan numbers are given by
C_n: = \frac{ \binom{2n}n}{n+1} = \binom{2n}n- \binom{2n}{n+1}\ \ (n\in{\Bbb N}). |
For
\frac{ \binom{2k}k}{2k-1} = \begin{cases}-1& \text{if}\ k = 0,\\2C_{k-1}& \text{if}\ k > 0.\end{cases} |
Thus, for any
\begin{align*} \sum\limits_{k = 0}^\infty \frac{(ak^2+bk+c) \binom{2k}k^3}{(2k-1)^3m^k} = &-c+\sum\limits_{k = 1}^\infty \frac{(ak^2+bk+c)(2C_{k-1})^3}{m^k} \\ = &-c+ \frac 8m\sum\limits_{k = 0}^\infty \frac{a(k+1)^2+b(k+1)+c}{m^k}C_k^3. \end{align*} |
For example, (1.2) has the equivalent form
\sum\limits_{k = 0}^\infty \frac{4k+3}{(-64)^k}C_k^3 = 8- \frac{16}{\pi}.\;\;\;\;\;\;\;\;(1.2') |
For any odd prime
\sum\limits_{k = 0}^{(p+1)/2} \frac{(4k-1) \binom{2k}k^3}{(2k-1)^3(-64)^k}{\equiv} p \left( \frac{-1}p \right)+p^3(E_{p-3}-2)\ ({\rm{mod}}\ {p^4}) |
(where
\sum\limits_{k = 0}^{(p-1)/2} \frac{4k+3}{(-64)^k}C_k^3{\equiv}8 \left(1-p \left( \frac{-1}p \right)-p^3(E_{p-3}-2) \right)\ ({\rm{mod}}\ {p^4}). |
Recently, C. Wang [50] proved that for any prime
\sum\limits_{k = 0}^{(p+1)/2} \frac{(3k-1) \binom{2k}k^3}{(2k-1)^216^k}{\equiv} p+2p^3 \left( \frac{-1}p \right)(E_{p-3}-3)\ ({\rm{mod}}\ {p^4}) |
and
\sum\limits_{k = 0}^{p-1} \frac{(3k-1) \binom{2k}k^3}{(2k-1)^216^k}{\equiv} p-2p^3\ ({\rm{mod}}\ {p^4}). |
(Actually, Wang stated his results only in the language of hypergeometric series.) These two congruences extend a conjecture of Guo and M. J. Schlosser [21].
We are also able to prove some other variants of Ramanujan-type series such as
\sum\limits_{k = 0}^\infty \frac{(56k^2+118k+61) \binom{2k}k^3}{(k+1)^24096^k} = \frac{192}{\pi} |
and
\sum\limits_{k = 0}^\infty \frac{(420k^2+992k+551) \binom{2k}k^3}{(k+1)^2(2k-1)4096^k} = - \frac{1728}{\pi}. |
Now we state our second theorem.
Theorem 1.2. We have the identities
\begin{align} \sum\limits_{k = 1}^\infty \frac{28k^2+31k+8}{(2k+1)^2k^3 \binom{2k}k^3}& = \frac{\pi^2-8}2, \end{align} | (1.73) |
\begin{align} \sum\limits_{k = 1}^\infty \frac{42k^2+39k+8}{(2k+1)^3k^3 \binom{2k}k^3}& = \frac{9\pi^2-88}2, \end{align} | (1.74) |
\begin{align} \sum\limits_{k = 1}^\infty \frac{(8k^2+5k+1)(-8)^k}{(2k+1)^2k^3 \binom{2k}k^3}& = 4-6G, \end{align} | (1.75) |
\begin{align} \sum\limits_{k = 1}^\infty \frac{(30k^2+33k+7)(-8)^k}{(2k+1)^3k^3 \binom{2k}k^3}& = 54G-52, \end{align} | (1.76) |
\begin{align} \sum\limits_{k = 1}^\infty \frac{(3k+1)16^k}{(2k+1)^2k^3 \binom{2k}k^3}& = \frac{\pi^2-8}2, \end{align} | (1.77) |
\begin{align} \sum\limits_{k = 1}^\infty \frac{(4k+1)(-64)^k}{(2k+1)^2k^2 \binom{2k}k^3}& = 4-8G, \end{align} | (1.78) |
\begin{align} \sum\limits_{k = 1}^\infty \frac{(4k+1)(-64)^k}{(2k+1)^3k^3 \binom{2k}k^3}& = 16G-16, \end{align} | (1.79) |
\begin{align} \sum\limits_{k = 1}^\infty \frac{(2k^2-11k-3)8^k}{(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k}& = \frac{48-5\pi^2}2, \end{align} | (1.80) |
\begin{align} \sum\limits_{k = 2}^\infty \frac{(178k^2-103k-39)8^k}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k}& = \frac{1125\pi^2-11096}{36}, \end{align} | (1.81) |
\begin{align} \sum\limits_{k = 1}^\infty \frac{(5k+1)(-27)^k}{(2k+1)(3k+1)k^2 \binom{2k}k^2 \binom{3k}k}& = 6-9K, \end{align} | (1.82) |
\begin{align} \sum\limits_{k = 2}^\infty \frac{(45k^2+5k-2)(-27)^{k-1}}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k}& = \frac{37-48K}{16}, \end{align} | (1.83) |
\begin{align} \sum\limits_{k = 1}^\infty \frac{(98k^2-21k-8)81^k}{(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}}& = 216-20\pi^2, \end{align} | (1.84) |
\begin{align} \sum\limits_{k = 2}^\infty \frac{(1967k^2-183k-104)81^k}{(k-1)(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}}& = \frac{20000\pi^2-190269}{120}, \end{align} | (1.85) |
\begin{align} \sum\limits_{k = 1}^\infty \frac{(46k^2+3k-1)(-144)^k}{(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}}& = 72- \frac{225}2K, \end{align} | (1.86) |
\begin{align} \sum\limits_{k = 2}^\infty \frac{(343k^2+18k-16)(-144)^k}{(k-1)(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}}& = \frac{9375K-7048}{10}, \end{align} | (1.87) |
where
G: = \sum\limits_{k = 0}^\infty \frac{(-1)^k}{(2k+1)^2}\ \ \mathit{\text{and}}\ \ K: = \sum\limits_{k = 0}^\infty \frac{( \frac k3)}{k^2}. |
For
(k-1)k \binom{2k}k = 2(2j+1)j \binom{2j}j. |
Thus, for any
\sum\limits_{j = 1}^\infty \frac{(aj^2+bj+c)m^j}{(2j+1)^3j^3 \binom{2j}j^3} = \frac 8m\sum\limits_{k = 2}^\infty \frac{(a(k-1)^2+b(k-1)+c)m^k}{(k-1)^3k^3 \binom{2k}k^3}. |
For example, (1.77) has the following equivalent form
\sum\limits_{k = 2}^\infty \frac{(2k-1)(3k-2)16^k}{(k-1)^3k^3 \binom{2k}k^3} = \pi^2-8.\;\;\;\;\;\;\;\;\;\;\;(1.77') |
In contrast with the Domb numbers, we introduce a new kind of numbers
S_n: = \sum\limits_{k = 0}^n \binom nk^2T_kT_{n-k}\ \ (n = 0,1,2,\ldots). |
The values of
1,\, 2,\, 10,\, 68,\, 586,\, 5252,\, 49204,\, 475400,\, 4723786,\, 47937812,\, 494786260 |
respectively. We may extend the numbers
S_n(b,c): = \sum\limits_{k = 0}^n \binom nk^2T_k(b,c)T_{n-k}(b,c)\ \ (n = 0,1,2,\ldots). |
Note that
Now we state our third theorem.
Theorem 1.3. We have
\begin{align} \sum\limits_{k = 0}^\infty \frac{7k+3}{24^k}S_k(1,-6)& = \frac{15}{\sqrt2\,\pi}, \end{align} | (1.88) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{12k+5}{(-28)^k}S_k(1,7)& = \frac{6\sqrt7}{\pi}, \end{align} | (1.89) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{84k+29}{80^k}S_k(1,-20)& = \frac{24\sqrt{15}}{\pi}, \end{align} | (1.90) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{3k+1}{(-100)^k}S_k(1,25)& = \frac{25}{8\pi}, \end{align} | (1.91) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{228k+67}{224^k}S_k(1,-56)& = \frac{80\sqrt7}{\pi}, \end{align} | (1.92) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{399k+101}{(-676)^k}S_k(1,169)& = \frac{2535}{8\pi}, \end{align} | (1.93) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{2604k+563}{2600^k}S_k(1,-650)& = \frac{850\sqrt{39}}{3\pi}, \end{align} | (1.94) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{39468k+7817}{(-6076)^k}S_k(1,1519)& = \frac{4410\sqrt{31}}{\pi}, \end{align} | (1.95) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{41667k+7879}{9800^k}S_k(1,-2450)& = \frac{40425\sqrt6}{4\pi}, \end{align} | (1.96) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{74613k+10711}{(-530^2)^k}S_k(1,265^2)& = \frac{1615175}{48\pi}. \end{align} | (1.97) |
Remark 1.1. The author found the 10 series in Theorem 1.3 in Nov. 2019.
We shall prove Theorems 1.1-1.3 in the next section. In Sections 3-10, we propose 117 new conjectural series for powers of
Type Ⅷ.
where
Unlike Ramanujan-type series given by others, all our series for
Motivated by the author's effective way to find new series for
Conjecture 1.1 (General Criterion for Rational Ramanujan-type Series for
\begin{equation} \sum\limits_{k = 0}^\infty \frac{bk+c}{m^k}a_k = \frac{\sum\limits_{i = 1}^r\lambda_i\sqrt{d_i}}{\pi} \end{equation} | (1.98) |
for some nonzero rational numbers
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{bk+c}{m^k}a_k{\equiv} p\left(\sum\limits_{i = 1}^rc_i \left( \frac{{\varepsilon}_id_i}p \right)+\sum\limits_{r < j \leq 3}c_j \left( \frac{d_j}p \right)\right) \ ({\rm{mod}}\ {p^2}), \end{equation} | (1.99) |
where
For a Ramanujan-type series of the form (1.98), we call
Conjecture 1.2. Let
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{bk+c}{m^k}a_k{\equiv} p \left(c_1 \left( \frac{d_1}p \right)+c_2 \left( \frac{d_2}p \right)+c_3 \left( \frac{d_3}p \right) \right)\ ({\rm{mod}}\ {p^2}) \end{equation} | (1.100) |
for all primes
\begin{equation*} \label{pn} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{bk+c}{m^k}a_k-p{\delta}\sum\limits_{k = 0}^{n-1} \frac{bk+c}{m^k}a_k\right)\in{\Bbb Z}_p \ \ \mathit{\text{for all}}\ n\in{\Bbb Z}^+. \end{equation*} |
Joint with the author's PhD student Chen Wang, we pose the following conjecture.
Conjecture 1.3 (Chen Wang and Z.-W. Sun). Let
Remark 1.2. The author [39,Conjecture 1.1(i)] conjectured that
\sum\limits_{k = 0}^{p-1}(8k+5)T_k^2{\equiv} 3p \left( \frac{-3}p \right)\ ({\rm{mod}}\ {p^2}) |
for any prime
All the new series and related congruences in Sections 3-9 support Conjectures 1.1-1.3. We discover the conjectural series for
Conjecture 1.4 (Duality Principle). Let
\begin{equation} a_{k}{\equiv} \left( \frac dp \right)D^ka_{p-1-k}\ ({\rm{mod}}\ p) \end{equation} | (1.101) |
for any prime
\sum\limits_{k = 0}^\infty \frac{bk+c}{m^k}a_k = \frac{\lambda_1\sqrt{d_1}+\lambda_2\sqrt{d_2}+\lambda_3\sqrt{d_3}}{\pi} |
for some
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{a_k}{m^k}{\equiv} \left( \frac dp \right)\sum\limits_{k = 0}^{p-1} \frac{a_k}{(D/m)^k}\ ({\rm{mod}}\ {p^2}) \end{equation} | (1.102) |
for any prime
Remark 1.3 (ⅰ) For any prime
(ⅱ) For any
\begin{equation} T_k(b,c){\equiv} \left( \frac{b^2-4c}p \right)(b^2-4c)^kT_{p-1-k}(b,c)\ ({\rm{mod}}\ p) \end{equation} | (1.103) |
for all
For a series
In Section 10, we pose two curious conjectural series for
Lemma 2.1. Let
\begin{align} \sum\limits_{k = 0}^n \frac{((64-m)k^3-32k^2-16k+8) \binom{2k}k^3}{(2k-1)^2m^k} = & \frac{8(2n+1)}{m^n} \binom{2n}n^3, \end{align} | (2.1) |
\begin{align} \sum\limits_{k = 0}^n \frac{((64-m)k^3-96k^2+48k-8) \binom{2k}k^3}{(2k-1)^3m^k} = & \frac 8{m^n} \binom{2n}n^3, \end{align} | (2.2) |
\begin{align} \sum\limits_{k = 0}^n \frac{((108-m)k^3-54k^2-12k+6) \binom{2k}k^2 \binom{3k}k}{(2k-1)(3k-1)m^k} = & \frac{6(3n+1)}{m^n} \binom{2n}n^2 \binom{3n}n, \end{align} | (2.3) |
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((108-m)k^3-(54+m)k^2-12k+6) \binom{2k}k^2 \binom{3k}k}{(k+1)(2k-1)(3k-1)m^k} \\&\qquad\qquad\qquad = \frac{6(3n+1)}{(n+1)m^n} \binom{2n}n^2 \binom{3n}n, \end{aligned} \end{equation} | (2.4) |
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((256-m)k^3-128k^2-16k+8) \binom{2k}k^2 \binom{4k}{2k}}{(2k-1)(4k-1)m^k} \\&\qquad\qquad\qquad = \frac{8(4n+1)}{m^n} \binom{2n}n^2 \binom{4n}{2n}, \end{aligned} \end{equation} | (2.5) |
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((256-m)k^3-(128+m)k^2-16k+8) \binom{2k}k^2 \binom{4k}{2k}}{(k+1)(2k-1)(4k-1)m^k} \\&\qquad\qquad\qquad = \frac{8(4n+1)}{(n+1)m^n} \binom{2n}n^2 \binom{4n}{2n}, \end{aligned} \end{equation} | (2.6) |
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((1728-m)k^3-864k^2-48k+24) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)m^k} \\&\qquad\qquad\qquad = \frac{24(6n+1)}{m^n} \binom{2n}n \binom{3n}n \binom{6n}{3n}, \end{aligned} \end{equation} | (2.7) |
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^n \frac{((1728-m)k^3-(864+m)k^2-48k+24) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)m^k} \\&\qquad\qquad\qquad = \frac{24(6n+1)}{(n+1)m^n} \binom{2n}n \binom{3n}n \binom{6n}{3n}. \end{aligned} \end{equation} | (2.8) |
Remark 2.1. The eight identities in Lemma 2.1 can be easily proved by induction on
\begin{gather} \binom{2n}n\sim \frac{4^n}{\sqrt{n\pi}},\ \ \binom{2n}n \binom{3n}n\sim \frac{\sqrt3\, 27^n}{2n\pi}, \end{gather} | (2.9) |
\begin{gather} \binom{2n}n \binom{4n}{2n}\sim \frac{64^n}{\sqrt2\, n\pi},\ \ \binom{3n}n \binom{6n}n\sim \frac{432^n}{2n\pi}. \end{gather} | (2.10) |
Proof of Theorem 1.1. Just apply Lemma 2.1 and the 36 known rational Ramanujan-type series listed in [16]. Let us illustrate the proofs by showing (1.1), (1.2), (1.71) and (1.72) in details.
By (2.1) with
\sum\limits_{k = 0}^\infty \frac{(16k^3-4k^2-2k+1) \binom{2k}k^3}{(2k-1)^2(-64)^k} = \lim\limits_{n\to+\infty} \frac{2n+1}{(-64)^n} \binom{2n}n^3 = 0. |
Note that
16k^3-4k^2-2k+1 = (4k+1)(2k-1)^2+2k(4k-1) |
and recall Bauer's series
\sum\limits_{k = 0}^\infty(4k+1) \frac{ \binom{2k}k^3}{(-64)^k} = \frac2{\pi}. |
So, we get
\sum\limits_{k = 0}^\infty \frac{k(4k-1) \binom{2k}k^3}{(2k-1)^2(-64)^k} = - \frac12\sum\limits_{k = 0}^\infty(4k+1) \frac{ \binom{2k}k^3}{(-64)^k} = - \frac1{\pi}. |
This proves (1.1). By (2.2) with
\sum\limits_{k = 0}^n \frac{(4k-1)(4k^2-2k+1) \binom{2k}k^3}{(2k-1)^3(-64)^k} = \frac{ \binom{2n}n^3}{(-64)^n} |
and hence
\sum\limits_{k = 0}^\infty \frac{(2k(2k-1)(4k-1)+4k-1) \binom{2k}k^3}{(2k-1)^3(-64)^k} = \lim\limits_{n\to+\infty} \frac{ \binom{2n}n^3}{(-64)^n} = 0. |
Combining this with
In view of (2.7) with
\begin{align*} &\sum\limits_{k = 0}^n \frac{(10939058860032072k^3-36k^2-2k+1) \binom{2k}k \binom{3k}k \binom{6k}{3k}} {(2k-1)(6k-1)(-640320)^{3k}} \\ = & \frac{6n+1}{(-640320)^{3n}} \binom{2n}n \binom{3n}n \binom{6n}{3n}. \end{align*} |
and hence
\sum\limits_{k = 0}^\infty \frac{(10939058860032072k^3-36k^2-2k+1) \binom{2k}k \binom{3k}k \binom{6k}{3k}} {(2k-1)(6k-1)(-640320)^{3k}} = 0. |
In 1987, D. V. Chudnovsky and G. V. Chudnovsky [8] got the formula
\sum\limits_{k = 0}^\infty \frac{545140134k+13591409}{(-640320)^{3k}} \binom{2k}k \binom{3k}k \binom{6k}{3k} = \frac{3\times53360^2}{2\pi\sqrt{10005}}, |
which enabled them to hold the world record for the calculation of
\begin{align*} &10939058860032072k^3-36k^2-2k+1 \\ = &1672209(2k-1)(6k-1)(545140134k+13591409) \\&+426880 (16444841148 k^2 - 1709536232 k-53241371 ) \end{align*} |
and hence
\begin{align*} &\sum\limits_{k = 0}^\infty \frac{(16444841148 k^2 - 1709536232 k-53241371 ) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(2k-1)(6k-1)(-640320)^{3k}} \\ = &- \frac{1672209}{426880}\times \frac{3\times53360^2}{2\pi\sqrt{10005}} = -1672209 \frac{\sqrt{10005}}{\pi}. \end{align*} |
This proves
By (2.8) with
\begin{align*} &\sum\limits_{k = 0}^n \frac{(10939058860032072k^3 +10939058860031964k^2-2k+1) \binom{2k}k \binom{3k}k \binom{6k}{3k}} {(k+1)(2k-1)(6k-1)(-640320)^{3k}} \\&\qquad\qquad = \frac{6n+1}{(n+1)(-640320)^{3n}} \binom{2n}n \binom{3n}n \binom{6n}{3n} \end{align*} |
and hence
\sum\limits_{k = 0}^\infty \frac{(10939058860032072k^3 +10939058860031964k^2-2k+1) \binom{2k}k \binom{3k}k \binom{6k}{3k}} {(k+1)(2k-1)(6k-1)(-640320)^{3k}} = 0. |
Note that
\begin{align*} &2802461(10939058860032072k^3 +10939058860031964k^2-2k+1) \\ = &1864188626454(k+1)(16444841148 k^2 - 1709536232 k-53241371)+5P(k). \end{align*} |
Therefore, with the help of
\begin{align*} &\sum\limits_{k = 0}^\infty \frac{P(k) \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-640320)^{3k}} \\ = &- \frac{1864188626454}5\times(-1672209) \frac{\sqrt{10005}}{\pi} = 18\times557403^3 \frac{\sqrt{10005}}{5\pi}. \end{align*} |
This proves
The identities (1.3)–(1.70) can be proved similarly.
Lemma 2.2. Let
\begin{align} \sum\limits_{k = 1}^n \frac{m^k((m-64)k^3-32k^2+16k+8)}{(2k+1)^2k^3 \binom{2k}k^3} = & \frac{m^{n+1}}{(2n+1)^2 \binom{2n}n^3}-m, \end{align} | (2.11) |
\begin{align} \sum\limits_{k = 1}^n \frac{m^k((m-64)k^3-96k^2-48k-8)}{(2k+1)^3k^3 \binom{2k}k^3} = & \frac{m^{n+1}}{(2n+1)^3 \binom{2n}n^3}-m, \end{align} | (2.12) |
\begin{align} \sum\limits_{k = 1}^n \frac{m^k((m-108)k^3-54k^2+12k+6)}{(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} = & \frac{m^{n+1}}{(2n+1)(3n+1) \binom{2n}n^2 \binom{3n}n}-m, \end{align} | (2.13) |
\begin{equation} \begin{aligned} &\sum\limits_{1 < k \leq n} \frac{m^k((m-108)k^3-(54+m)k^2+12k+6)}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} \\&\qquad\qquad = \frac{m^{n+1}}{n(2n+1)(3n+1) \binom{2n}n^2 \binom{3n}n}- \frac{m^2}{144}, \end{aligned} \end{equation} | (2.14) |
\begin{equation} \begin{aligned} &\sum\limits_{k = 1}^n \frac{m^k((m-256)k^3-128k^2+16k+8)}{(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}} \\&\qquad\qquad = \frac{m^{n+1}}{(2n+1)(4n+1) \binom{2n}n^2 \binom{4n}{2n}}-m, \end{aligned} \end{equation} | (2.15) |
\begin{equation} \begin{aligned} &\sum\limits_{1 < k \leq n} \frac{m^k((m-256)k^3-(128+m)k^2+16k+8)}{(k-1)(2k+1)(4k+1)k^3 \binom{2k}k^2 \binom{4k}{2k}} \\&\qquad\qquad\quad = \frac{m^{n+1}}{n(2n+1)(4n+1) \binom{2n}n^2 \binom{4n}{2n}}- \frac{m^2}{360}. \end{aligned} \end{equation} | (2.16) |
Remark 2.2. This can be easily proved by induction on
Proof of Theorem 1.2. We just apply Lemma 2.2 and use the known identities:
\begin{gather*} \sum\limits_{k = 1}^\infty \frac{21k-8}{k^3 \binom{2k}k^3} = \frac{\pi^2}6, \ \ \sum\limits_{k = 1}^\infty \frac{(4k-1)(-64)^k}{k^3 \binom{2k}k^3} = -16G, \\\sum\limits_{k = 1}^\infty \frac{(3k-1)(-8)^k}{k^3 \binom{2k}k^3} = -2G, \ \ \sum\limits_{k = 1}^\infty \frac{(3k-1)16^k}{k^3 \binom{2k}k^3} = \frac{\pi^2}2, \\\sum\limits_{k = 1}^\infty \frac{(15k-4)(-27)^{k-1}}{k^3 \binom{2k}k^2 \binom{3k}k} = K,\ \ \sum\limits_{k = 1}^\infty \frac{(5k-1)(-144)^k}{k^3 \binom{2k}k^2 \binom{4k}{2k}} = - \frac{45}2K, \\\sum\limits_{k = 1}^\infty \frac{(11k-3)64^k}{k^2 \binom{2k}k^2 \binom{3k}k} = 8\pi^2,\ \sum\limits_{k = 1}^\infty \frac{(10k-3)8^k}{k^3 \binom{2k}k^2 \binom{3k}k} = \frac{\pi^2}2,\ \ \sum\limits_{k = 1}^\infty \frac{(35k-8)81^k}{k^3 \binom{2k}k^2 \binom{4k}{2k}} = 12\pi^2. \end{gather*} |
Here, the first identity was found and proved by D. Zeilberger [52] in 1993. The second, third and fourth identities were obtained by J. Guillera [17] in 2008. The fifth identity on
Let us illustrate our proofs by proving (1.77)-(1.79) and (1.82)-(1.83) in details.
In view of (2.11) with
\sum\limits_{k = 1}^n \frac{16^k(-48k^3-32k^2+16k+8)}{(2k+1)^2k^3 \binom{2k}k^3} = \frac{16^{n+1}}{(2n+1)^2 \binom{2n}n^3}-16 |
for all
\sum\limits_{k = 1}^\infty \frac{16^k(6k^3+4k^2-2k-1)}{(2k+1)^2k^3 \binom{2k}k^3} = \lim\limits_{n\to+\infty} \left( \frac{-2\times16^n}{(2n+1)^2 \binom{2n}n^3}+2 \right) = 2. |
Notice that
2(6k^3+4k^2-2k-1) = (2k+1)^2(3k-1)-(3k+1). |
So we have
-\sum\limits_{k = 1}^\infty \frac{(3k+1)16^k}{(2k+1)^2k^3 \binom{2k}k^3} = 2\times2-\sum\limits_{k = 1}^\infty \frac{(3k-1)16^k}{k^3 \binom{2k}k^3} = 4- \frac{\pi^2}2 |
and hence (1.77) holds.
By (2.11) with
\sum\limits_{k = 1}^n \frac{(-64)^k(-128k^3-32k^2+16k+8)}{(2k+1)^2k^3 \binom{2k}k^3} = \frac{(-64)^{n+1}}{(2n+1)^2 \binom{2n}n^3}+64 |
for all
\sum\limits_{k = 1}^\infty \frac{(-64)^k(16k^3+4k^2-2k-1)}{(2k+1)^2k^3 \binom{2k}k^3} = -8+\lim\limits_{n\to+\infty} \frac{8(-64)^n}{(2n+1)^2 \binom{2n}n^3} = -8. |
Since
\sum\limits_{k = 1}^\infty \frac{(4k-1)(-64)^k}{k^3 \binom{2k}k^3} = -16G, |
we see that
-16G-2\sum\limits_{k = 1}^\infty \frac{(4k+1)(-64)^k}{(2k+1)^2k^2 \binom{2k}k^3} = -8 |
and hence (1.78) holds. In light of (2.12) with
\sum\limits_{k = 1}^n \frac{(-64)^k(-128k^3-96k^2-48k-8)}{(2k+1)^3k^3 \binom{2k}k^3} = \frac{(-64)^{n+1}}{(2n+1)^3 \binom{2n}n^3}+64 |
for all
\sum\limits_{k = 1}^\infty \frac{(-64)^k(16k^3+12k^2+6k+1)}{(2k+1)^3k^3 \binom{2k}k^3} = -8+\lim\limits_{n\to+\infty} \frac{8(-64)^n}{(2n+1)^3 \binom{2n}n^3} = -8. |
Since
\sum\limits_{k = 1}^\infty \frac{(4k+1)(-64)^k}{(2k+1)^3k^3 \binom{2k}k^3} = -8-2(4-8G) = 16G-16. |
This proves (1.79).
By (2.13) with
\sum\limits_{k = 1}^\infty \frac{(45k^3+18k^2-4k-2)(-27)^k}{(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} = -9. |
As
2(45k^3+18k^2-4k-2) = (15k-4)(2k+1)(3k+1)-3k(5k+1) |
and
\sum\limits_{k = 1}^\infty \frac{(15k-4)(-27)^k}{k^3 \binom{2k}k^2 \binom{3k}k} = -27K, |
we see that (1.82) follows. By (2.14) with
-3\sum\limits_{k = 2}^\infty \frac{(-27)^k(45k^3+9k^2-4k-2)}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} = - \frac{(-27)^2}{144} |
and hence
\sum\limits_{k = 2}^\infty \frac{(-27)^k(45k^3+9k^2-4k-2)}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} = \frac{27}{16}. |
As
45k^3+9k^2-4k-2 = 9(k-1)k(5k+1)+(45k^2+5k-2), |
with the aid of (1.82) we get
\begin{align*} &\sum\limits_{k = 2}^\infty \frac{(-27)^k(45k^2+5k-2)}{(k-1)(2k+1)(3k+1)k^3 \binom{2k}k^2 \binom{3k}k} \\ = & \frac{27}{16}-9 \left(6-9K- \frac{6(-27)}{12^2} \right) = \frac{27}{16}(48K-37) \end{align*} |
and hence
Other identities in Theorem 1.2 can be proved similarly.
For integers
\begin{equation} s_{n,k}: = \frac1{ \binom nk}\sum\limits_{i = 0}^{k} \binom n{2i} \binom n{2(k-i)} \binom{2i}i \binom{2(k-i)}{k-i}. \end{equation} | (2.17) |
For
\begin{equation} t_n: = \sum\limits_{0 < k \leq n} \binom{n-1}{k-1}(-1)^k4^{n-k}s_{n+k,k}. \end{equation} | (2.18) |
Lemma 2.3. For any
\begin{equation} \sum\limits_{k = 0}^n \binom nk(-1)^k4^{n-k}s_{n+k,k} = f_n \end{equation} | (2.19) |
and
\begin{equation} (2n+1)t_{n+1}+8nt_n = (2n+1)f_{n+1}-4(n+1)f_n, \end{equation} | (2.20) |
where
Proof. For
F(n,i,k) = {n \choose k} \frac{(-1)^k 4^{n-k}}{ \binom{n+k}k} {n+k \choose 2i}{2i \choose i} {n+k \choose 2(k-i)}{2(k-i) \choose k-i}. |
By the telescoping method for double summation [7], for
{\mathcal F}(n,i,k): = F(n,i,k) + \frac{7n^2+21n+16}{8(n+1)^2} F(n+1,i,k) - \frac{(n+2)^2}{8(n+1)^2} F(n+2,i,k) |
with
{\mathcal F}(n,i,k) = (G_1(n,i+1,k)-G_1(n,i,k)) + (G_2(n,i,k+1)-G_2(n,i,k)), |
where
G_1(n,i,k): = \frac{i^2(-k+i-1)(-1)^{k+1} 4^{n-k} n!^2 (n+k)! p(n,i,k)}{(2n+3)(n-k+2)!(n+k+2-2i)!(n-k+2i)!(i!(k-i+1)!)^2} |
and
G_2(n,i,k): = \frac{ 2(k-i)(-1)^{k} 4^{n-k} n!^2 (n+k)! q(n,i,k)} {(2n+3) (n-k+2)! (n+k-2i+1)! (n-k+2i+2)! (i!(k-i)!)^2}, |
with
\begin{align*} &-10n^4+(i-10k-68)n^3+(-24i^2+(32k+31)i+2k^2-67k-172)n^2\\ &+(36i^3+(-68k-124)i^2+(39k^2+149k+104)i+2k^3-8k^2-145k-192)n\\ &+60i^3+(-114k-140)i^2+(66k^2+160k+92)i+3k^3-19k^2-102k-80 \end{align*} |
and
\begin{align*} &10(i-k)n^4+(-20i^2+(46k+47)i-6k^2-47k)n^3\\ +&(72i^3+(-60k-38)i^2+(22k^2+145k+90)i+4k^3-11k^2-90k)n^2\\ +&(72k+156)i^3n+(-72k^2-60k-10)i^2n+(18k^3+4k^2+165k+85)in \\+&(22k^3-5k^2-85k)n +(120k+60)i^3+(-120k^2+68k-4)i^2 \\+&(30k^3-56k^2+86k+32)i+26k^3-6k^2-32k \end{align*} |
respectively. Therefore
\begin{align*} &\sum\limits_{k = 0}^{n+2} \sum\limits_{i = 0}^k {\mathcal F}(n,i,k) \\ = & \sum\limits_{k = 0}^{n+2} (G_1(n,k+1,k)-G_1(n,0,k)) + \sum\limits_{i = 0}^{n+2} (G_2(n,i,n+3)-G_2(n,i,i)) \\ = &\sum\limits_{k = 0}^{n+2}(0-0)+\sum\limits_{i = 0}^{n+2}(0-0) = 0, \end{align*} |
and hence
u(n): = \sum\limits_{k = 0}^n {n \choose k} (-1)^k 4^{n-k} s_{n+k,k} |
satisfies the recurrence relation
8(n+1)^2 u(n) + (7n^2+21n+16) u(n+1) - (n+2)^2 u(n+2) = 0. |
As pointed out by J. Franel [14], the Franel numbers satisfy the same recurrence. Note also that
The identity (2.20) can be proved similarly. In fact, if we use
\begin{align*} &8(n+1)(n+2)(18n^3+117n^2+249n+172)v(n) \\&+(126n^5+1197n^4+4452n^3+8131n^2+7350n+2656)v(n+1) \\ = &(n+3)^2(18n^3+63n^2+69n+22)v(n+2). \end{align*} |
In view of the above, we have completed the proof of Lemma 2.3.
Lemma 2.4. For any
\begin{equation} S_n(4,c) = \sum\limits_{k = 0}^{\lfloor n/2\rfloor} \binom{n-k}k \binom{2(n-k)}{n-k}c^k4^{n-2k}s_{n,k}. \end{equation} | (2.21) |
Proof. For each
\begin{align*} T_k(4,c)T_{n-k}(4,c) = &\sum\limits_{i = 0}^{\lfloor k/2\rfloor} \binom {k}{2i} \binom{2i}i4^{k-2i}c^i \sum\limits_{j = 0}^{\lfloor(n-k)/2\rfloor} \binom{n-k}{2j} \binom{2j}j4^{n-k-2j}c^j \\ = &\sum\limits_{r = 0}^{\lfloor n/2\rfloor}c^r4^{n-2r}\sum\limits_{i,j\in{\Bbb N}\atop i+j = r} \binom k{2i} \binom{n-k}{2j} \binom{2i}i \binom{2j}j. \end{align*} |
If
\begin{align*} \sum\limits_{k = 0}^n \binom nk^2 \binom{k}{2i} \binom{n-k}{2j} = & \binom n{2i} \binom n{2j}\sum\limits_{k = 2i}^{n-2j} \binom{n-2i}{k-2i} \binom{n-2j}{n-k-2j} \\ = & \binom n{2i} \binom n{2j} \binom{2n-2(i+j)}{n-2(i+j)} = \binom{2n-2r}n \binom n{2i} \binom n{2j} \end{align*} |
with the aid of the Chu-Vandermonde identity. Therefore
\begin{align*} S_n(4,c) = &\sum\limits_{k = 0}^{\lfloor n/2\rfloor}c^k4^{n-2k} \binom{2n-2k}{n} \binom nks_{n,k} \\ = &\sum\limits_{k = 0}^{\lfloor n/2\rfloor}c^k4^{n-2k} \binom{2n-2k}{n-k} \binom{n-k}ks_{n,k}. \end{align*} |
This proves (2.21).
Lemma 2.5. For
\begin{equation} s_{k+l,k} \leq (2k+1)4^kl \binom{k+l}l. \end{equation} | (2.22) |
Proof. Let
\begin{align*} \binom nks_{n,k} \leq&\sum\limits_{i,j\in{\Bbb N}\atop i+j = k} \binom n{2i} \binom n{2j}\sum\limits_{i,j\in{\Bbb N}\atop i+j = k} \binom{2i}i \binom{2j}j \\ \leq&\sum\limits_{s,t\in{\Bbb N}\atop s+t = 2k} \binom ns \binom nt\sum\limits_{i,j\in{\Bbb N}\atop i+j = k}4^i4^j = \binom{2n}{2k}(k+1)4^k \end{align*} |
and
\begin{align*} \frac{ \binom{2n}{2k}}{ \binom nk} = & \frac{ \binom{2n}{2l}}{ \binom nl} = \prod\limits_{j = 0}^{l-1} \frac{2(j+k)+1}{2j+1} \\ \leq&(2k+1)\prod\limits_{0 < j < l} \frac{2(j+k)}{2j} \\ = &(2k+1) \binom{k+l-1}{l-1}. \end{align*} |
Hence
s_{k+l,k} \leq (k+1)4^k(2k+1) \frac{l}{k+l} \binom{k+l}l \leq (2k+1)4^kl \binom{k+l}l. |
This proves (2.22).
To prove Theorem 1.3, we need an auxiliary theorem.
Theorem 2.6. Let
\begin{equation} \sum\limits_{n = 0}^\infty(an+b) \frac{S_n(4,-m)}{m^n} = \frac1{m+16}\sum\limits_{n = 0}^\infty(2a(m+4)n-8a+b(m+16)) \frac{ \binom{2n}nf_n}{m^n}. \end{equation} | (2.23) |
Proof. Let
\begin{align*} \sum\limits_{n = 0}^N \frac{S_n(4,-m)}{m^n} = &\sum\limits_{n = 0}^N \frac1{m^n}\sum\limits_{k = 0}^{\lfloor n/2\rfloor}(-m)^k4^{n-2k} \binom{2n-2k}{n-k} \binom{n-k}ks_{n,k} \end{align*} |
\begin{align*} = &\sum\limits_{l = 0}^N \frac{ \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \binom lk(-1)^k4^{l-k}s_{l+k,k} \\ = &\sum\limits_{l = 0}^{\lfloor N/2\rfloor} \frac{ \binom{2l}l}{m^l}\sum\limits_{k = 0}^l \binom lk(-1)^k4^{l-k}s_{l+k,k} \\&+\sum\limits_{N/2 < l \leq N} \frac{ \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \binom lk(-1)^k4^{l-k}s_{l+k,k} \end{align*} |
and similarly
\begin{align*} \sum\limits_{n = 0}^N \frac{nS_n(4,-m)}{m^n} = &\sum\limits_{l = 0}^N \frac{ \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \binom lk(-1)^k4^{l-k}(k+l)s_{l+k,k} \\ = &\sum\limits_{l = 0}^N \frac{l \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \left( \binom lk+ \binom{l-1}{k-1} \right)(-1)^k4^{l-k}s_{l+k,k} \\ = &\sum\limits_{l = 0}^{\lfloor N/2\rfloor} \frac{l \binom{2l}l}{m^l}\sum\limits_{k = 0}^{l} \left( \binom lk+ \binom{l-1}{k-1} \right)(-1)^k4^{l-k}s_{l+k,k} \\&+\sum\limits_{N/2 < l \leq N} \frac{l \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \left( \binom lk+ \binom{l-1}{k-1} \right)(-1)^k4^{l-k}s_{l+k,k}, \end{align*} |
where we consider
If
\begin{align*} & \bigg|\sum\limits_{k = 0}^{N-l} \binom lk(-1)^k4^{l-k}s_{l+k,k} \bigg| \\ \leq&\sum\limits_{k = 0}^l \binom lk4^{l-k}s_{l+k,k} \leq\sum\limits_{k = 0}^l \binom lk4^{l-k}(2k+1)4^kl \binom{k+l}l \\ \leq& l(2l+1)4^l\sum\limits_{k = 0}^l \binom lk \binom{l+k}k = l(2l+1)4^lP_l(3), \end{align*} |
where
\begin{align*} & \bigg|\sum\limits_{N/2 < l \leq N} \frac{ \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \binom lk(-1)^k4^{l-k}s_{l+k,k} \bigg| \\ \leq&\sum\limits_{N/2 < l \leq N}l(2l+1) \left( \frac{16}m \right)^lP_l(3) \leq\sum\limits_{l > N/2}l(2l+1)P_l(3) \left( \frac{16}m \right)^l \end{align*} |
and
\begin{align*} & \bigg|\sum\limits_{N/2 < l \leq N} \frac{l \binom{2l}l}{m^l}\sum\limits_{k = 0}^{N-l} \left( \binom lk+ \binom{l-1}{k-1} \right)(-1)^k4^{l-k}s_{l+k,k} \bigg| \\ \leq&\sum\limits_{N/2 < l \leq N} \frac{l4^l}{m^l}\sum\limits_{k = 0}^l 2 \binom lk4^{l-k}s_{l+k,k} \end{align*} |
\begin{align*} \leq&\sum\limits_{N/2 < l \leq N}2l^2(2l+1) \left( \frac{16}m \right)^lP_l(3) \\ \leq&2\sum\limits_{l > N/2}l^2(2l+1)P_l(3) \left( \frac{16}m \right)^l. \end{align*} |
Recall that
P_l(3) = T_l(3,2)\sim \frac{(3+2\sqrt2)^{l+1/2}}{2\root4\of{2}\sqrt{l\pi}}\ \ \text{as}\ l\to+\infty. |
As
\sum\limits_{l = 0}^\infty l^2(2l+1)P_l(3) \left( \frac{16}m \right)^l |
converges. Thus
\lim\limits_{N\to+\infty}\sum\limits_{l > N/2}l(2l+1)P_l(3) \left( \frac{16}m \right)^l = 0 = \lim\limits_{N\to+\infty}\sum\limits_{l > N/2}l^2(2l+1)P_l(3) \left( \frac{16}m \right)^l |
and hence by the above we have
\sum\limits_{n = 0}^\infty \frac{S_n(4,-m)}{m^n} = \sum\limits_{l = 0}^{\infty} \frac{ \binom{2l}l}{m^l}\sum\limits_{k = 0}^l \binom lk(-1)^k4^{l-k}s_{l+k,k} |
and
\sum\limits_{n = 0}^\infty \frac{nS_n(4,-m)}{m^n} = \sum\limits_{l = 0}^{\infty} \frac{l \binom{2l}l}{m^l}\sum\limits_{k = 0}^{l} \left( \binom lk+ \binom{l-1}{k-1} \right)(-1)^k4^{l-k}s_{l+k,k}. |
Therefore, with the aid of (2.19), we obtain
\begin{equation} \sum\limits_{n = 0}^\infty \frac{S_n(4,-m)}{m^n} = \sum\limits_{n = 0}^\infty \frac{ \binom{2n}n}{m^n}f_n \end{equation} | (2.24) |
and
\begin{equation} \sum\limits_{n = 0}^\infty \frac{nS_n(4,-m)}{m^n} = \sum\limits_{n = 0}^\infty \frac{n \binom{2n}n}{m^n}(f_n+t_n). \end{equation} | (2.25) |
In view of (2.25) and (2.20),
\begin{align*} &(m+16)\sum\limits_{n = 0}^\infty \frac{nS_n(4,-m)}{m^n} \\ = &\sum\limits_{n = 1}^\infty \frac{n \binom{2n}n}{m^{n-1}}(f_n+t_n)+16\sum\limits_{n = 0}^\infty \frac{n \binom{2n}n}{m^n}(f_n+t_n) \\ = &\sum\limits_{n = 0}^\infty \frac{(n+1) \binom{2n+2}{n+1}(f_{n+1}+t_{n+1})+16n \binom{2n}n(f_n+t_n)}{m^n} \\ = &2\sum\limits_{n = 0}^\infty \frac{ \binom{2n}n}{m^n} \left((2n+1)(f_{n+1}+t_{n+1})+8n(f_n+t_n) \right) \end{align*} |
\begin{align*} = &2\sum\limits_{n = 0}^\infty \frac{ \binom{2n}n}{m^n} \left(2(2n+1)f_{n+1}+4(n-1)f_n \right) \\ = &2\sum\limits_{n = 0}^\infty \frac{(n+1) \binom{2n+2}{n+1}f_{n+1}}{m^n}+8\sum\limits_{n = 0}^\infty \frac{(n-1) \binom{2n}nf_n}{m^n} \\ = &2\sum\limits_{n = 0}^\infty \frac{n \binom{2n}nf_n}{m^{n-1}}+8\sum\limits_{n = 0}^\infty \frac{(n-1) \binom{2n}nf_n}{m^n} = 2\sum\limits_{n = 0}^\infty((m+4)n-4) \frac{ \binom{2n}nf_n}{m^n}. \end{align*} |
Combining this with (2.24), we immediately obtain the desired (2.23).
Proof of Theorem 1.3. Let
4^nT_n(1,m) = \sum\limits_{k = 0}^{\lfloor n/2\rfloor} \binom n{2k} \binom{2k}k4^{n-2k}(16m)^k = T_n(4,16m) |
for any
\begin{align*} &\sum\limits_{n = 0}^\infty(an+b) \frac{S_n(1,m)}{(-4m)^n} \\ = &\sum\limits_{n = 0}^\infty(an+b) \frac{S_n(4,16m)}{(-16m)^n} \\ = & \frac1{16-16m}\sum\limits_{n = 0}^\infty(2a(4-16m)n-8a+(16-16m)b) \frac{ \binom{2n}nf_n}{(-16m)^n} \\ = & \frac1{2(m-1)}\sum\limits_{n = 0}^\infty(a(4m-1)n+a+2b(m-1)) \frac{ \binom{2n}nf_n}{(-16m)^n}. \end{align*} |
Therefore
\begin{align*} \sum\limits_{k = 0}^\infty \frac{7k+3}{24^k}S_k(1,-6) = & \frac52\sum\limits_{k = 0}^\infty \frac{5k+1}{96^k} \binom{2k}kf_k, \\\sum\limits_{k = 0}^\infty \frac{12k+5}{(-28)^k}S_k(1,7) = &3\sum\limits_{k = 0}^\infty \frac{9k+2}{(-112)^k} \binom{2k}kf_k,\\ \sum\limits_{k = 0}^\infty \frac{84k+29}{80^k}S_k(1,-20) = &27\sum\limits_{k = 0}^\infty \frac{6k+1}{320^k} \binom{2k}kf_k, \\\sum\limits_{k = 0}^\infty \frac{3k+1}{(-100)^k}S_k(1,25) = & \frac1{16}\sum\limits_{k = 0}^\infty \frac{99k+17}{(-400)^k} \binom{2k}kf_k, \\\sum\limits_{k = 0}^\infty \frac{228k+67}{224^k}S_k(1,-56) = &5\sum\limits_{k = 0}^\infty \frac{90k+13}{896^k} \binom{2k}kf_k, \\\sum\limits_{k = 0}^\infty \frac{399k+101}{(-676)^k}S_k(1,169) = & \frac{15}{16}\sum\limits_{k = 0}^\infty \frac{855k+109}{(-2704)^k} \binom{2k}kf_k, \\ \sum\limits_{k = 0}^\infty \frac{2604k+563}{2600^k}S_k(1,-650) = &51\sum\limits_{k = 0}^\infty \frac{102k+11}{10400^k} \binom{2k}kf_k,\\ \sum\limits_{k = 0}^\infty \frac{39468k+7817}{(-6076)^k}S_k(1,1519) = &135\sum\limits_{k = 0}^\infty \frac{585k+58}{(-24304)^k} \binom{2k}kf_k, \\\sum\limits_{k = 0}^\infty \frac{41667k+7879}{9800^k}S_k(1,-2450) = & \frac{297}2\sum\limits_{k = 0}^\infty \frac{561k+53}{39200^k} \binom{2k}kf_k, \\\sum\limits_{k = 0}^\infty \frac{74613k+10711}{(-530^2)^k}S_k(1,265^2) = & \frac{23}{32}\sum\limits_{k = 0}^\infty \frac{207621k+14903}{(-1060^2)^k} \binom{2k}kf_k. \end{align*} |
\begin{gather*} \sum\limits_{k = 0}^\infty \frac{5k+1}{96^k} \binom{2k}kf_k = \frac{3\sqrt2}{\pi}, \ \ \sum\limits_{k = 0}^\infty \frac{9k+2}{(-112)^k} \binom{2k}kf_k = \frac{2\sqrt7}{\pi}, \\ \sum\limits_{k = 0}^\infty \frac{6k+1}{320^k} \binom{2k}kf_k = \frac{8\sqrt{15}}{9\pi},\ \ \sum\limits_{k = 0}^\infty \frac{99k+17}{(-400)^k} \binom{2k}kf_k = \frac{50}{\pi}, \\\sum\limits_{k = 0}^\infty \frac{90k+13}{896^k} \binom{2k}kf_k = \frac{16\sqrt7}{\pi},\ \ \sum\limits_{k = 0}^\infty \frac{855k+109}{(-2704)^k} \binom{2k}kf_k = \frac{338}{\pi}, \\\sum\limits_{k = 0}^\infty \frac{102k+11}{10400^k} \binom{2k}kf_k = \frac{50\sqrt{39}}{9\pi},\ \ \sum\limits_{k = 0}^\infty \frac{585k+58}{(-24304)^k} \binom{2k}kf_k = \frac{98\sqrt{31}}{3\pi}, \\\sum\limits_{k = 0}^\infty \frac{561k+53}{39200^k} \binom{2k}kf_k = \frac{1225\sqrt6}{18\pi}, \ \ \sum\limits_{k = 0}^\infty \frac{207621k+14903}{(-1060^2)^k} \binom{2k}kf_k = \frac{140450}{3\pi}. \end{gather*} |
So we get the identities (1.88)-(1.97) finally.
Now we pose a conjecture related to the series (Ⅰ1)-(Ⅰ4) of Sun [34,40].
Conjecture 3.1. We have the following identities:
\sum\limits_{k = 0}^\infty \frac{50k+1}{(-256)^k} \binom{2k}k \binom{2k}{k+1}T_k(1,16) = \frac{8}{3\pi},\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}1') |
\sum\limits_{k = 0}^\infty \frac{(100k^2-4k-7) \binom{2k}k^2T_k(1,16)}{(2k-1)^2(-256)^k} = - \frac{24}{\pi},\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}1'') |
\sum\limits_{k = 0}^\infty \frac{30k+23}{(-1024)^k} \binom{2k}k \binom{2k}{k+1}T_k(34,1) = - \frac{20}{3\pi},\;\;\;\;\;\;\;(\mathrm{Ⅰ}2') |
\sum\limits_{k = 0}^\infty \frac{(36k^2-12k+1) \binom{2k}k^2T_k(34,1)}{(2k-1)^2(-1024)^k} = - \frac{6}{\pi},\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}2'') |
\sum\limits_{k = 0}^\infty \frac{110k+103}{4096^k} \binom{2k}k \binom{2k}{k+1}T_k(194,1) = \frac{304}{\pi},\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}3') |
\sum\limits_{k = 0}^\infty \frac{(20k^2+28k-11) \binom{2k}k^2T_k(194,1)}{(2k-1)^2 4096^k} = - \frac{6}{\pi},\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}3'') |
\sum\limits_{k = 0}^\infty \frac{238k+263}{4096^k} \binom{2k}k \binom{2k}{k+1}T_k(62,1) = \frac{112\sqrt3}{3\pi},\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}4') |
\sum\limits_{k = 0}^\infty \frac{(44k^2+4k-5) \binom{2k}k^2T_k(62,1)}{(2k-1)^2 4096^k} = - \frac{4\sqrt3}{\pi},\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}4'') |
\sum\limits_{k = 0}^\infty \frac{6k+1}{256^k} \binom{2k}k^2T_k(8,-2) = \frac{2}{\pi}\sqrt{8+6\sqrt2},\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}5) |
\sum\limits_{k = 0}^\infty \frac{2k+3}{256^k} \binom{2k}k \binom{2k}{k+1}T_k(8,-2) = \frac{6\sqrt{8+6\sqrt2}-16\root4\of{2}}{3\pi},\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}5') |
\sum\limits_{k = 0}^\infty \frac{(4k^2+2k-1) \binom{2k}k^2T_k(8,-2)}{(2k-1)^2 256^k} = - \frac{3\root4\of{2}}{4\pi}.\;\;\;\;\;\;\;\;\;(\mathrm{Ⅰ}5'') |
Remark 3.1. For each
((1+\lambda_0-\lambda_1)k+\lambda_0)C_k = (k+\lambda_0) \binom{2k}k-(k+\lambda_1) \left(\begin{array}{c}2 k \\ k+1\end{array}\right) |
since
\sum\limits_{k = 0}^\infty \frac{26k+5}{(-256)^k} \binom{2k}kC_kT_k(1,16) = \frac{16}{\pi}, |
and (I5) and (I5
\sum\limits_{k = 0}^\infty \frac{2k-1}{256^k} \binom{2k}kC_kT_k(8,-2) = \frac4{\pi} \left(\sqrt{8+6\sqrt2}-4\root4\of2 \right). |
For the conjectural identities in Conjecture 3.1, we have conjectures for the corresponding
\sum\limits_{k = 0}^{p-1} \frac{30k+23}{(-1024)^k} \binom{2k}k \binom{2k}{k+1}T_k(34,1){\equiv} \frac p3 \left(21 \left( \frac 2p \right)-10 \left( \frac{-1}p \right)-11 \right)\ ({\rm{mod}}\ {p^2}) |
and
\sum\limits_{k = 0}^{p-1} \frac{2k+1}{(-1024)^k} \binom{2k}kC_kT_k(34,1){\equiv} \frac p3 \left(2-3 \left( \frac 2p \right)+4 \left( \frac{-1}p \right) \right)\ ({\rm{mod}}\ {p^2}). |
Concerning (I5) and (I5
\frac1{2^{\lfloor n/2\rfloor+1}n \binom{2n}n}\sum\limits_{k = 0}^{n-1}(6k+1) \binom{2k}k^2T_k(8,-2)256^{n-1-k}\in{\Bbb Z}^+ |
and
\frac1{ \binom{2n-2}{n-1}}\sum\limits_{k = 0}^{n-1} \frac{(1-2k-4k^2) \binom{2k}k^2T_k(8,-2)}{(2k-1)^2 256^k}\in{\Bbb Z}^+ |
for each
\sum\limits_{k = 0}^{p-1} \frac{ \binom{2k}k^2T_k(8,-2)}{256^k} {\equiv}\left\{ \begin{array}{l} (-1)^{y/2}(4x^2-2p)\ ({\rm{mod}}\ {p^2})& \text{if}\ p{\equiv}1\ ({\rm{mod}}\ 8),\\ (-1)^{(xy-1)/2}8xy\ ({\rm{mod}}\ {p^2})& \text{if}\ p{\equiv}5\ ({\rm{mod}}\ 8), \end{array} \right. |
and
\sum\limits_{k = 0}^{p-1} \frac{(4k^2+2k-1) \binom{2k}k^2T_k(8,-2)}{(2k-1)^2256^k}{\equiv}0\ ({\rm{mod}}\ {p^2}). |
By [40,Theorem 5.1], we have
\sum\limits_{k = 0}^{p-1} \frac{ \binom{2k}k^2T_k(8,-2)}{256^k}{\equiv}0\ ({\rm{mod}}\ {p^2}) |
for any prime
Next we pose a conjecture related to the series (Ⅱ1)-(Ⅱ7) and (Ⅱ10)-(Ⅱ12) of Sun [34,40].
Conjecture 3.2. We have the following identities:
\sum\limits_{k = 0}^\infty \frac{3k+4}{972^k} \binom{2k}{k+1} \binom{3k}kT_k(18,6) = \frac{63\sqrt3}{40\pi},\;\;\;\;\;\;\;\;(Ⅱ1') |
\sum\limits_{k = 0}^\infty \frac{91k+107}{10^{3k}} \binom{2k}{k+1} \binom{3k}kT_k(10,1) = \frac{275\sqrt3}{18\pi},\;\;\;\;\;\;\;(Ⅱ2') |
\sum\limits_{k = 0}^\infty \frac{195k+83}{18^{3k}} \binom{2k}{k+1} \binom{3k}{k}T_k(198,1) = \frac{9423\sqrt3}{10\pi},\;\;\;\;\;\;\;(Ⅱ3') |
\sum\limits_{k = 0}^\infty \frac{483k-419}{30^{3k}} \binom{2k}{k+1} \binom{3k}kT_k(970,1) = \frac{6550\sqrt3}{\pi},\;\;\;\;\;\;\;(Ⅱ4') |
\sum\limits_{k = 0}^\infty \frac{666k+757}{30^{3k}} \binom{2k}{k+1} \binom{3k}{k}T_k(730,729) = \frac{3475\sqrt3}{4\pi},\;\;\;\;\;\;\;(Ⅱ5') |
\sum\limits_{k = 0}^\infty \frac{8427573k+8442107}{102^{3k}} \binom{2k}{k+1} \binom{3k}kT_k(102,1) = \frac{125137\sqrt6}{20\pi},\;\;\;\;\;\;\;(Ⅱ6') |
\sum\limits_{k = 0}^\infty \frac{959982231k+960422503}{198^{3k}} \binom{2k}{k+1} \binom{3k}kT_k(198,1) = \frac{5335011\sqrt3}{20\pi},\;\;\;\;\;\;\;(Ⅱ7') |
\sum\limits_{k = 0}^\infty \frac{99k+1}{24^{3k}} \binom{2k}{k+1} \binom{3k}kT_k(26,729) = \frac{16(289\sqrt{15}-645\sqrt3)}{15\pi},\;\;\;\;\;\;\;(Ⅱ10') |
\sum\limits_{k = 0}^\infty \frac{45k+1}{(-5400)^k} \binom{2k}{k+1} \binom{3k}kT_k(70,3645) = \frac{345\sqrt3-157\sqrt{15}}{6\pi},\;\;\;\;\;\;\;(Ⅱ11') |
\sum\limits_{k = 0}^\infty \frac{252k-1}{(-13500)^k} \binom{2k}{k+1} \binom{3k}kT_k(40,1458) = \frac{25(1212\sqrt3-859\sqrt6)}{24\pi},\;\;\;\;\;\;\;(Ⅱ12') |
\sum\limits_{k = 0}^\infty \frac{9k+2}{(-675)^k} \binom{2k}{k} \binom{3k}kT_k(15,-5) = \frac{7\sqrt{15}}{8\pi},\;\;\;\;\;\;\;(Ⅱ13) |
\sum\limits_{k = 0}^\infty \frac{45k+31}{(-675)^k} \binom{2k}{k+1} \binom{3k}kT_k(15,-5) = - \frac{19\sqrt{15}}{8\pi},\;\;\;\;\;\;\;(Ⅱ13') |
\sum\limits_{k = 0}^\infty \frac{39k+7}{(-1944)^k} \binom{2k}{k} \binom{3k}kT_k(18,-3) = \frac{9\sqrt{3}}{\pi},\;\;\;\;\;\;\;(Ⅱ14) |
\sum\limits_{k = 0}^\infty \frac{312k+263}{(-1944)^k} \binom{2k}{k+1} \binom{3k}kT_k(18,-3) = - \frac{45\sqrt{3}}{2\pi}.\;\;\;\;\;\;\;(Ⅱ14') |
Remark 3.2. We also have conjectures on related congruences. For example, concerning (Ⅱ), for any prime
\sum\limits_{k = 0}^{p-1} \frac{39k+7}{(-1944)^k} \binom{2k}k \binom{3k}kT_k(18,-3){\equiv} \frac p2 \left(13 \left( \frac p3 \right)+1 \right) \ ({\rm{mod}}\ {p^2}) |
and that
\begin{align*} &\sum\limits_{k = 0}^{p-1} \frac{ \binom{2k}k \binom{3k}kT_k(18,-3)}{(-1944)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \text{if}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p7) = 1\ &\ p = x^2+21y^2, \\2p-2x^2\ ({\rm{mod}}\ {p^2})& \text{if}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p7) = 1\ &\ 2p = x^2+21y^2, \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \text{if}\ ( \frac{-1}p) = ( \frac p7) = -1,\ ( \frac p3) = 1\ &\ p = 3x^2+7y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \text{if}\ ( \frac{-1}p) = 1,\ ( \frac p3) = ( \frac p7) = -1\ &\ 2p = 3x^2+7y^2, \\0\ ({\rm{mod}}\ {p^2})& \text{if}\ ( \frac{-21}p) = -1, \end{cases} \end{align*} |
where
The following conjecture is related to the series (Ⅲ1)-(Ⅲ10) and (Ⅲ12) of Sun [34,40].
Conjecture 3.3. We have the following identities:
\sum\limits_{k = 0}^\infty \frac{17k+18}{66^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(52,1) = \frac{77\sqrt{33}}{12\pi},\;\;\;\;\;\;\;\;(Ⅲ1') |
\sum\limits_{k = 0}^\infty \frac{4k+3}{(-96^2)^k} \binom{2k}{k+1} \binom{4k}{2k}T_k(110,1) = - \frac{\sqrt6}{3\pi},\;\;\;\;\;\;\;\;(Ⅲ2') |
\sum\limits_{k = 0}^\infty \frac{8k+9}{112^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(98,1) = \frac{154\sqrt{21}}{135\pi},\;\;\;\;\;\;\;\;(Ⅲ3') |
\sum\limits_{k = 0}^\infty \frac{3568k+4027}{264^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(257,256) = \frac{869\sqrt{66}}{10\pi},\;\;\;\;\;\;\;\;(Ⅲ4') |
\sum\limits_{k = 0}^\infty \frac{144k+1}{(-168^2)^{k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(7,4096) = \frac{7(1745\sqrt{42}-778\sqrt{210})}{120\pi},\;\;\;\;\;\;\;\;(Ⅲ5') |
\sum\limits_{k = 0}^\infty \frac{3496k+3709}{336^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(322,1) = \frac{182\sqrt7}{\pi},\;\;\;\;\;\;\;\;(Ⅲ6') |
\sum\limits_{k = 0}^\infty \frac{286k+229}{336^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(1442,1) = \frac{1113\sqrt{210}}{20\pi},\;\;\;\;\;\;\;\;(Ⅲ7') |
\sum\limits_{k = 0}^\infty \frac{8426k+8633}{912^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(898,1) = \frac{703\sqrt{114}}{20\pi},\;\;\;\;\;\;\;\;(Ⅲ8') |
\sum\limits_{k = 0}^\infty \frac{1608k+79}{912^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(12098,1) = \frac{67849\sqrt{399}}{105\pi},\;\;\;\;\;\;\;\;(Ⅲ9') |
\sum\limits_{k = 0}^\infty \frac{134328722k+134635283}{10416^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(10402,1) = \frac{93961\sqrt{434}}{4\pi},\;\;\;\;\;\;\;\;(Ⅲ10') |
and
\begin{equation*} \begin{aligned}&\sum\limits_{k = 0}^\infty \frac{39600310408k+39624469807}{39216^{2k}} \binom{2k}{k+1} \binom{4k}{2k}T_k(39202,1) \\&\qquad\qquad = \frac{1334161\sqrt{817}}{\pi}.\end{aligned}\;\;\;\;\;\;\;\;(Ⅲ12') \end{equation*} |
The following conjecture is related to the series (Ⅳ1)-(Ⅳ21) of Sun [34,40].
Conjecture 3.4. We have the following identities:
\sum\limits_{k = 0}^\infty \frac{(356k^2+288k+7) \binom{2k}k^2T_{2k}(7,1)}{(k+1)(2k-1)(-48^2)^k} = - \frac{304}{3\pi},\;\;\;\;\;\;\;\;(Ⅳ1') |
\sum\limits_{k = 0}^\infty \frac{(172k^2+141k-1) \binom{2k}k^2T_{2k}(62,1)}{(k+1)(2k-1)(-480^2)^k} = - \frac{80}{3\pi},\;\;\;\;\;\;\;\;(Ⅳ2') |
\sum\limits_{k = 0}^\infty \frac{(782k^2+771k+19) \binom{2k}k^2T_{2k}(322,1)}{(k+1)(2k-1)(-5760^2)^k} = - \frac{90}{\pi},\;\;\;\;\;\;\;\;(Ⅳ3') |
\sum\limits_{k = 0}^\infty \frac{(34k^2+45k+5) \binom{2k}k^2T_{2k}(10,1)}{(k+1)(2k-1)96^{2k}} = - \frac{20\sqrt2}{3\pi},\;\;\;\;\;\;\;\;(Ⅳ4') |
\sum\limits_{k = 0}^\infty \frac{(106k^2+193k+27) \binom{2k}k^2T_{2k}(38,1)}{(k+1)(2k-1)240^{2k}} = - \frac{10\sqrt6}{\pi},\;\;\;\;\;\;\;\;(Ⅳ5') |
\sum\limits_{k = 0}^\infty \frac{(214166k^2+221463k+7227) \binom{2k}k^2T_{2k}(198,1)}{(k+1)(2k-1)39200^{2k}} = - \frac{9240\sqrt6}{\pi},\;\;\;\;\;\;\;\;(Ⅳ6') |
\sum\limits_{k = 0}^\infty \frac{(112k^2+126k+9) \binom{2k}k^2T_{2k}(18,1)}{(k+1)(2k-1)320^{2k}} = - \frac{6\sqrt{15}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ7') |
\sum\limits_{k = 0}^\infty \frac{(926k^2+995k+55) \binom{2k}k^2T_{2k}(30,1)}{(k+1)(2k-1)896^{2k}} = - \frac{60\sqrt7}{\pi},\;\;\;\;\;\;\;\;(Ⅳ8') |
\sum\limits_{k = 0}^\infty \frac{(1136k^2+2962k+503) \binom{2k}k^2T_{2k}(110,1)}{(k+1)(2k-1)24^{4k}} = - \frac{90\sqrt7}{\pi},\;\;\;\;\;\;\;\;(Ⅳ9') |
\sum\limits_{k = 0}^\infty \frac{(5488k^2+8414k+901) \binom{2k}k^2T_{2k}(322,1)}{(k+1)(2k-1)48^{4k}} = - \frac{294\sqrt7}{\pi},\;\;\;\;\;\;\;\;(Ⅳ10') |
\sum\limits_{k = 0}^\infty \frac{(170k^2+193k+11) \binom{2k}k^2T_{2k}(198,1)}{(k+1)(2k-1)2800^{2k}} = - \frac{6\sqrt{14}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ11') |
\sum\limits_{k = 0}^\infty \frac{(104386k^2+108613k+4097) \binom{2k}k^2T_{2k}(102,1)}{(k+1)(2k-1)10400^{2k}} = - \frac{2040\sqrt{39}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ12') |
\sum\limits_{k = 0}^\infty \frac{(7880k^2+8217k+259) \binom{2k}k^2T_{2k}(1298,1)}{(k+1)(2k-1)46800^{2k}} = - \frac{144\sqrt{26}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ13') |
\sum\limits_{k = 0}^\infty \frac{(6152k^2+45391k+9989) \binom{2k}k^2T_{2k}(1298,1)}{(k+1)(2k-1)5616^{2k}} = - \frac{663\sqrt3}{\pi},\;\;\;\;\;\;\;\;(Ⅳ14') |
\sum\limits_{k = 0}^\infty \frac{(147178k^2+2018049k+471431) \binom{2k}k^2T_{2k}(4898,1)}{(k+1)(2k-1)20400^{2k}} = -3740 \frac{\sqrt{51}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ15') |
\sum\limits_{k = 0}^\infty \frac{(1979224k^2+5771627k+991993) \binom{2k}k^2T_{2k}(5778,1)}{(k+1)(2k-1)28880^{2k}} = -73872 \frac{\sqrt{10}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ16') |
\sum\limits_{k = 0}^\infty \frac{(233656k^2+239993k+5827) \binom{2k}k^2T_{2k}(5778,1)}{(k+1)(2k-1)439280^{2k}} = -4080 \frac{\sqrt{19}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ17') |
\sum\limits_{k = 0}^\infty \frac{(5890798k^2+32372979k+6727511) \binom{2k}k^2T_{2k}(54758,1)}{(k+1)(2k-1)243360^{2k}} = -600704 \frac{\sqrt{95}}{9\pi},\;\;\;\;\;\;\;\;(Ⅳ18') |
\sum\limits_{k = 0}^\infty \frac{(148k^2+272k+43) \binom{2k}k^2T_{2k}(10,-2)}{(k+1)(2k-1)4608^{k}} = -28 \frac{\sqrt{6}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ19') |
\sum\limits_{k = 0}^\infty \frac{(3332k^2+17056k+3599) \binom{2k}k^2T_{2k}(238,-14)}{(k+1)(2k-1)1161216^{k}} = -744 \frac{\sqrt{2}}{\pi},\;\;\;\;\;\;\;\;(Ⅳ20') |
\sum\limits_{k = 0}^\infty \frac{(11511872k^2+10794676k+72929) \binom{2k}k^2T_{2k}(9918,-19)}{(k+1)(2k-1)(-16629048064)^{k}} = -390354 \frac{\sqrt{7}}{\pi}.\;\;\;\;\;\;\;\;(Ⅳ21') |
For the five open conjectural series (Ⅵ1), (Ⅵ2), (Ⅵ3), (ⅥI2) and (ⅥI7) of Sun [34,40], we make the following conjecture on related supercongruences.
Conjecture 3.5. Let
\begin{equation*} \label{VI1} \sum\limits_{k = 0}^{pn-1} \frac{66k+17}{(2^{11}3^3)^k}T_k(10,11^2)^3 -p \left( \frac{-2}p \right)\sum\limits_{k = 0}^{n-1} \frac{66k+17}{(2^{11}3^3)^k}T_k(10,11^2)^3 \end{equation*} |
divided by
\begin{equation*} \label{VI2} \sum\limits_{k = 0}^{pn-1} \frac{126k+31}{(-80)^{3k}}T_k(22,21^2)^3 -p \left( \frac{-5}p \right)\sum\limits_{k = 0}^{n-1} \frac{126k+31}{(-80)^{3k}}T_k(22,21^2)^3 \end{equation*} |
divided by
\begin{equation*} \label{VI3} \sum\limits_{k = 0}^{pn-1} \frac{3990k+1147}{(-288)^{3k}}T_k(62,95^2)^3 -p \left( \frac{-2}p \right)\sum\limits_{k = 0}^{n-1} \frac{3990k+1147}{(-288)^{3k}}T_k(62,95^2)^3 \end{equation*} |
divided by
\begin{equation*} \label{VII2} \sum\limits_{k = 0}^{pn-1} \frac{24k+5}{28^{2k}} \binom{2k}kT_k(4,9)^2 -p \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1} \frac{24k+5}{28^{2k}} \binom{2k}kT_k(4,9)^2 \end{equation*} |
divided by
\sum\limits_{k = 0}^{pn-1} \frac{2800512k+435257}{434^{2k}} \binom{2k}kT_k(73,576)^2 \\ -p\sum\limits_{k = 0}^{n-1} \frac{2800512k+435257}{434^{2k}} \binom{2k}kT_k(73,576)^2 |
divided by
Now we pose four conjectural series for
Conjecture 3.6. We have
\sum\limits_{k = 0}^\infty \frac{40k+13}{(-50)^k}T_k(4,1)T_k(1,-1)^2 = \frac{55\sqrt{15}}{9\pi},\;\;\;\;\;\;\;\;(Ⅷ1) |
\sum\limits_{k = 0}^\infty \frac{1435k+113}{3240^k}T_k(7,1)T_k(10,10)^2 = \frac{1452\sqrt{5}}{\pi},\;\;\;\;\;\;\;\;(Ⅷ2) |
\sum\limits_{k = 0}^\infty \frac{840k+197}{(-2430)^k}T_k(8,1)T_k(5,-5)^2 = \frac{189\sqrt{15}}{2\pi},\;\;\;\;\;\;\;\;(Ⅷ3) |
\sum\limits_{k = 0}^\infty \frac{39480k+7321}{(-29700)^k}T_k(14,1)T_k(11,-11)^2 = \frac{6795\sqrt{5}}{\pi}.\;\;\;\;\;\;\;\;(Ⅷ4) |
Remark 3.3. The author found the identity (Ⅷ1) on Nov. 3, 2019. The identities (Ⅷ2), (Ⅷ3) and (Ⅷ4) were formulated on Nov. 4, 2019.
Below we present some conjectures on congruences related to Conjecture 3.6.
Conjecture 3.7. (ⅰ) For each
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(40k+13)(-1)^k50^{n-1-k}T_k(4,1)T_k(1,-1)^2\in{\Bbb Z}^+, \end{equation} | (3.1) |
and this number is odd if and only if
(ⅱ) Let
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{40k+13}{(-50)^k}T_k(4,1)T_k(1,-1)^2{\equiv} \frac p3 \left(12+5 \left( \frac3p \right)+22 \left( \frac {-15}p \right) \right) \ ({\rm{mod}}\ {p^2}). \end{equation} | (3.2) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{40k+13}{(-50)^k}T_k(4,1)T_k(1,-1)^2-p\sum\limits_{k = 0}^{n-1} \frac{40k+13}{(-50)^k}T_k(4,1)T_k(1,-1)^2\right) \in{\Bbb Z}_p \end{equation} | (3.3) |
for all
(ⅲ) Let
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{T_k(4,1)T_k(1,-1)^2}{(-50)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,9\ ({\rm{mod}}\ {20})\ &\ p = x^2+5y^2\ (x,y\in{\Bbb Z}), \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3,7\ ({\rm{mod}}\ {20})\ &\ 2p = x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-5}p) = -1.\end{cases} \end{aligned} \end{equation} | (3.4) |
Remark 3.4. The imaginary quadratic field
Conjecture 3.8. (ⅰ) For any
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(40k+27)(-6)^{n-1-k}T_k(4,1)T_k(1,-1)^2\in{\Bbb Z}, \end{equation} | (3.5) |
and the number is odd if and only if
(ⅱ) Let
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{40k+27}{(-6)^k}T_k(4,1)T_k(1,-1)^2{\equiv} \frac p9 \left(55 \left( \frac{-5}p \right)+198 \left( \frac 3p \right)-10 \right) \ ({\rm{mod}}\ {p^2}). \end{equation} | (3.6) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{40k+27}{(-6)^k}T_k(4,1)T_k(1,-1)^2-p\sum\limits_{k = 0}^{n-1} \frac{40k+27}{(-6)^k}T_k(4,1)T_k(1,-1)^2\right) \in{\Bbb Z}_p \end{equation} | (3.7) |
for all
(ⅲ) Let
\begin{equation} \begin{aligned}& \left( \frac p3 \right)\sum\limits_{k = 0}^{p-1} \frac{T_k(4,1)T_k(1,-1)^2}{(-6)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,9\ ({\rm{mod}}\ {20})\ &\ p = x^2+5y^2\ (x,y\in{\Bbb Z}), \\2p-2x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3,7\ ({\rm{mod}}\ {20})\ &\ 2p = x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-5}p) = -1.\end{cases} \end{aligned} \end{equation} | (3.8) |
Remark 3.5. This conjecture can be viewed as the dual of Conjecture 3.7. Note that the series
Conjecture 3.9. (ⅰ) For each
\begin{equation} \frac1{n10^{n-1}}\sum\limits_{k = 0}^{n-1}(1435k+113) 3240^{n-1-k}T_k(7,1)T_k(10,10)^2\in{\Bbb Z}^+. \end{equation} | (3.9) |
(ⅱ) Let
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{1435k+113}{3240^k}T_k(7,1)T_k(10,10)^2 \\{\equiv}& \frac p9 \left(2420 \left( \frac{-5}p \right)+105 \left( \frac5p \right)-1508 \right) \ ({\rm{mod}}\ {p^2}).\end{aligned} \end{equation} | (3.10) |
If
\begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{1435k+113}{3240^k}T_k(7,1)T_k(10,10)^2 -p\sum\limits_{k = 0}^{n-1} \frac{1435k+113}{3240^k}T_k(7,1)T_k(10,10)^2 \end{equation} | (3.11) |
divided by
(ⅲ) Let
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{T_k(7,1)T_k(10,10)^2}{3240^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-15}p) = -1.\end{cases} \end{aligned} \end{equation} | (3.12) |
Remark 3.6. The imaginary quadratic field
Conjecture 3.10. (ⅰ) For each
\begin{equation} \frac3{2n10^{n-1}}\sum\limits_{k = 0}^{n-1}(1435k+1322) 50^{n-1-k}T_k(7,1)T_k(10,10)^2\in{\Bbb Z}^+. \end{equation} | (3.13) |
(ⅱ) Let
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{1435k+1322}{50^k}T_k(7,1)T_k(10,10)^2 \\{\equiv}& \frac p3 \left(3432 \left( \frac{5}p \right)+968 \left( \frac{-1}p \right)-434 \right) \ ({\rm{mod}}\ {p^2}).\end{aligned} \end{equation} | (3.14) |
If
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{pn-1} \frac{1435k+1322}{50^k}T_k(7,1)T_k(10,10)^2 -p\sum\limits_{k = 0}^{n-1} \frac{1435k+1322}{50^k}T_k(7,1)T_k(10,10)^2 \end{aligned} \end{equation} | (3.15) |
divided by
(ⅲ) Let
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{T_k(7,1)T_k(10,10)^2}{50^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-15}p) = -1.\end{cases} \end{aligned} \end{equation} | (3.16) |
Remark 3.7. This conjecture can be viewed as the dual of Conjecture 3.9. Note that the series
\sum\limits_{k = 0}^\infty \frac{1435k+1322}{50^k}T_k(7,1)T_k(10,10)^2 |
diverges.
Conjecture 3.11. (ⅰ) For each
\begin{equation} \frac1{n5^{n-1}}\sum\limits_{k = 0}^{n-1}(840k+197)(-1)^k 2430^{n-1-k}T_k(8,1)T_k(5,-5)^2\in{\Bbb Z}^+. \end{equation} | (3.17) |
(ⅱ) Let
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{840k+197}{(-2430)^k}T_k(8,1)T_k(5,-5)^2 {\equiv} p \left(140 \left( \frac{-15}p \right)+5 \left( \frac{15}p \right)+52 \right) \ ({\rm{mod}}\ {p^2}).\end{aligned} \end{equation} | (3.18) |
If
\begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{840k+197}{(-2430)^k}T_k(8,1)T_k(5,-5)^2 -p\sum\limits_{k = 0}^{n-1} \frac{840k+197}{(-2430)^k}T_k(8,1)T_k(5,-5)^2 \end{equation} | (3.19) |
divided by
(ⅲ Let
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{T_k(8,1)T_k(5,-5)^2}{(-2430)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p7) = 1,\ p = x^2+105y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p7) = 1,\ ( \frac p3) = ( \frac p5) = -1,\ 2p = x^2+105y^2, \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p7) = -1,\ p = 3x^2+35y^2, \\6x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p7) = -1,\ ( \frac p3) = ( \frac p5) = 1,\ 2p = 3x^2+35y^2, \\2p-20x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p5) = 1,\ ( \frac p5) = ( \frac p7) = -1,\ p = 5x^2+21y^2, \\2p-10x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = 1,\ ( \frac p5) = ( \frac p7) = -1,\ 2p = 5x^2+21y^2, \\28x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p5) = -1,\ ( \frac p3) = ( \frac p7) = 1,\ p = 7x^2+15y^2, \\14x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p5) = ( \frac p7) = 1,\ 2p = 7x^2+15y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-105}p) = -1,\end{cases} \end{aligned} \end{equation} | (3.20) |
where
Remark 3.8. Note that the imaginary quadratic field
Conjecture 3.12. (ⅰ) For each
\begin{equation} \frac1{n}\sum\limits_{k = 0}^{n-1}(39480k+7321)(-1)^k 29700^{n-1-k}T_k(14,1)T_k(11,-11)^2\in{\Bbb Z}^+, \end{equation} | (3.21) |
and this number is odd if and only if
(ⅱ) Let
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{39480k+7321}{(-29700)^k}T_k(14,1)T_k(11,-11)^2 \\{\equiv}& p \left(5738 \left( \frac{-5}p \right)+70 \left( \frac3p \right)+1513 \right) \ ({\rm{mod}}\ {p^2}).\end{aligned} \end{equation} | (3.22) |
If
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{pn-1} \frac{39480k+7321}{(-29700)^k}T_k(14,1)T_k(11,-11)^2 \\&-p\sum\limits_{k = 0}^{n-1} \frac{39480k+7321}{(-29700)^k}T_k(14,1)T_k(11,-11)^2 \end{aligned} \end{equation} | (3.23) |
divided by
(ⅲ) Let
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{T_k(14,1)T_k(11,-11)^2}{(-29700)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p{11}) = 1,\ p = x^2+165y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p{11}) = -1,\ 2p = x^2+165y^2, \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p5) = -1,\ ( \frac p3) = ( \frac p{11}) = 1,\ p = 3x^2+55y^2, \\6x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p5) = 1,\ ( \frac p3) = ( \frac p{11}) = -1,\ 2p = 3x^2+55y^2, \\2p-20x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{11}) = 1,\ ( \frac p3) = ( \frac p5) = -1,\ p = 5x^2+33y^2, \\2p-10x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{11}) = -1,\ ( \frac p3) = ( \frac p5) = 1,\ 2p = 5x^2+33y^2, \\44x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p5) = ( \frac p{11}) = 1,\ p = 11x^2+15y^2, \\22x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = 1,\ ( \frac p5) = ( \frac p{11}) = -1,\ 2p = 11x^2+15y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-165}p) = -1,\end{cases} \end{aligned} \end{equation} | (3.24) |
where
Remark 3.9. Note that the imaginary quadratic field
Conjectures 4.1–4.14 below provide congruences related to (1.88)–(1.97).
Conjecture 4.1. (ⅰ) For any
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(7k+3)S_k(1,-6)24^{n-1-k}\in{\Bbb Z}^+. \end{equation} | (4.1) |
(ⅱ) Let
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{7k+3}{24^k}S_k(1,-6){\equiv} \frac p2 \left(5 \left( \frac{-2}p \right)+ \left( \frac 6p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (4.2) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{7k+3}{24^k}S_k(1,-6)-p \left( \frac{-2}p \right)\sum\limits_{k = 0}^{n-1} \frac{7k+3}{24^k}S_k(1,-6)\right) \in{\Bbb Z}_p \end{equation} | (4.3) |
for all
(ⅲ) For any prime
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-6)}{24^k} \\{\equiv}&\begin{cases}( \frac p3)(4x^2-2p)\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,3\ ({\rm{mod}}\ 8)\ &\ p = x^2+2y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}5,7\ ({\rm{mod}}\ 8). \end{cases}\end{aligned} \end{equation} | (4.4) |
Conjecture 4.2. (ⅰ) For any
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(12k+5)S_k(1,7)(-1)^k28^{n-1-k}\in{\Bbb Z}^+, \end{equation} | (4.5) |
and this number is odd if and only if
(ⅱ) Let
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{12k+5}{(-28)^k}S_k(1,7){\equiv}5p \left( \frac p7 \right)\ ({\rm{mod}}\ {p^2}), \end{equation} | (4.6) |
and moreover
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{12k+5}{(-28)^k}S_k(1,7)-p \left( \frac p7 \right)\sum\limits_{k = 0}^{n-1} \frac{12k+5}{(-28)^k}S_k(1,7)\right)\in{\Bbb Z}_p \end{equation} | (4.7) |
for all
(ⅲ) For any prime
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,7)}{(-28)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p7) = 1\ &\ p = x^2+21y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p7) = 1\ &\ 2p = x^2+21y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p7) = -1,\ ( \frac p3) = 1\ &\ p = 3x^2+7y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = 1,\ ( \frac p3) = ( \frac p7) = -1\ &\ 2p = 3x^2+7y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-21}p) = -1, \end{cases}\end{aligned} \end{equation} | (4.8) |
where
Conjecture 4.3. (ⅰ) For any
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(84k+29)S_k(1,-20)80^{n-1-k}\in{\Bbb Z}^+, \end{equation} | (4.9) |
and this number is odd if and only if
(ⅱ) Let
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{84k+29}{80^k}S_k(1,-20){\equiv} p \left(2 \left( \frac 5p \right)+27 \left( \frac{-15}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (4.10) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{84k+29}{80^k}S_k(1,-20)- p \left( \frac p5 \right)\sum\limits_{k = 0}^{n-1} \frac{84k+29}{80^k}S_k(1,-20)\right)\in{\Bbb Z}_p \end{equation} | (4.11) |
for all
(ⅲ) For any prime
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-20)}{80^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p5) = 1\ &\ p = x^2+30y^2, \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = 1,\ ( \frac p3) = ( \frac p5) = -1\ &\ p = 2x^2+15y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p3) = 1,\ ( \frac 2p) = ( \frac p5) = -1\ &\ p = 3x^2+10y^2, \\20x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}5) = 1,\ ( \frac 2p) = ( \frac p3) = -1\ &\ p = 5x^2+6y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-30}p) = -1, \end{cases}\end{aligned} \end{equation} | (4.12) |
where
Conjecture 4.4. (ⅰ) For any
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(3k+1)(-1)^k100^{n-1-k}S_k(1,25)\in{\Bbb Z}^+. \end{equation} | (4.13) |
(ⅱ) Let
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{3k+1}{(-100)^k}S_k(1,25)- p \left( \frac{-1}p \right)\sum\limits_{k = 0}^{n-1} \frac{3k+1}{(-100)^k}S_k(1,25)\right)\in{\Bbb Z}_p \end{equation} | (4.14) |
for all
(ⅲ) For any prime
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,25)}{(-100)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p{11}) = 1\ &\ p = x^2+33y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}{p}) = 1,\ ( \frac p3) = ( \frac p{11}) = -1\ &\ 2p = x^2+33y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{11}) = 1,\ ( \frac {-1}p) = ( \frac p3) = -1\ &\ p = 3x^2+11y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}3) = 1,\ ( \frac {-1}p) = ( \frac p{11}) = -1\ &\ 2p = 3x^2+11y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-33}p) = -1, \end{cases}\end{aligned} \end{equation} | (4.15) |
where
Conjecture 4.5. (ⅰ) For any
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(228k+67)S_k(1,-56)224^{n-1-k}\in{\Bbb Z}^+, \end{equation} | (4.16) |
and this number is odd if and only if
(ⅱ) Let
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{228k+67}{224^k}S_k(1,-56){\equiv} p \left(65 \left( \frac{-7}p \right)+2 \left( \frac {14}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (4.17) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{228k+67}{224^k}S_k(1,-56)- p \left( \frac p7 \right)\sum\limits_{k = 0}^{n-1} \frac{228k+67}{224^k}S_k(1,-56)\right)\in{\Bbb Z}_p \end{equation} | (4.18) |
for all
(ⅲ) For any prime
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-56)}{224^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p3) = ( \frac p7) = 1\ &\ p = x^2+42y^2, \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}7) = 1,\ ( \frac {-2}p) = ( \frac p3) = -1\ &\ p = 2x^2+21y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac {-2}p) = 1,\ ( \frac p3) = ( \frac p7) = -1\ &\ p = 3x^2+14y^2, \\2p-24x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}5) = 1,\ ( \frac 2p) = ( \frac p3) = -1\ &\ p = 6x^2+7y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-42}p) = -1, \end{cases}\end{aligned} \end{equation} | (4.19) |
where
Conjecture 4.6. (ⅰ) For any
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(399k+101)(-1)^k676^{n-1-k}S_k(1,169)\in{\Bbb Z}^+. \end{equation} | (4.20) |
(ⅱ) Let
\begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{399k+101}{(-676)^k}S_k(1,169)- p \left( \frac{-1}p \right)\sum\limits_{k = 0}^{n-1} \frac{399k+101}{(-676)^k}S_k(1,169) \end{equation} | (4.21) |
divided by
(ⅲ) For any prime
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,169)}{(-676)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p{19}) = 1\ &\ p = x^2+57y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}{p}) = 1,\ ( \frac p3) = ( \frac p{19}) = -1\ &\ 2p = x^2+57y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{3}) = 1,\ ( \frac {-1}p) = ( \frac p{19}) = -1\ &\ p = 3x^2+19y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}{19}) = 1,\ ( \frac {-1}p) = ( \frac p{3}) = -1\ &\ 2p = 3x^2+19y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-57}p) = -1, \end{cases}\end{aligned} \end{equation} | (4.22) |
where
Conjecture 4.7. (ⅰ) For any
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(2604k+563)S_k(1,-650)2600^{n-1-k}\in{\Bbb Z}^+, \end{equation} | (4.23) |
and this number is odd if and only if
(ⅱ) Let
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{2604k+563}{2600^k}S_k(1,-650){\equiv} p \left(561 \left( \frac{-39}p \right)+2 \left( \frac {26}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (4.24) |
If
\begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{2604k+563}{2600^k}S_k(1,-650)- p \left( \frac{26}p \right)\sum\limits_{k = 0}^{n-1} \frac{2604k+563}{2600^k}S_k(1,-650) \end{equation} | (4.25) |
divided by
(ⅲ)For any odd prime
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-650)}{2600^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p{13}) = 1\ &\ p = x^2+78y^2, \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac2{p}) = 1,\ ( \frac p3) = ( \frac p{13}) = -1\ &\ p = 2x^2+39y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{13}) = 1,\ ( \frac 2p) = ( \frac p3) = -1\ &\ p = 3x^2+26y^2, \\2p-24x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}3) = 1,\ ( \frac 2p) = ( \frac p{13}) = -1\ &\ p = 6x^2+13y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-78}p) = -1, \end{cases}\end{aligned} \end{equation} | (4.26) |
where
Conjecture 4.8. (ⅰ) For any
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(39468k+7817)(-1)^k6076^{n-1-k}S_k(1,1519)\in{\Bbb Z}^+, \end{equation} | (4.27) |
and this number is odd if and only if
(ⅱ) Let
\begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{39468k+7817}{(-6076)^k}S_k(1,1519)- p \left( \frac{-31}p \right)\sum\limits_{k = 0}^{n-1} \frac{39468k+7817}{(-6076)^k}S_k(1,1519) \end{equation} | (4.28) |
divided by
(ⅲ) For any prime
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,1519)}{(-6076)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p{31}) = 1\ &\ p = x^2+93y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}{31}) = 1,\ ( \frac {-1}p) = ( \frac p{3}) = -1\ &\ 2p = x^2+93y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{3}) = 1,\ ( \frac {-1}p) = ( \frac p{31}) = -1\ &\ p = 3x^2+31y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = 1,\ ( \frac {p}3) = ( \frac p{31}) = -1\ &\ 2p = 3x^2+31y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-93}p) = -1, \end{cases}\end{aligned} \end{equation} | (4.29) |
where
Conjecture 4.9. (ⅰ) For any
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(41667k+7879)9800^{n-1-k}S_k(1,-2450)\in{\Bbb Z}^+. \end{equation} | (4.30) |
(ⅱ) Let
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{41667k+7879}{9800^k}S_k(1,-2450) {\equiv} \frac p2 \left(15741 \left( \frac{-6}p \right)+17 \left( \frac 2p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (4.31) |
If
\begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{41667k+7879}{9800^k}S_k(1,-2450)- p \left( \frac{2}p \right)\sum\limits_{k = 0}^{n-1} \frac{41667k+7879}{9800^k}S_k(1,-2450) \end{equation} | (4.32) |
divided by
(ⅲ) For any prime
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-2450)}{9800^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p{17}) = 1\ &\ p = x^2+102y^2, \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}{17}) = 1,\ ( \frac {2}p) = ( \frac p{3}) = -1\ &\ p = 2x^2+51y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{3}) = 1,\ ( \frac {2}p) = ( \frac p{17}) = -1\ &\ p = 3x^2+34y^2, \\2p-24x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = 1,\ ( \frac {p}3) = ( \frac p{17}) = -1\ &\ p = 6x^2+17y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-102}p) = -1, \end{cases}\end{aligned} \end{equation} | (4.33) |
where
Conjecture 4.10. (ⅰ) For any
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(74613k+10711)(-1)^k530^{2(n-1-k)}S_k(1,265^2)\in{\Bbb Z}^+. \end{equation} | (4.34) |
\begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{74613k+10711}{(-530^2)^k}S_k(1,265^2)- p \left( \frac{-1}p \right)\sum\limits_{k = 0}^{n-1} \frac{74613k+10711}{(-530^2)^k}S_k(1,265^2) \end{equation} | (4.35) |
divided by
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,265^2)}{(-530^2)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p{59}) = 1\ &\ p = x^2+177y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}{p}) = 1,\ ( \frac {p}3) = ( \frac p{59}) = -1\ &\ 2p = x^2+177y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{59}) = 1,\ ( \frac {-1}p) = ( \frac p{3}) = -1\ &\ p = 3x^2+59y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{p}3) = 1,\ ( \frac {-1}p) = ( \frac p{59}) = -1\ &\ 2p = 3x^2+59y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-177}p) = -1, \end{cases}\end{aligned} \end{equation} | (4.36) |
where
Conjecture 4.11. For any odd prime
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k}{(-4)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ 12\mid p-1\ &\ p = x^2+y^2\ (x,y\in{\Bbb Z}\ &\ 3\nmid x), \\4xy\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ 12\mid p-5\ &\ p = x^2+y^2\ (x,y\in{\Bbb Z}\ &\ 3\mid x-y), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3\ ({\rm{mod}}\ 4). \end{cases}\end{aligned} \end{equation} | (4.37) |
Also, for any prime
\begin{equation} \sum\limits_{k = 0}^{p-1}(8k+5) \frac{S_k}{(-4)^k}{\equiv}4p\ ({\rm{mod}}\ {p^2}). \end{equation} | (4.38) |
Conjecture 4.12.
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(4k+3)4^{n-1-k}S_k(1,-1)\in{\Bbb Z}, \end{equation} | (4.39) |
and this number is odd if and only if
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{4k+3}{4^k}S_k(1,-1)- p\sum\limits_{k = 0}^{n-1} \frac{4k+3}{4^k}S_k(1,-1)\right)\in{\Bbb Z}_p. \end{equation} | (4.40) |
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-1)}{4^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,9\ ({\rm{mod}}\ {20})\ &\ p = x^2+5y^2\ (x,y\in{\Bbb Z}), \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3,7\ ({\rm{mod}}\ {20})\ &\ 2p = x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-5}p) = -1. \end{cases}\end{aligned} \end{equation} | (4.41) |
Conjecture 4.13.
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(33k+25)S_k(1,-6)(-6)^{n-1-k}\in{\Bbb Z}, \end{equation} | (4.42) |
and this number is odd if and only if
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{33k+25}{(-6)^k}S_k(1,-6){\equiv} p \left(35-10 \left( \frac 3p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (4.43) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{33k+25}{(-6)^k}S_k(1,-6) -p\sum\limits_{k = 0}^{n-1} \frac{33k+25}{(-6)^k}S_k(1,-6)\right)\in{\Bbb Z}_p \end{equation} | (4.44) |
for all
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{S_k(1,-6)}{(-6)^k}{\equiv}\begin{cases}( \frac{-1}p)(4x^2-2p)\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p = x^2+3y^2\ (x,y\in{\Bbb Z}),\\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2\ ({\rm{mod}}\ 3).\end{cases} \end{equation} | (4.45) |
Conjecture 4.14.
\begin{equation} n\ \ \bigg| \ \sum\limits_{k = 0}^{n-1}(18k+13)S_k(2,9)8^{n-1-k}. \end{equation} | (4.46) |
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{18k+13}{8^k}S_k(2,9) {\equiv} p \left(1+12 \left( \frac p3 \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (4.47) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{18k+13}{8^k}S_k(2,9)-p\sum\limits_{k = 0}^{n-1} \frac{18k+13}{8^k}S_k(2,9)\right)\in{\Bbb Z}_p \end{equation} | (4.48) |
for all
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{S_k(1,-2)}{8^k}{\equiv} \left( \frac p3 \right)\sum\limits_{k = 0}^{p-1} \frac{S_k(2,9)}{8^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,7\ ({\rm{mod}}\ {24})\ &\ p = x^2+6y^2\ \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}5,11\ ({\rm{mod}}\ {24})\ &\ p = 2x^2+3y^2,\\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-6}p) = -1,\end{cases} \end{aligned} \end{equation} | (4.49) |
where
Conjecture 4.15. Let
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{S_k(3,1)}{4^k} {\equiv}\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,9\ ({\rm{mod}}\ {20})\ &\ p = x^2+5y^2,\ \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3,7\ ({\rm{mod}}\ {20})\ &\ 2p = x^2+5y^2,\\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}11,13,17,19\ ({\rm{mod}}\ {20}),\end{cases} \end{equation} | (4.50) |
where
\sum\limits_{k = 0}^{p-1} \frac{40k+29}{4^k}S_k(3,1){\equiv} 18p\ ({\rm{mod}}\ {p^2}). |
Remark 4.1. We also have some similar conjectures involving
\begin{gather*} \sum\limits_{k = 0}^{p-1} \frac{S_k(5,4)}{4^k},\ \sum\limits_{k = 0}^{p-1} \frac{S_k(4,-5)}{4^k}, \ \sum\limits_{k = 0}^{p-1} \frac{S_k(7,6)}{6^k}, \\ \sum\limits_{k = 0}^{p-1} \frac{S_k(10,-2)}{32^k}, \ \sum\limits_{k = 0}^{p-1} \frac{S_k(14,9)}{72^k},\ \sum\limits_{k = 0}^{p-1} \frac{S_k(19,9)}{36^k} \end{gather*} |
modulo
Motivated by Theorem 2.6, we pose the following general conjecture.
Conjecture 4.16. For any odd prime
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{S_k(4,-m)}{m^k}{\equiv}\sum\limits_{k = 0}^{p-1} \frac{ \binom{2k}kf_k}{m^k}\ ({\rm{mod}}\ {p^2}). \end{equation} | (4.51) |
and
\begin{equation} \frac{m+16}2\sum\limits_{k = 0}^{p-1} \frac{kS_k(4,-m)}{m^k} -\sum\limits_{k = 0}^{p-1}((m+4)k-4) \frac{ \binom{2k}kf_k}{m^k}{\equiv}4p \left( \frac mp \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (4.52) |
Remark 4.2 We have checked this conjecture via \mathsf{Mathematica}. In view of the proof of Theorem 2.6, both (4.51) and (4.52) hold modulo
The numbers
Z_n: = \sum\limits_{k = 0}^n \binom nk \binom{2k}k \binom{2(n-k)}{n-k}\ \ (n = 0,1,2,\ldots) |
were first introduced by D. Zagier in his paper [51] the preprint of which was released in 2002. Thus we name such numbers as Zagier numbers. As pointed out by the author [41,Remark 4.3], for any
{\mathcal P}_n: = 2^n\sum\limits_{k = 1}^{\lfloor n/2\rfloor} \binom n{2k} \binom{2k}k^24^{n-2k} = \sum\limits_{k = 0}^n \frac{ \binom{2k}k^2 \binom{2(n-k)}{n-k}^2}{ \binom nk}. |
Let
{\mathcal P}_k{\equiv} \left( \frac{-1}p \right)128^k{\mathcal P}_{p-1-k}\ ({\rm{mod}}\ p) |
by F. Jarvis and H.A. Verrill [24,Corollary 2.2], and hence
Z_k = \frac{{\mathcal P}_k}{2^k}{\equiv} \left( \frac{-1}p \right)64^k(2^{p-1-k}Z_{p-1-k}){\equiv} \left( \frac{-1}p \right)32^kZ_{p-1-k}\ ({\rm{mod}}\ p). |
Combining this with Remark 1.3(ⅱ), we see that
\begin{align*} \sum\limits_{k = 0}^{p-1} \frac{Z_kT_k(b,c)}{m^k}{\equiv}& \left( \frac{4c-b^2}p \right)\sum\limits_{k = 0}^{p-1} \left( \frac{32(b^2-4c)}m \right)^kZ_{p-1-k}T_{p-1-k}(b,c) \\{\equiv}& \left( \frac{4c-b^2}p \right)\sum\limits_{k = 0}^{p-1} \frac{Z_kT_k(b,c)}{(32(b^2-4c)/m)^k}\ ({\rm{mod}}\ p) \end{align*} |
for any
J. Wan and Zudilin [49] obtained the following irrational series for
\sum\limits_{k = 0}^\infty(15k+4-2\sqrt6)Z_kP_k \left( \frac{24-\sqrt6}{15\sqrt2} \right) \left( \frac{4-\sqrt6}{10\sqrt3} \right)^k = \frac{6}{\pi}(7+3\sqrt6). |
Via our congruence approach (including Conjecture 1.4), we find 24 rational series for
Conjecture 5.1. We have the following identities for
\begin{align} \sum\limits_{k = 1}^\infty \frac{5k+1}{32^k}T_kZ_k& = \frac{8(2+\sqrt5)}{3\pi}, \end{align} | (5.1) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{21k+5}{(-252)^k}T_k(1,16)Z_k& = \frac{6\sqrt7}{\pi}, \end{align} | (5.2) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{3k+1}{36^k}T_k(1,-2)Z_k& = \frac{3}{\pi}, \end{align} | (5.3) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{k}{192^k}T_k(14,1)Z_k& = \frac{8}{3\pi}, \end{align} | (5.4) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{30k+11}{(-192)^k}T_k(14,1)Z_k& = \frac{12}{\pi}, \end{align} | (5.5) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{15k+1}{480^k}T_k(22,1)Z_k& = \frac{6\sqrt{10}}{\pi}, \end{align} | (5.6) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{7k+2}{(-672)^k}T_k(26,1)Z_k& = \frac{2\sqrt{21}}{3\pi}, \end{align} | (5.7) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{21k+2}{1152^k}T_k(34,1)Z_k& = \frac{18}{\pi}, \end{align} | (5.8) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{30k-7}{640^k}T_k(62,1)Z_k& = \frac{160}{\pi}, \end{align} | (5.9) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{195k+34}{(-9600)^k}T_k(98,1)Z_k& = \frac{80}{\pi}, \end{align} | (5.10) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{195k+22}{11232^k}T_k(106,1)Z_k& = \frac{27\sqrt{13}}{\pi}, \end{align} | (5.11) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{42k+17}{(-1440)^k}T_k(142,1)Z_k& = \frac{33}{\sqrt5\,\pi}, \end{align} | (5.12) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{2k-1}{1792^k}T_k(194,1)Z_k& = \frac{56}{3\pi}, \end{align} | (5.13) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{1785k+254}{(-37632)^k}T_k(194,1)Z_k& = \frac{672}{\pi}, \end{align} | (5.14) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{210k+23}{40800^k}T_k(202,1)Z_k& = \frac{15\sqrt{34}}{\pi}, \end{align} | (5.15) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{210k-1}{4608^k}T_k(254,1)Z_k& = \frac{288}{\pi}, \end{align} | (5.16) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{21k-5}{5600^k}T_k(502,1)Z_k& = \frac{105}{\sqrt2\,\pi}, \end{align} | (5.17) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{7410k+1849}{(-36992)^k}T_k(1154,1)Z_k& = \frac{2992}{\pi}, \end{align} | (5.18) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{1326k+101}{57760^k}T_k(1442,1)Z_k& = \frac{2014}{\sqrt5\,\pi}, \end{align} | (5.19) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{78k-131}{20800^k}T_k(2498,1)Z_k& = \frac{2600}{\pi}, \end{align} | (5.20) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{62985k+11363}{(-394272)^k}T_k(5474,1)Z_k& = \frac{7659\sqrt{10}}{\pi}, \end{align} | (5.21) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{358530k+33883}{486720^k}T_k(6082,1)Z_k& = \frac{176280}{\pi}, \end{align} | (5.22) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{510k-1523}{78400^k}T_k(9602,1)Z_k& = \frac{33320}{\pi}, \end{align} | (5.23) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{570k-457}{93600^k}T_k(10402,1)Z_k& = \frac{1590\sqrt{13}}{\pi}. \end{align} | (5.24) |
Below we present some conjectures on congruences related to
Conjecture 5.2. (ⅰ) For any
\begin{equation} n\ \bigg|\ \sum\limits_{k = 0}^{n-1}(5k+1)T_kZ_k32^{n-1-k}. \end{equation} | (5.25) |
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{5k+1}{32^k}T_kZ_k{\equiv} \frac p3 \left(5 \left( \frac{-5}p \right)-2 \left( \frac{-1}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (5.26) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{5k+1}{32^k}T_kZ_k-p \left( \frac{-1}p \right)\sum\limits_{k = 0}^{n-1} \frac{5k+1}{32^k}T_kZ_k\right) \in{\Bbb Z}_p \end{equation} | (5.27) |
for all
\begin{equation} \begin{aligned}& \left( \frac{-1}p \right)\sum\limits_{k = 0}^{p-1} \frac{T_kZ_k}{32^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-15}p) = -1. \end{cases} \end{aligned} \end{equation} | (5.28) |
Conjecture 5.3. (ⅰ) For any
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(-1)^k(21k+5)T_k(1,16)Z_k252^{n-1-k}\in{\Bbb Z}^+. \end{equation} | (5.29) |
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{21k+5}{(-252)^k}T_k(1,16)Z_k{\equiv} \frac p3 \left(16 \left( \frac{-7}p \right)- \left( \frac{-1}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (5.30) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{21k+5}{(-252)^k}T_k(1,16)Z_k-p \left( \frac{-1}p \right) \sum\limits_{k = 0}^{n-1} \frac{21k+5}{(-252)^k}T_k(1,16)Z_k\right)\in{\Bbb Z}_p \end{equation} | (5.31) |
for all
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{T_k(1,16)Z_k}{(-252)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,2,4\ ({\rm{mod}}\ {7})\ &\ p = x^2+7y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv} 3,5,6\ ({\rm{mod}}\ 7). \end{cases} \end{aligned} \end{equation} | (5.32) |
Conjecture 5.4.
\begin{equation} n\ \bigg|\ \sum\limits_{k = 0}^{n-1}kT_k(14,1)Z_k192^{n-1-k}. \end{equation} | (5.33) |
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{k}{192^k}T_k(14,1)Z_k{\equiv} \frac p9 \left( \left( \frac{-1}p \right)- \left( \frac{2}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (5.34) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{k}{192^k}T_k(14,1)Z_k-p \left( \frac{-1}p \right) \sum\limits_{k = 0}^{n-1} \frac{k}{192^k}T_k(14,1)\right)\in{\Bbb Z}_p \end{equation} | (5.35) |
for all
\begin{equation} \begin{aligned}& \left( \frac 3p \right)\sum\limits_{k = 0}^{p-1} \frac{T_k(14,1)Z_k}{192^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,3\ ({\rm{mod}}\ {8})\ &\ p = x^2+2y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv} 5,7\ ({\rm{mod}}\ 8). \end{cases} \end{aligned} \end{equation} | (5.36) |
Conjecture 5.5.
\begin{equation} n\ \bigg|\ \sum\limits_{k = 0}^{n-1}(30k-7)T_k(62,1)Z_k640^{n-1-k}. \end{equation} | (5.37) |
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{30k-7}{640^k}T_k(62,1)Z_k{\equiv} p \left(2 \left( \frac{-1}p \right)-9 \left( \frac{15}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (5.38) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{30k-7}{640^k}T_k(62,1)Z_k -p \left( \frac{-1}p \right)\sum\limits_{k = 0}^{n-1} \frac{30k-7}{640^k}T_k(62,1)Z_k\right)\in{\Bbb Z}_p \end{equation} | (5.39) |
for all
\begin{equation} \begin{aligned}& \left( \frac{-1}p \right)\sum\limits_{k = 0}^{p-1} \frac{T_k(62,1)Z_k}{640^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac 2p) = ( \frac p3) = ( \frac p5) = 1\ &\ p = x^2+30y^2, \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac 2p) = 1,\ ( \frac p3) = ( \frac p5) = -1\ &\ p = 2x^2+15y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p3) = 1,\ ( \frac 2p) = ( \frac p5) = -1\ &\ p = 3x^2+10y^2, \\20x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p5) = 1,\ ( \frac 2p) = ( \frac p3) = -1\ &\ p = 5x^2+6y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-30}p) = -1, \end{cases} \end{aligned} \end{equation} | (5.40) |
where
Sun [36,37] obtained some supercongruences involving the Franel numbers
Let
\begin{align*} \sum\limits_{k = 0}^{p-1} \frac{f_kT_k(b,c)}{m^k}{\equiv}& \left( \frac{b^2-4c}p \right)\sum\limits_{k = 0}^{p-1} \left( \frac{-8(b^2-4c)}m \right)^kf_{p-1-k}T_{p-1-k}(b,c) \\{\equiv}& \left( \frac{b^2-4c}p \right)\sum\limits_{k = 0}^{p-1} \frac{f_kT_k(b,c)}{(8(4c-b^2)/m)^k}\ ({\rm{mod}}\ p) \end{align*} |
for any
Wan and Zudilin [49] deduced the following irrational series for
\sum\limits_{k = 0}^\infty(18k+7-2\sqrt3)f_kP_k \left( \frac{1+\sqrt3}{\sqrt6} \right) \left( \frac{2-\sqrt3}{2\sqrt6} \right)^k = \frac{27+11\sqrt3}{\sqrt2\,\pi}. |
Via our congruence approach (including Conjecture 1.4), we find
Conjecture 6.1. We have
\begin{align} \sum\limits_{k = 0}^\infty \frac{3k+1}{(-48)^k}f_kT_k(4,-2)& = \frac{4\sqrt2}{3\pi}, \end{align} | (6.1) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{99k+23}{(-288)^k}f_kT_k(8,-2)& = \frac{39\sqrt2}{\pi}, \end{align} | (6.2) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{105k+17}{480^k}f_kT_k(8,1)& = \frac{92\sqrt5}{3\pi}, \end{align} | (6.3) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{45k-2}{441^k}f_kT_k(47,1)& = \frac{483\sqrt5}{4\pi}, \end{align} | (6.4) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{165k+46}{(-2352)^k}f_kT_k(194,1)& = \frac{112\sqrt5}{3\pi}, \end{align} | (6.5) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{42k+5}{11616^k}f_kT_k(482,1)& = \frac{374\sqrt2}{15\pi}, \end{align} | (6.6) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{990k+31}{11200^k}f_kT_k(898,1)& = \frac{680\sqrt7}{\pi}, \end{align} | (6.7) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{585k+172}{(-13552)^k}f_kT_k(1454,1)& = \frac{110\sqrt7}{\pi}, \end{align} | (6.8) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{90k+11}{101568^k}f_kT_k(2114,1)& = \frac{92\sqrt{15}}{7\pi}, \end{align} | (6.9) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{94185k+17014}{(-105984)^k}f_kT_k(2302,1)& = \frac{8520\sqrt{23}}{\pi}, \end{align} | (6.10) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{5355k+1381}{(-61952)^k}f_kT_k(4354,1)& = \frac{968\sqrt{7}}{\pi}, \end{align} | (6.11) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{210k+23}{475904^k}f_kT_k(16898,1)& = \frac{2912\sqrt{231}}{297\pi}. \end{align} | (6.12) |
We now present a conjecture on congruence related to
Conjecture 6.2.
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(105k+17)480^{n-1-k}f_kT_k(8,1)\in{\Bbb Z}^+. \end{equation} | (6.13) |
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{105k+17}{480^k}f_kT_k(8,1) {\equiv} \frac p9 \left(161 \left( \frac{-5}p \right)-8 \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (6.14) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{105k+17}{480^k}f_kT_k(8,1) -p\sum\limits_{k = 0}^{n-1} \frac{105k+17}{480^k}f_kT_k(8,1)\right)\in{\Bbb Z}_p \end{equation} | (6.15) |
for all
\begin{equation} \begin{aligned}& \left( \frac{-1}p \right)\sum\limits_{k = 0}^{p-1} \frac{f_kT_k(8,1)}{480^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-15}p) = -1.\end{cases} \end{aligned} \end{equation} | (6.16) |
Remark 6.1 This conjecture was formulated by the author on Oct. 25, 2019.
Conjecture 6.3. For any
\begin{equation} \frac1{4n}\sum\limits_{k = 0}^{n-1}(-1)^{n-1-k}(105k+88)f_kT_k(8,1)\in{\Bbb Z}^+. \end{equation} | (6.17) |
\begin{equation} \sum\limits_{k = 0}^{p-1}(-1)^k(105k+88)f_kT_k(8,1) {\equiv} \frac 83p \left(23 \left( \frac {-3}p \right)+10 \left( \frac{15}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (6.18) |
If
\begin{equation} \sum\limits_{k = 0}^{pn-1}(-1)^k(105k+88)f_kT_k(8,1)-p \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1}(-1)^k(105k+88)f_kT_k(8,1) \end{equation} | (6.19) |
divided by
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1}(-1)^kf_kT_k(8,1) \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-15}p) = -1.\end{cases} \end{aligned} \end{equation} | (6.20) |
Remark 6.2. This conjecture is the dual of Conjecture 6.2.
The following conjecture is related to the identity
Conjecture 6.4.
\begin{equation} \frac1{2n}\sum\limits_{k = 0}^{n-1}(-1)^k(585k+172)13552^{n-1-k}f_kT_k(1454,1)\in{\Bbb Z}^+. \end{equation} | (6.21) |
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{585k+172}{(-13552)^k}f_kT_k(1454,1){\equiv} \frac p{11} \left(1580 \left( \frac{-7}p \right) +312 \left( \frac{273}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (6.22) |
If
\begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{585k+172}{(-13552)^k}f_kT_k(1454,1) -p \left( \frac p7 \right)\sum\limits_{k = 0}^{n-1} \frac{585k+172}{(-13552)^k}f_kT_k(1454,1) \end{equation} | (6.23) |
divided by
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{f_kT_k(1454,1)}{(-13552)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p7) = ( \frac p{13}) = 1,\ p = x^2+273y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{7}) = 1,\ ( \frac p3) = ( \frac p{13}) = -1,\ 2p = x^2+273y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p7) = -1,\ ( \frac p3) = ( \frac p{13}) = 1,\ p = 3x^2+91y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p7) = ( \frac p{13}) = -1,\ 2p = 3x^2+91y^2, \\28x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{13}) = -1,\ ( \frac p3) = ( \frac p{7}) = 1,\ p = 7x^2+39y^2, \\14x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p7) = ( \frac p{13}) = 1,\ 2p = 7x^2+39y^2, \\52x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{3}) = 1,\ ( \frac p7) = ( \frac p{13}) = -1,\ p = 13x^2+21y^2, \\26x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{13}) = 1,\ ( \frac p3) = ( \frac p{7}) = -1,\ 2p = 13x^2+21y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-273}p) = -1, \end{cases} \end{aligned} \end{equation} | (6.24) |
where
Remark 6.3. Note that the imaginary quadratic field
The following conjecture is related to the identity
Conjecture 6.5.
\begin{equation} \frac1{2n}\sum\limits_{k = 0}^{n-1}(-1)^k(94185k+17014)105984^{n-1-k}f_kT_k(2302,1)\in{\Bbb Z}^+. \end{equation} | (6.25) |
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{94185k+17014}{(-105984)^k}f_kT_k(2302,1) \\{\equiv}& \frac p{16} \left(22659+249565 \left( \frac{-23}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{aligned} \end{equation} | (6.26) |
If
\begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{94185k+17014}{(-105984)^k}f_kT_k(2302,1) -p\sum\limits_{k = 0}^{n-1} \frac{94185k+17014}{(-105984)^k}f_kT_k(2302,1) \end{equation} | (6.27) |
divided by
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{f_kT_k(2302,1)}{(-105984)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p{23}) = 1,\ p = x^2+345y^2, \\2x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{23}) = 1,\ ( \frac p3) = ( \frac p{5}) = -1,\ 2p = x^2+345y^2, \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p5) = -1,\ ( \frac p3) = ( \frac p{23}) = 1,\ p = 3x^2+115y^2, \\6x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p5) = ( \frac p{23}) = 1,\ 2p = 3x^2+115y^2, \\2p-20x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{5}) = 1,\ ( \frac p3) = ( \frac p{23}) = -1,\ p = 5x^2+69y^2, \\2p-10x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = 1,\ ( \frac p5) = ( \frac p{23}) = -1,\ 2p = 5x^2+69y^2, \\2p-60x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{3}) = ( \frac p5) = ( \frac p{23}) = -1,\ p = 15x^2+23y^2, \\2p-30x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{23}) = -1,\ ( \frac p3) = ( \frac p{5}) = 1,\ 2p = 15x^2+23y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-345}p) = -1, \end{cases} \end{aligned} \end{equation} | (6.28) |
where
Remark 6.4. Note that the imaginary quadratic field
The following conjecture is related to the identity
Conjecture 6.6.
\begin{equation} \frac1{n}\sum\limits_{k = 0}^{n-1}(210k+23)475904^{n-1-k}f_kT_k(16898,1)\in{\Bbb Z}^+. \end{equation} | (6.29) |
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{210k+23}{475904^k}f_kT_k(16898,1) \\{\equiv}& \frac p{1287} \left(40621 \left( \frac{-231}p \right)-11020 \left( \frac{66}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{aligned} \end{equation} | (6.30) |
If
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{pn-1} \frac{210k+23}{475904^k}f_kT_k(16898,1) \\&-p \left( \frac{66}p \right)\sum\limits_{k = 0}^{n-1} \frac{210k+23}{475904^k}f_kT_k(16898,1) \end{aligned} \end{equation} | (6.31) |
divided by
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{f_kT_k(16898,1)}{475904^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p7) = ( \frac p{11}) = 1\ &\ p = x^2+462y^2, \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p7) = 1,\ ( \frac p3) = ( \frac p{11}) = -1\ &\ p = 2x^2+231y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p7) = -1,\ ( \frac p3) = ( \frac p{11}) = 1\ &\ p = 3x^2+154y^2, \\2p-24x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p7) = ( \frac p{11}) = -1\ &\ p = 6x^2+77y^2, \\28x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p{3}) = 1,\ ( \frac p7) = ( \frac p{11}) = -1\ &\ p = 7x^2+66y^2, \\44x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = -1,\ ( \frac p7) = ( \frac p{11}) = 1\ &\ p = 11x^2+42y^2, \\56x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p{11}) = 1,\ ( \frac p3) = ( \frac p7) = -1\ &\ p = 14x^2+33y^2, \\2p-84x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p{11}) = -1,\ ( \frac p3) = ( \frac p{7}) = 1\ &\ p = 21x^2+22y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-462}p) = -1, \end{cases} \end{aligned} \end{equation} | (6.32) |
where
Remark 6.5. Note that the imaginary quadratic field
The identities
{\Bbb Q}(\sqrt{-165}),\ {\Bbb Q}(\sqrt{-210}),\ {\Bbb Q}(\sqrt{-210}),\ {\Bbb Q}(\sqrt{-330}),\ {\Bbb Q}(\sqrt{-357}) |
(with class number
For
g_n: = \sum\limits_{k = 0}^n \binom nk^2 \binom{2k}k. |
It is known that
Let
g_k{\equiv} \left( \frac{-3}p \right)9^kg_{p-1-k}\ ({\rm{mod}}\ p) |
by [24,Lemma 2.7(ⅱ)]. Combining this with Remark 1.3(ⅱ), we see that
\begin{align*} \sum\limits_{k = 0}^{p-1} \frac{g_kT_k(b,c)}{m^k}{\equiv}& \left( \frac{-3(b^2-4c)}p \right)\sum\limits_{k = 0}^{p-1} \left( \frac{9(b^2-4c)}m \right)^kg_{p-1-k}T_{p-1-k}(b,c) \\{\equiv}& \left( \frac{3(4c-b^2)}p \right)\sum\limits_{k = 0}^{p-1} \frac{g_kT_k(b,c)}{(9(b^2-4c)/m)^k}\ ({\rm{mod}}\ p) \end{align*} |
for any
Wan and Zudilin [49] obtained the following irrational series for
\sum\limits_{k = 0}^\infty(22k+7-3\sqrt3)g_kP_k \left( \frac{\sqrt{14\sqrt3-15}}3 \right) \left( \frac{\sqrt{2\sqrt3-3}}{9} \right)^k = \frac{9}{2\pi}(9+4\sqrt3). |
Using our congruence approach (including Conjecture 1.4), we find 12 rational series for
Conjecture 7.1. We have the following identities.
\begin{align} \sum\limits_{k = 0}^\infty \frac{8k+3}{(-81)^k}g_kT_k(7,-8)& = \frac{9\sqrt3}{4\pi}, \end{align} | (7.1) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{4k+1}{(-1089)^k}g_kT_k(31,-32)& = \frac{33}{16\pi}, \end{align} | (7.2) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{7k-1}{540^k}g_kT_k(52,1)& = \frac{30\sqrt3}{\pi}, \end{align} | (7.3) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{20k+3}{3969^k}g_kT_k(65,64)& = \frac{63\sqrt3}{8\pi}, \end{align} | (7.4) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{280k+93}{(-1980)^k}g_kT_k(178,1)& = \frac{20\sqrt{33}}{\pi}, \end{align} | (7.5) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{176k+15}{12600^k}g_kT_k(502,1)& = \frac{25\sqrt{42}}{\pi}, \end{align} | (7.6) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{560k-23}{13068^k}g_kT_k(970,1)& = \frac{693\sqrt3}{\pi}, \end{align} | (7.7) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{12880k+1353}{105840^k}g_kT_k(2158,1)& = \frac{4410\sqrt3}{\pi}, \end{align} | (7.8) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{299k+59}{(-101430)^k}g_kT_k(2252,1)& = \frac{735\sqrt{115}}{64\pi}, \end{align} | (7.9) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{385k+118}{(-53550)^k}g_kT_k(4048,1)& = \frac{2415\sqrt{17}}{64\pi}, \end{align} | (7.10) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{385k-114}{114264^k}g_kT_k(10582,1)& = \frac{15939\sqrt3}{16\pi}, \end{align} | (7.11) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{16016k+1273}{510300^k}g_kT_k(17498,1)& = \frac{14175\sqrt3}{2\pi}. \end{align} | (7.12) |
Now we present a conjecture on congruences related to
Conjecture 7.2.
\begin{equation} \frac1{3n}\sum\limits_{k = 0}^{n-1}(176k+15)12600^{n-1-k}g_kT_k(502,1)\in{\Bbb Z}^+, \end{equation} | (7.13) |
and this number is odd if and only if
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{176k+15}{12600^k}g_kT_k(502,1){\equiv} p \left(26 \left( \frac{-42}p \right)-11 \left( \frac{21}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (7.14) |
If
\begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{176k+15}{12600^k}g_kT_k(502,1) -p \left( \frac{21}p \right)\sum\limits_{k = 0}^{n-1} \frac{176k+15}{12600^k}g_kT_k(502,1) \end{equation} | (7.15) |
divided by
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{g_kT_k(502,1)}{12600^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p3) = ( \frac p5) = ( \frac p7) = 1\ &\ p = x^2+210y^2, \\2p-8x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p7) = 1,\ ( \frac p3) = ( \frac p5) = -1\ &\ p = 2x^2+105y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p3) = 1,\ ( \frac p5) = ( \frac p7) = -1\ &\ p = 3x^2+70y^2, \\2p-20x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p3) = ( \frac p5) = ( \frac p7) = -1\ &\ p = 5x^2+42y^2, \\24x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p5) = 1,\ ( \frac p3) = ( \frac p7) = -1\ &\ p = 6x^2+35y^2, \\28x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p5) = -1,\ ( \frac p3) = ( \frac p7) = 1\ &\ p = 7x^2+30y^2, \\40x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p7) = -1,\ ( \frac p3) = ( \frac p5) = 1\ &\ p = 10x^2+21y^2, \\56x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-2}p) = ( \frac p3) = -1,\ ( \frac p5) = ( \frac p7) = 1\ &\ p = 14x^2+15y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-210}p) = -1, \end{cases} \end{aligned} \end{equation} | (7.16) |
where
Remark 7.1. Note that the imaginary quadratic field
The following conjecture is related to the identity
Conjecture 7.3.
\begin{equation} \frac1{3n}\sum\limits_{k = 0}^{n-1}(12880k+1353)105840^{n-1-k}g_kT_k(2158,1)\in{\Bbb Z}^+, \end{equation} | (7.17) |
and this number is odd if and only if
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{12880k+1353}{105840^k}g_kT_k(2158,1) \\{\equiv}& \frac p{2} \left(3419 \left( \frac {-3}p \right)-713 \left( \frac{5}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{aligned} \end{equation} | (7.18) |
If
\begin{equation} \sum\limits_{k = 0}^{pn-1} \frac{12880k+1353}{105840^k}g_kT_k(2158,1) -p \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1} \frac{12880k+1353}{105840^k}g_kT_k(2158,1) \end{equation} | (7.19) |
divided by
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{g_kT_k(2158,1)}{105840^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p{11}) = 1,\ p = x^2+330y^2, \\2p-8x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p{11}) = -1,\ p = 2x^2+165y^2, \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{11}) = 1,\ ( \frac p3) = ( \frac p5) = -1,\ p = 3x^2+110y^2, \\2p-20x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p5) = ( \frac p{11}) = 1,\ p = 5x^2+66y^2, \\24x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{11}) = -1,\ ( \frac p3) = ( \frac p5) = 1,\ p = 6x^2+55y^2, \\40x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = 1,\ ( \frac p5) = ( \frac p{11}) = -1,\ p = 10x^2+33y^2, \\44x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{5}) = 1,\ ( \frac p3) = ( \frac p{11}) = -1,\ p = 11x^2+30y^2, \\60x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p5) = -1,\ ( \frac p3) = ( \frac p{11}) = 1,\ p = 15x^2+22y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-330}p) = -1, \end{cases} \end{aligned} \end{equation} | (7.20) |
where
Remark 7.2. Note that the imaginary quadratic field
Now we pose a conjecture related to the identity
Conjecture 7.4.
\begin{equation} \frac1{2n}\sum\limits_{k = 0}^{n-1}(-1)^k(385k+118)53550^{n-1-k}g_kT_k(4048,1)\in{\Bbb Z}^+. \end{equation} | (7.21) |
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{385k+118}{(-53550)^k}g_kT_k(4048,1) \\{\equiv}& \frac p{320} \left(29279 \left( \frac{-17}p \right)+8481 \left( \frac{7}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{aligned} \end{equation} | (7.22) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{385k+118}{(-53550)^k}g_kT_k(4048,1) -p \left( \frac{7}p \right)\sum\limits_{k = 0}^{n-1} \frac{385k+118}{(-53550)^k}g_kT_k(4048,1)\right) \end{equation} | (7.23) |
is a
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{g_kT_k(4048,1)}{(-53550)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p7) = ( \frac p{17}) = 1,\ p = x^2+357y^2, \\2p-2x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = -1,\ ( \frac p7) = ( \frac p{17}) = 1,\ 2p = x^2+357y^2, \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = ( \frac p5) = ( \frac p7) = -1,\ p = 3x^2+119y^2, \\2p-6x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p3) = 1,\ ( \frac p7) = ( \frac p{17}) = -1,\ 2p = 3x^2+119y^2, \\28x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{17}) = -1,\ ( \frac p3) = ( \frac p7) = 1,\ p = 7x^2+51y^2, \\2p-14x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p7) = 1,\ ( \frac p3) = ( \frac p{17}) = -1,\ 2p = 7x^2+51y^2, \\2p-68x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p{17}) = 1,\ ( \frac p3) = ( \frac p7) = -1,\ p = 17x^2+21y^2, \\34x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-1}p) = ( \frac p7) = -1,\ ( \frac p3) = ( \frac p{17}) = 1,\ 2p = 17x^2+21y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-357}p) = -1, \end{cases} \end{aligned} \end{equation} | (7.24) |
where
Remark 7.3. Note that the imaginary quadratic field
Now we pose a conjecture related to the identity
Conjecture 7.5.
\begin{equation} \frac1{n}\sum\limits_{k = 0}^{n-1}(16016k+1273)510300^{n-1-k}g_kT_k(17498,1)\in{\Bbb Z}^+, \end{equation} | (7.25) |
and this number is odd if and only if
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{p-1} \frac{16016k+1273}{510300^k}g_kT_k(17498,1) \\{\equiv}& \frac p{3} \left(6527 \left( \frac{-3}p \right)-2708 \left( \frac{42}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{aligned} \end{equation} | (7.26) |
If
\begin{equation} \begin{aligned} &\sum\limits_{k = 0}^{pn-1} \frac{16016k+1273}{510300^k}g_kT_k(17498,1) \\&-p \left( \frac{p}3 \right)\sum\limits_{k = 0}^{n-1} \frac{16016k+1273}{510300^k}g_kT_k(17498,1) \end{aligned} \end{equation} | (7.27) |
divided by
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{g_kT_k(17498,1)}{510300^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p7) = ( \frac p{11}) = 1\ &\ p = x^2+462y^2, \\2p-8x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p7) = 1,\ ( \frac p3) = ( \frac p{11}) = -1\ &\ p = 2x^2+231y^2, \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p7) = -1,\ ( \frac p3) = ( \frac p{11}) = 1\ &\ p = 3x^2+154y^2, \\2p-24x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = ( \frac p7) = ( \frac p{11}) = -1\ &\ p = 6x^2+77y^2, \\28x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p{3}) = 1,\ ( \frac p7) = ( \frac p{11}) = -1\ &\ p = 7x^2+66y^2, \\44x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p3) = -1,\ ( \frac p7) = ( \frac p{11}) = 1\ &\ p = 11x^2+42y^2, \\2p-56x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p{11}) = 1,\ ( \frac p3) = ( \frac p7) = -1\ &\ p = 14x^2+33y^2, \\84x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{2}p) = ( \frac p{11}) = -1,\ ( \frac p3) = ( \frac p{7}) = 1\ &\ p = 21x^2+22y^2, \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-462}p) = -1, \end{cases} \end{aligned} \end{equation} | (7.28) |
where
Remark 7.4. Note that the imaginary quadratic field
The identities
To conclude this section, we confirm an open series for
Theorem 7.1. We have
\begin{equation} \sum\limits_{n = 0}^\infty \frac{16n+5}{324^n} \binom{2n}ng_n(-20) = \frac{189}{25\pi}, \end{equation} | (7.29) |
where
g_n(x): = \sum\limits_{k = 0}^n \binom nk^2 \binom{2k}kx^k. |
Proof. The Franel numbers of order
f_n^{(4)} \leq\left(\sum\limits_{k = 0}^n \binom nk^2\right)^2 = \binom{2n}n^2 \leq ((1+1)^{2n})^2 = 16^n. |
By [11,(8.1)], for
\begin{equation} \begin{aligned}&\sum\limits_{n = 0}^\infty \binom{2n}n \frac{(an+b)x^n}{(1+2x)^{2n}}\sum\limits_{k = 0}^n \binom nk^2 \binom{2(n-k)}{n-k}x^k \\ = &(1+2x)\sum\limits_{n = 0}^\infty \left( \frac{4a(1-x)(1+2x)n+6ax(2-x)}{5(1-4x)}+b \right)f_n^{(4)}x^n. \end{aligned} \end{equation} | (7.30) |
Since
\begin{align*} & \frac{x^n}{(1+2x)^{2n}}\sum\limits_{k = 0}^n \binom nk^2 \binom{2n-2k}{n-k}x^k \\ = & \frac{x^n}{(1+2x)^{2n}}\sum\limits_{k = 0}^n \binom nk^2 \binom{2k}{k}x^{n-k} = (2+x^{-1})^{-2n}g_n(x^{-1}), \end{align*} |
putting
\sum\limits_{n = 0}^\infty \frac{16n+5}{18^{2n}} \binom{2n}ng_n(-20) = \frac{378}{125}\sum\limits_{n = 0}^\infty \frac{3n+1}{(-20)^n}f_n^{(4)}. |
As
\sum\limits_{n = 0}^\infty \frac{3n+1}{(-20)^n}f_n^{(4)} = \frac{5}{2\pi} |
by Cooper [9], we finally get
\sum\limits_{n = 0}^\infty \frac{16n+5}{18^{2n}} \binom{2n}ng_n(-20) = \frac{378}{125}\times \frac 5{2\pi} = \frac{189}{25\pi}. |
This concludes the proof of (7.29).
Recall that the numbers
\beta_n: = \sum\limits_{k = 0}^n \binom nk^2 \binom{n+k}k\ \ \ (n = 0,1,2,\ldots) |
are a kind of Apéry numbers. Let
\beta_k{\equiv}(-1)^k\beta_{p-1-k}\ ({\rm{mod}}\ p) |
by [24,Lemma 2.7(ⅰ)]. Combining this with Remark 1.3(ⅱ), we see that
\begin{align*} \sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(b,c)}{m^k}{\equiv}& \left( \frac{b^2-4c}p \right)\sum\limits_{k = 0}^{p-1} \left( \frac{-(b^2-4c)}m \right)^k\beta_{p-1-k}T_{p-1-k}(b,c) \\{\equiv}& \left( \frac{b^2-4c}p \right)\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(b,c)}{((4c-b^2)/m)^k}\ ({\rm{mod}}\ p) \end{align*} |
for any
Wan and Zudilin [49] obtained the following irrational series for
\sum\limits_{k = 0}^\infty(60k+16-5\sqrt{10})\beta_kP_k \left( \frac{5\sqrt2+17\sqrt5}{45} \right) \left( \frac{5\sqrt2-3\sqrt5} 5 \right)^k = \frac{135\sqrt2+81\sqrt5}{\sqrt2\,\pi}. |
Using our congruence approach (including Conjecture 1.4), we find one rational series for
Conjecture 8.1. (ⅰ) We have
\begin{equation} \sum\limits_{k = 0}^\infty \frac{145k+9}{900^k}\beta_kT_k(52,1) = \frac{285}{\pi}. \end{equation} | (8.1) |
Also, for any
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(145k+9)900^{n-1-k}\beta_kT_k(52,1)\in{\Bbb Z}^+. \end{equation} | (8.2) |
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{145k+9}{900^k}\beta_kT_k(52,1) {\equiv} \frac p5 \left(133 \left( \frac{-1}p \right)-88 \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (8.3) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{145k+9}{900^k}\beta_kT_k(52,1) -p\sum\limits_{k = 0}^{n-1} \frac{145k+9}{900^k}\beta_kT_k(52,1)\right) \in{\Bbb Z}_p \end{equation} | (8.4) |
for all
\begin{equation} \begin{aligned}& \left( \frac{-1}p \right)\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(52,1)}{900^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\2p-12x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-15}p) = -1.\end{cases} \end{aligned} \end{equation} | (8.5) |
Remark 8.1. This conjecture was formulated by the author on Oct. 27, 2019.
Conjecture 8.2.
\begin{equation} \frac1{2n}\sum\limits_{k = 0}^{n-1}(-1)^k(15k+8)\beta_kT_k(4,-1)\in{\Bbb Z}, \end{equation} | (8.6) |
and this number is odd if and only if
\begin{equation} \sum\limits_{k = 0}^{p-1}(-1)^k(15k+8)\beta_kT_k(4,-1) {\equiv} \frac p4 \left(27 \left( \frac p3 \right)+5 \left( \frac p5 \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (8.7) |
If
\begin{equation} \sum\limits_{k = 0}^{pn-1}(-1)^k(15k+8)\beta_kT_k(4,-1) -p \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1}(-1)^k(15k+8)\beta_kT_k(2,2) \end{equation} | (8.8) |
divided by
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1}(-1)^k\beta_kT_k(4,-1) \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1,4\ ({\rm{mod}}\ {15})\ &\ p = x^2+15y^2\ (x,y\in{\Bbb Z}), \\12x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}2,8\ ({\rm{mod}}\ {15})\ &\ p = 3x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac {-15}p) = -1.\end{cases} \end{aligned} \end{equation} | (8.9) |
Remark 8.2. This conjecture was formulated by the author on Nov. 13, 2019.
Conjecture 8.3.
\begin{equation} \frac3{n2^{\lfloor n/2\rfloor}}\sum\limits_{k = 0}^{n-1}(2k+1)(-2)^{n-1-k}\beta_kT_k(2,2)\in{\Bbb Z}^+, \end{equation} | (8.10) |
and this number is odd if and only if
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{2k+1}{(-2)^k}\beta_kT_k(2,2) {\equiv} \frac p3 \left(1+2 \left( \frac{-1}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (8.11) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{2k+1}{(-2)^k}\beta_kT_k(2,2) -p\sum\limits_{k = 0}^{n-1} \frac{2k+1}{(-2)^k}\beta_kT_k(2,2)\right)\in{\Bbb Z}_p \end{equation} | (8.12) |
for all
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(2,2)}{(-2)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1\ ({\rm{mod}}\ 4)\ &\ p = x^2+4y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3\ ({\rm{mod}}\ 4).\end{cases} \end{aligned} \end{equation} | (8.13) |
Remark 8.3. This conjecture was formulated by the author on Nov. 13, 2019.
Conjecture 8.4.
\begin{equation} \frac1{n2^{\lfloor(n+1)/2\rfloor}}\sum\limits_{k = 0}^{n-1}(3k+2)(-2)^{n-1-k}\beta_kT_k(20,2)\in{\Bbb Z}^+, \end{equation} | (8.14) |
and this number is odd if and only if
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{3k+2}{(-2)^k}\beta_kT_k(20,2) {\equiv}2p \left( \frac 2p \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (8.15) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{3k+2}{(-2)^k}\beta_kT_k(20,2) -p\sum\limits_{k = 0}^{n-1} \frac{3k+2}{(-2)^k}\beta_kT_k(20,2)\right)\in{\Bbb Z}_p \end{equation} | (8.16) |
for all
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(20,2)}{(-2)^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}1\ ({\rm{mod}}\ 4)\ &\ p = x^2+4y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3\ ({\rm{mod}}\ 4).\end{cases} \end{aligned} \end{equation} | (8.17) |
Conjecture 8.5.
\begin{equation} \frac3n\sum\limits_{k = 0}^{n-1}(5k+3)4^{n-1-k}\beta_kT_k(14,-1)\in{\Bbb Z}. \end{equation} | (8.18) |
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{5k+3}{4^k}\beta_kT_k(14,-1) {\equiv} \frac p3 \left(4 \left( \frac{-2}p \right)+5 \left( \frac 2p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (8.19) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{5k+3}{4^k}\beta_kT_k(14,-1) -p \left( \frac 2p \right)\sum\limits_{k = 0}^{n-1} \frac{5k+3}{4^k}\beta_kT_k(14,-1)\right)\in{\Bbb Z}_p \end{equation} | (8.20) |
for all
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(14,-1)}{4^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac {-2}p) = ( \frac 5p) = 1\ &\ p = x^2+10y^2\ (x,y\in{\Bbb Z}), \\8x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac {-2}p) = ( \frac 5p) = -1\ &\ p = 2x^2+5y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac {-10}p) = -1.\end{cases} \end{aligned} \end{equation} | (8.21) |
Conjecture 8.6.
\begin{equation} \frac1{3n}\sum\limits_{k = 0}^{n-1}(22k+15)(-4)^{n-1-k}\beta_kT_k(46,1)\in{\Bbb Z}^+, \end{equation} | (8.22) |
and this number is odd if and only if
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{22k+15}{(-4)^k}\beta_kT_k(46,1) {\equiv} \frac p4 \left(357-297 \left( \frac{33}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (8.23) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{22k+15}{(-4)^k}\beta_kT_k(46,1) -p\sum\limits_{k = 0}^{n-1} \frac{22k+15}{(-4)^k}\beta_kT_k(46,1)\right)\in{\Bbb Z}_p \end{equation} | (8.24) |
for all
\begin{equation} \begin{aligned}&\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(46,1)}{(-4)^k} \\{\equiv}&\begin{cases}x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{11}) = 1\ &\ 4p = x^2+11y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{11}) = -1.\end{cases} \end{aligned} \end{equation} | (8.25) |
Conjecture 8.7.
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(190k+91)(-60)^{n-1-k}\beta_kT_k(82,1)\in{\Bbb Z}^+, \end{equation} | (8.26) |
and this number is odd if and only if
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{190k+91}{(-60)^k}\beta_kT_k(82,1) {\equiv} \frac p4 \left(111+253 \left( \frac{-15}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (8.27) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{190k+91}{(-60)^k}\beta_kT_k(82,1) -p\sum\limits_{k = 0}^{n-1} \frac{190k+91}{(-60)^k}\beta_kT_k(82,1)\right)\in{\Bbb Z}_p \end{equation} | (8.28) |
for all
\begin{equation} \begin{aligned}& \left( \frac p3 \right)\sum\limits_{k = 0}^{p-1} \frac{\beta_kT_k(82,1)}{(-60)^k} \\{\equiv}&\begin{cases}x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{5}) = ( \frac p7) = 1\ &\ 4p = x^2+35y^2\ (x,y\in{\Bbb Z}), \\2p-5x^2\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p{5}) = ( \frac p7) = -1\ &\ 4p = 5x^2+7y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac {-35}p) = -1.\end{cases} \end{aligned} \end{equation} | (8.29) |
The numbers
w_n: = \sum\limits_{k = 0}^{\lfloor n/3\rfloor}(-1)^k3^{n-3k} \binom n{3k} \binom{3k}k \binom{2k}k\ \ \ (n = 0,1,2,\ldots) |
were first introduced by Zagier [51] during his study of Apéry-like integer sequences, who noted the recurrence
(n+1)^2w_{n+1} = (9n(n+1)+3)w_n-27n^2w_{n-1}\ (n = 1,2,3,\ldots). |
Lemma 9.1. Let
w_k{\equiv} \left( \frac{-3}p \right)27^kw_{p-1-k}\ ({\rm{mod}}\ p)\quad \mathit{\text{for all}}\ k = 0,\ldots,p-1. |
Proof. Note that
\begin{align*} w_{p-1} = &\sum\limits_{k = 0}^{\lfloor(p-1)/3\rfloor}(-1)^k3^{p-1-3k} \binom{p-1}{3k} \binom{3k}k \binom{2k}k \\{\equiv}&\sum\limits_{k = 0}^{p-1} \frac{ \binom{2k}k \binom{3k}k}{27^k}{\equiv} \left( \frac p3 \right)\ ({\rm{mod}}\ p) \end{align*} |
with the help of the known congruence
\begin{align*} w_{p-2} = &\sum\limits_{k = 0}^{\lfloor(p-2)/3\rfloor}(-1)^k3^{p-2-3k} \binom{p-2}{3k} \binom{3k}k \binom{2k}k \\ = &\sum\limits_{k = 0}^{\lfloor(p-2)/3\rfloor}(-1)^k3^{p-2-3k} \frac{3k+1}{p-1} \binom{p-1}{3k+1} \binom{3k}k \binom{2k}k \\{\equiv}& \frac19\sum\limits_{k = 0}^{p-1}(9k+3) \frac{ \binom{2k}k \binom{3k}k}{27^k} {\equiv} \frac19 \left( \frac p3 \right)+ \frac19\sum\limits_{k = 0}^{p-1}(9k+2) \frac{ \binom{2k}k \binom{3k}k}{27^k}\ ({\rm{mod}}\ p). \end{align*} |
By induction,
\sum\limits_{k = 0}^n(9k+2) \frac{ \binom{2k}k \binom{3k}k}{27^k} = (3n+1)(3n+2) \frac{ \binom{2n}n \binom{3n}n}{27^n} |
for all
\sum\limits_{k = 0}^{p-1}(9k+2) \frac{ \binom{2k}k \binom{3k}k}{27^k} = \frac{(3p-2)(3p-1)}{27^{p-1}}pC_{p-1} \binom{3p-3}{p-1}{\equiv}0\ ({\rm{mod}}\ p). |
So we have
Now let
w_j{\equiv} \left( \frac{-3}p \right)27^jw_{p-1-j}\quad\text{for all}\ j = 0,\ldots,k. |
Then
\begin{align*} &(k+1)^2w_{k+1} = (9k(k+1)+3)w_k-27k^2w_{k-1} \\{\equiv}&(9(p-k)(p-k-1)+3) \left( \frac{-3}p \right)27^kw_{p-1-k} -27(p-k)^2 \left( \frac{-3}p \right)27^{k-1}w_{p-1-(k-1)} \\ = & \left( \frac{-3}p \right)27^k\times 27(p-k-1)^2w_{p-k-2}\ ({\rm{mod}}\ p) \end{align*} |
and hence
w_{k+1}{\equiv} \left( \frac{-3}p \right)27^{k+1}w_{p-1-(k+1)}\ ({\rm{mod}}\ p). |
In view of the above, we have proved the desired result by induction.
For Lemma 9.1 one may also consult [31,Corollary 3.1]. Let
\begin{align*} \sum\limits_{k = 0}^{p-1} \frac{w_kT_k(b,c)}{m^k}{\equiv}& \left( \frac{-3(b^2-4c)}p \right)\sum\limits_{k = 0}^{p-1} \left( \frac{27(b^2-4c)}m \right)^kw_{p-1-k}T_{p-1-k}(b,c) \\{\equiv}& \left( \frac{-3(b^2-4c)}p \right)\sum\limits_{k = 0}^{p-1} \frac{w_kT_k(b,c)}{(27(b^2-4c)/m)^k}\ ({\rm{mod}}\ p) \end{align*} |
for any
Wan and Zudilin [49] obtained the following irrational series for
\sum\limits_{k = 0}^\infty(14k+7-\sqrt{21})w_kP_k \left( \frac{\sqrt{21}}{5} \right) \left( \frac{7\sqrt{21}-27} {90} \right)^k = \frac{5\sqrt{7(7\sqrt{21}+27)}}{4\sqrt2\,\pi}. |
Using our congruence approach (including Conjecture 1.4), we find five rational series for
Conjecture 9.1. We have
\begin{align} \sum\limits_{k = 0}^\infty \frac{13k+3}{100^k}w_kT_k(14,-1)& = \frac{30\sqrt2}{\pi}, \end{align} | (9.1) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{14k+5}{108^k}w_kT_k(18,1)& = \frac{27\sqrt3}{\pi}, \end{align} | (9.2) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{19k+2}{486^k}w_kT_k(44,-2)& = \frac{81\sqrt3}{4\pi}, \end{align} | (9.3) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{91k+32}{(-675)^k}w_kT_k(52,1)& = \frac{45\sqrt3}{2\pi}, \end{align} | (9.4) |
\begin{align} \sum\limits_{k = 0}^\infty \frac{182k+37}{756^k}w_kT_k(110,1)& = \frac{315\sqrt3}{\pi}. \end{align} | (9.5) |
Below we present our conjectures on congruences related to the identities (9.2) and (9.5).
Conjecture 9.2.
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(14k+5)108^{n-1-k}w_kT_k(18,1)\in{\Bbb Z}^+, \end{equation} | (9.6) |
and this number is odd if and only if
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{14k+5}{108^k}w_kT_k(18,1){\equiv} \frac p4 \left(27 \left( \frac {-3}p \right)-7 \left( \frac {21}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (9.7) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{14k+5}{108^k}w_kT_k(18,1)- \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1} \frac{14k+5}{108^k}w_kT_k(18,1)\right)\in{\Bbb Z}_p \end{equation} | (9.8) |
for all
\begin{equation} \begin{aligned}& \left( \frac p3 \right)\sum\limits_{k = 0}^{p-1} \frac{w_kT_k(18,1)}{108^k} \\{\equiv}&\begin{cases}x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p5) = ( \frac p7) = 1\ &\ 4p = x^2+35y^2\ (x,y\in{\Bbb Z}), \\5x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p5) = ( \frac p7) = -1\ &\ 4p = 5x^2+7y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-35}p) = -1. \end{cases}\end{aligned} \end{equation} | (9.9) |
Conjecture 9.3.
\begin{equation} \frac1n\sum\limits_{k = 0}^{n-1}(182k+37)756^{n-1-k}w_kT_k(110,1)\in{\Bbb Z}^+, \end{equation} | (9.10) |
and this number is odd if and only if
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{182k+37}{756^k}w_kT_k(110,1){\equiv} \frac p4 \left(265 \left( \frac {-3}p \right)-117 \left( \frac {21}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (9.11) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{182k+37}{756^k}w_kT_k(110,1)- \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1} \frac{182k+37}{756^k}w_kT_k(110,1)\right)\in{\Bbb Z}_p \end{equation} | (9.12) |
for all
\begin{equation} \begin{aligned}& \left( \frac p3 \right)\sum\limits_{k = 0}^{p-1} \frac{w_kT_k(110,1)}{756^k} \\{\equiv}&\begin{cases}x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p7) = ( \frac p{13}) = 1\ &\ 4p = x^2+91y^2\ (x,y\in{\Bbb Z}), \\7x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac p7) = ( \frac p{13}) = -1\ &\ 4p = 7x^2+13y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ ( \frac{-91}p) = -1. \end{cases}\end{aligned} \end{equation} | (9.13) |
Now we give one more conjecture in this section.
Conjecture 9.4.
\begin{equation} \frac1{3n2^{\lfloor(n+1)/2\rfloor}} \sum\limits_{k = 0}^{n-1}(2k+1)54^{n-1-k}w_kT_k(10,-2)\in{\Bbb Z}^+. \end{equation} | (9.14) |
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{2k+1}{54^k}w_kT_k(10,-2){\equiv} p \left( \frac p3 \right)+ \frac{p}2(2^{p-1}-1) \left(5 \left( \frac p3 \right)+3 \left( \frac 3p \right) \right)\ ({\rm{mod}}\ {p^3}). \end{equation} | (9.15) |
If
\begin{equation} \frac1{(pn)^2}\left(\sum\limits_{k = 0}^{pn-1} \frac{2k+1}{54^k}w_kT_k(10,-2)- \left( \frac p3 \right)\sum\limits_{k = 0}^{n-1} \frac{2k+1}{54^k}w_kT_k(10,-2)\right)\in{\Bbb Z}_p \end{equation} | (9.16) |
for all
\begin{equation} \begin{aligned}& \left( \frac p3 \right)\sum\limits_{k = 0}^{p-1} \frac{w_kT_k(10,-2)}{54^k} \\{\equiv}&\begin{cases}4x^2-2p\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ 4\mid p-1\ &\ p = x^2+4y^2\ (x,y\in{\Bbb Z}), \\0\ ({\rm{mod}}\ {p^2})& \mathit{\text{if}}\ p{\equiv}3\ ({\rm{mod}}\ 4). \end{cases}\end{aligned} \end{equation} | (9.17) |
Remark 9.1. For primes
Let
\begin{align*} \sum\limits_{k = 1}^{p-1} \frac{a+dk}{m^k} \binom{2k}k^2T_k(b,c) {\equiv}&\sum\limits_{k = 1}^{(p-1)/2} \frac{a+dk}{k^2m^k} \left(k \binom{2k}k \right)^2T_k(b,c) \\{\equiv}&\sum\limits_{k = 1}^{(p-1)/2} \frac{a+dk}{k^2m^k} \left(- \frac{2p}{ \binom{2(p-k)}{p-k}} \right)^2T_k(b,c)\ ({\rm{mod}}\ p) \end{align*} |
with the aid of [33,Lemma 2.1]. Thus
\begin{align*} &\sum\limits_{k = 1}^{p-1} \frac{a+dk}{m^k} \binom{2k}k^2T_k(b,c) \\{\equiv}&4p^2\sum\limits_{k = 1}^{(p-1)/2} \frac{a+dk}{k^2m^k}\times \frac{T_k(b,c)}{ \binom{2(p-k)}{p-k}^2} \\{\equiv}&4p^2\sum\limits_{p/2 < k < p} \frac{a+d(p-k)}{(p-k)^2m^{p-k}}\times \frac{T_{p-k}(b,c)}{ \binom{2k}k^2} \\{\equiv}&4p^2\sum\limits_{k = 1}^{p-1} \frac{(a-dk)m^{k-1}}{k^2 \binom{2k}k^2} \left( \frac{b^2-4c}p \right)(b^2-4c)^{p-k}T_{p-1-(p-k)}(b,c) \\{\equiv}& \left( \frac{b^2-4c}p \right)4p^2\sum\limits_{k = 1}^{p-1} \frac{(a-dk)T_{k-1}(b,c)}{k^2 \binom{2k}k^2} \left( \frac{m}{b^2-4c} \right)^{k-1} \ ({\rm{mod}}\ p) \end{align*} |
in view of Remark 1.3(ⅱ).
Let
\sum\limits_{k = 0}^{p-1}(105k+44)(-1)^k \binom{2k}k^2T_k{\equiv} p \left(20+24 \left( \frac p3 \right)(2-3^{p-1}) \right)\ ({\rm{mod}}\ {p^3}) |
implies that
p^2\sum\limits_{k = 1}^{p-1} \frac{(105k-44)T_{k-1}}{k^2 \binom{2k}k^23^{k-1}}{\equiv} 11 \left( \frac p3 \right)\ ({\rm{mod}}\ p). |
Motivated by this, we pose the following curious conjecture.
Conjecture 10.1. We have the following identities:
\begin{align} \sum\limits_{k = 1}^\infty \frac{(105k-44)T_{k-1}}{k^2 \binom{2k}k^23^{k-1}} = & \frac{5\pi}{\sqrt3}+6\log3, \end{align} | (10.1) |
\begin{align} \sum\limits_{k = 2}^\infty \frac{(5k-2)T_{k-1}}{(k-1)k^2 \binom{2k}k^23^{k-1}} = & \frac{21-2\sqrt3\,\pi-9\log3}{12}. \end{align} | (10.2) |
Remark 10.1. The two identities were conjectured by the author on Dec. 7, 2019. One can easily check them numerically via \mathsf{Mathematica} as the two series converge fast.
Now we state our related conjectures on congruences.
Conjecture 10.2. For any prime
\begin{equation} p^2\sum\limits_{k = 1}^{p-1} \frac{(105k-44)T_{k-1}}{k^2 \binom{2k}k^23^{k-1}} {\equiv} 11 \left( \frac p3 \right)+ \frac p2 \left(13-35 \left( \frac p3 \right) \right)\ ({\rm{mod}}\ {p^2}) \end{equation} | (10.3) |
and
\begin{equation} p^2\sum\limits_{k = 2}^{p-1} \frac{(5k-2)T_{k-1}}{(k-1)k^2 \binom{2k}k^23^{k-1}} {\equiv}- \frac12 \left( \frac p3 \right)- \frac p8 \left(7+ \left( \frac p3 \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (10.4) |
Conjecture 10.3. (ⅰ) We have
\frac1{n \binom{2n}n}\sum\limits_{k = 0}^{n-1}(-1)^{n-1-k}(5k+2) \binom{2k}kC_kT_k\in{\Bbb Z}^+ |
for all
\sum\limits_{k = 0}^{p-1}(-1)^k(5k+2) \binom{2k}kC_kT_k{\equiv} 2p \left(1- \left( \frac p3 \right)(3^p-3) \right)\ ({\rm{mod}}\ {p^3}) |
for each prime
\begin{equation} \begin{aligned}& \frac{\sum\limits_{k = 0}^{pn-1}(-1)^k(5k+2) \binom{2k}kC_kT_k-p\sum\limits_{k = 0}^{n-1}(-1)^k(5k+2) \binom{2k}kC_kT_k} {(pn)^2 \binom{2n}n^2} \\\quad\qquad&{\equiv} \left( \frac p3 \right) \frac{3^p-3}{2p}(-1)^nT_{n-1}\ ({\rm{mod}}\ {p}). \end{aligned} \end{equation} | (10.5) |
Remark 10.2. See also [45,Conjecture 67] for a similar conjecture.
Let
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{8k+3}{(-16)^k} \binom{2k}k^2T_k(3,-4){\equiv} p \left(1+2 \left( \frac{-1}p \right) \right)\ ({\rm{mod}}\ {p^2}) \end{equation} | (10.6) |
and
\begin{equation} \sum\limits_{k = 0}^{p-1} \frac{33k+14}{4^k} \binom{2k}k^2T_k(8,-2){\equiv} p \left(6 \left( \frac{-1}p \right)+8 \left( \frac{2}p \right) \right)\ ({\rm{mod}}\ {p^2}). \end{equation} | (10.7) |
Though (10.6) implies the congruence
p^2\sum\limits_{k = 1}^{p-1} \frac{(8k-3)T_{k-1}(3,-4)}{k^2 \binom{2k}k^2} \left(- \frac{16}{25} \right)^{k-1}{\equiv} \frac 34\ ({\rm{mod}}\ p), |
and (10.7) with
p^2\sum\limits_{k = 1}^{p-1} \frac{(33k-14)T_{k-1}(8,-2)}{k^2 \binom{2k}k^218^{k-1}}{\equiv} \frac 7{2} \left( \frac 2p \right)\ ({\rm{mod}}\ p), |
we are unable to find the exact values of the two converging series
\sum\limits_{k = 1}^\infty \frac{(8k-3)T_{k-1}(3,-4)}{k^2 \binom{2k}k^2} \left(- \frac{16}{25} \right)^{k-1} \ \ \text{and}\ \ \sum\limits_{k = 1}^{\infty} \frac{(33k-14)T_{k-1}(8,-2)}{k^2 \binom{2k}k^218^{k-1}}. |
The author would like to thank Prof. Qing-Hu Hou at Tianjin Univ. for his helpful comments on the proof of Lemma 2.3.
[1] |
Sach VJ, Buchner E, Schmieder M (2020) Enigmatic earthquake-generated large-scale clastic dyke in the Biberach area (SW Germany). Sediment Geol 398: 105571. https://doi.org/10.1016/j.sedgeo.2019.105571 doi: 10.1016/j.sedgeo.2019.105571
![]() |
[2] |
Buchner E, Sach VJ, Schmieder M (2020) New discovery of two seismite horizons challenges the Ries–Steinheim double‑impact theory. Sci Rep 10: 22143. https://doi.org/10.1038/s41598-020-79032-4 doi: 10.1038/s41598-020-79032-4
![]() |
[3] |
Stöffler D, Artemieva NA, Wünnemann K, et al. (2013) Ries crater and suevite revisited—Observations and modeling. Part Ⅰ: Observations. Meteorit Planet Sci 48: 515–589. https://doi.org/10.1111/maps.12086 doi: 10.1111/maps.12086
![]() |
[4] |
Artemieva NA, Wünnemann K, Krien F, et al. (2013) Ries crater and suevite revisited—Observations and modeling. Part Ⅱ: Modeling. Meteorit Planet Sci 48: 590–627. https://doi.org/10.1111/maps.12085 doi: 10.1111/maps.12085
![]() |
[5] |
Schmieder M, Kennedy T, Jourdan F, et al. (2018) A high-precision 40Ar/39Ar age for the Nördlinger Ries impact crater, Germany, and implications for the accurate dating of terrestrial impact events. Geochim Cosmochim Acta 220: 146–157. https://doi.org/10.1016/j.gca.2017.09.036 doi: 10.1016/j.gca.2017.09.036
![]() |
[6] |
Schmieder M, Kennedy T, Jourdan F (2018) Response to comment on "A high-precision 40Ar/39Ar age for the Nördlinger Ries impact crater, Germany, and implications for the accurate dating of terrestrial impact events" by Schmieder et al. (Geochim. Cosmochim. Acta 220 (2018) 146–157). Geochim Cosmochim Acta 238: 602–605. https://doi.org/10.1016/j.gca.2018.07.025 doi: 10.1016/j.gca.2018.07.025
![]() |
[7] | Sach VJ (1999) Litho- und biostratigraphische Untersuchungen in der Oberen Süßwassermolasse des Landkreises Biberach an der Riß (Oberschwaben). Stuttgarter Beitr Naturk B 276: 1–167. |
[8] | Sach VJ (2014) Strahlenkalke (Shatter-Cones) aus dem Brockhorizont der Oberen Süßwassermolasse in Oberschwaben (Südwestdeutschland)—Fernauswürflinge des Nördlinger-Ries-Impaktes, Pfeil Verlag, München, 1–17. |
[9] | Hofmann B, Hofmann F (1992) An impactite horizon in the upper freshwater molasse in Eastern Switzerland: Distal Ries ejecta. Eclogae Geol Helv 85: 788–789. |
[10] |
Letsch D (2017) Diamictites and soft sediment deformation related to the Ries (ca. 14.9 Ma) meteorite impact: the "Blockhorizont" of Bernhardzell (Eastern Switzerland). Int J Earth Sci 107: 1379–1380. https://doi.org/10.1007/s00531-017-1542-1 doi: 10.1007/s00531-017-1542-1
![]() |
[11] |
Holm-Alwmark S, Alwmark C, Ferrière L, et al. (2021) Shocked quartz in distal ejecta from the Ries impact event (Germany) found at ~ 180 km distance, near Bernhardzell, eastern Switzerland. Sci Rep 11: 7438. https://doi.org/10.1038/s41598-021-86685-2 doi: 10.1038/s41598-021-86685-2
![]() |
[12] |
Buchner E, Sach VJ, Schmieder M (2022) Event- and biostratigraphic evidence for two independent Ries and Steinheim asteroid impacts in the Middle Miocene. Sci Rep 12: 18603. https://doi.org/10.1038/s41598-022-21409-8 doi: 10.1038/s41598-022-21409-8
![]() |
[13] |
Collins G, Melosh HJ, Marcus R (2005) Earth impact effects program: a web-based computer program for calculating the regional environmental consequences of a meteoroid impact on Earth. Meteorit Planet Sci 40: 817–840. https://doi.org/10.1111/j.1945-5100.2005.tb00157.x doi: 10.1111/j.1945-5100.2005.tb00157.x
![]() |
[14] |
Schmieder M, Sach VJ, Buchner E (2021) The Chöpfi pinnacles near Winterthur, Switzerland: Long-distance effects of the Ries impact-earthquake? Int J Earth Sci 111: 145–147. https://doi.org/10.1007/s00531-021-02082-0 doi: 10.1007/s00531-021-02082-0
![]() |
[15] |
Buchner E, Sach VJ, Schmieder M (2021) Sand spikes pinpoint powerful palaeoseismicity. Nat Commun 12: 6731. https://doi.org/10.1038/s41467-021-27061-6 doi: 10.1038/s41467-021-27061-6
![]() |
[16] |
Maurer H, Buchner E (2007) Rekonstruktion fluvialer Systeme der Oberen Süßwassermolasse im Nordalpinen Vorlandbecken SW-Deutschlands. German J Geosci (ZdGG) 158: 249–270. https://doi.org/10.1127/1860-1804/2007/0158-0249 doi: 10.1127/1860-1804/2007/0158-0249
![]() |
[17] | Heider J, Wegele A, Amstutz GC (1976) Beobachtungen über Sandrosen und Zapfensande aus der Süßwassermolasse Südwürttembergs. Der Aufschluß 27: 297–307. |
[18] | Sanborn WB (1976) Oddities of the Mineral World. Van Nostrand Reinhold Company, NY, USA. 142. Available from: http://allanmccollum.net/amcimages/sanborn.html, (last accessed February 19, 2025). |
[19] |
Akçiz SO, Grant Ludwig L, Arrowsmith JR, et al. (2010) Century-long average time intervals between earthquake ruptures of the San Andreas fault in the Carrizo Plain, California. Geology 38: 787–790. https://doi.org/10.1130/G30995.1 doi: 10.1130/G30995.1
![]() |
[20] |
McBride EF, Picard MD, Folk RL (1994) Orientated Concretions, Ionian Coast, Italy: Evidence of Groundwater flow direction. J Sediment Res A64: 535–540. https://doi.org/10.1306/D4267DFC-2B26-11D7-8648000102C1865D doi: 10.1306/D4267DFC-2B26-11D7-8648000102C1865D
![]() |
[21] | McCullough LN, Ritter JB, Zaleha MJ, et al. (2003) Habit, formation, and implications of elongeate, calcite concretions, Victoria, Australia. Department of Geology, Wittenberg University, Ohio, USA. Published Senior Honors Thesis. 25. |
[22] | Grant JA, Wilson SA (2018) Possible Geomorphic and Crater Density Evidence for Late Aqueous Activity in Gale Crater. LPI Contrib. 49th Annual Lunar and Planetary Science Conference. |
[23] |
Metz J, Grotzinger J, Okubo C, et al. (2010) Thin‐skinned deformation of sedimentary rocks in Valles Marineris, Mars. J Geophys Res Planets 115: E11004. https://doi.org/10.1029/2010JE003593 doi: 10.1029/2010JE003593
![]() |
[24] | NASA Mars Science Laboratory, Curiosity Rover, 2024. Available from: https://mars.nasa.gov/msl/mission/science/. |
[25] |
Wray JJ (2013) Gale Crater: The Mars Science Laboratory/Curiosity rover landing site. Int J Astrobiol 12: 25–38. https://doi.org/10.1017/S1473550412000328 doi: 10.1017/S1473550412000328
![]() |
[26] |
Grotzinger JP, Sumner DY, Kah LC, et al. (2014) A habitable fluvio-lacustrine environment at Yellowknife Bay, Gale Crater, Mars. Science 343: 6169. https://doi.org/10.1126/science.1242777 doi: 10.1126/science.1242777
![]() |
[27] |
Grotzinger JP, Gupta S, Malin MC, et al. (2015) Deposition, exhumation, and paleoclimate of an ancient lake deposit, Gale Crater, Mars. Science 350: 6257. https://doi.org/10.1126/science.aac7575 doi: 10.1126/science.aac7575
![]() |
[28] |
Buz J, Ehlmann BL, Pan L, et al. (2017) Mineralogy and stratigraphy of the Gale crater rim, wall, and floor units. J Geophys Res Planets 122: 1090–1118. https://doi.org/10.1002/2016JE005163 doi: 10.1002/2016JE005163
![]() |
[29] |
Schwenzer SP, Abramov O, Allen CC, et al. (2012) Gale Crater: Formation and post-impact hydrous environments. Planet Space Sci 70: 84–95. https://doi.org/10.1016/j.pss.2012.05.014 doi: 10.1016/j.pss.2012.05.014
![]() |
[30] |
Montenat C, Barrier P, Ott d'Estevou P, et al. (2007) Seismites: An attempt at critical analysis and classification. Sediment Geol 196: 5–30. https://doi.org/10.1016/j.sedgeo.2006.08.004 doi: 10.1016/j.sedgeo.2006.08.004
![]() |
[31] | Hargitai H, Levi T (2015) Clastic dikes, Encyclopedia of Planetary Landforms, Hargitai H and Kereszturi A, Eds., Encyclopedia of Planetary Landforms, Springer, NY, USA. |
[32] |
Sleep NH, Olds EP (2018) Remote faulting triggered by strong seismic waves from the Cretaceous-Paleogene asteroid impact. Seismol Res Lett 89: 570–576. https://doi.org/10.1785/0220170223 doi: 10.1785/0220170223
![]() |
[33] |
DePalma RA, Smit J, Burnham DA, et al. (2019) A seismically induced onshore surge deposit at the K-Pg. boundary, North Dakota. PNAS 116: 8190–8199. https://doi.org/10.1073/pnas.1817407116 doi: 10.1073/pnas.1817407116
![]() |
[34] |
Vaniman DT, Bish DL, Ming DW, et al. (2014) Mineralogy of a mudstone at Yellowknife Bay, Gale Crater, Mars. Science 343: 6169. https://doi.org/10.1126/science.1243480 doi: 10.1126/science.1243480
![]() |
[35] |
Ehlmann BL, Buz J (2015) Mineralogy and fluvial history of the watersheds of Gale, Knobel, and Sharp craters: A regional context for MSL Curiosity's exploration. Geophys Res Lett 42: 264–273. https://doi.org/10.1002/2014GL062553 doi: 10.1002/2014GL062553
![]() |
[36] |
Carter J, Viviano-Beck C, Loizeau D, et al. (2015) Orbital detection and implications of akaganéite on Mars. Icarus 253: 296–310. https://doi.org/10.1016/j.icarus.2015.01.020 doi: 10.1016/j.icarus.2015.01.020
![]() |
[37] |
Tohver E, Schmieder M, Lana C, et al. (2018) End-Permian impactogenic earthquake and tsunami deposits in the intracratonic Paraná Basin of Brazil. GSA Bull 130: 1099–1120. https://doi.org/10.1130/B31626.1 doi: 10.1130/B31626.1
![]() |
[38] |
Weatherley DK, Henley RW (2013) Flash vaporization during earthquakes evidenced by gold deposits. Nature Geosci 6: 294–298. https://doi.org/10.1038/ngeo1759 doi: 10.1038/ngeo1759
![]() |
[39] |
Simms JM (2003) Uniquely extensive seismite from the latest Triassic of the United Kingdom: evidence for bolide impact? Geology 31: 557–560. https://doi.org/10.1130/0091-7613(2003)031<0557:UESFTL>2.0.CO;2 doi: 10.1130/0091-7613(2003)031<0557:UESFTL>2.0.CO;2
![]() |
[40] |
Banham SG, Gupta S, Rubin DM, et al. (2018) Ancient Martian aeolian processes and palaeomorphology reconstructed from the Stimson formation on the lower slope of Aeolis Mons, Gale crater, Mars. Sedimentology 65: 993–1042. https://doi.org/10.1111/sed.12469 doi: 10.1111/sed.12469
![]() |
[41] |
Bohacs KM, Carrol AR, Neal JE (2003) Lessons from large lake systems—Thresholds, nonlinearity, and strange attractors. Special Papers-Geological Society of America, 75–90. https://doi.org/10.1130/0-8137-2370-1.75 doi: 10.1130/0-8137-2370-1.75
![]() |
[42] |
Brož P, Oehler D, Mazzini A, et al. (2023) An overview of sedimentary volcanism on Mars. Earth Surf Dynam 11: 633–661. https://doi.org/10.5194/egusphere-2022-1458 doi: 10.5194/egusphere-2022-1458
![]() |
[43] |
Sturm S, Wulf G, Jung D, et al. (2013) The Ries impact, a double-layer rampart crater on Earth. Geology 41: 531–534. https://doi.org/10.1130/G33934.1 doi: 10.1130/G33934.1
![]() |
[44] |
Wilson SA, Morgan AM, Howard AD, et al. (2021) The global distribution of craters with alluvial fans and deltas on Mars. Geophys Res Lett 48: e2020GL091653. https://doi.org/10.1029/2020GL091653 doi: 10.1029/2020GL091653
![]() |
[45] |
Grotzinger JP, Crisp J, Vasavada AR, et al. (2012) Mars Science Laboratory mission and science investigation. Space Sci Rev 170: 5–56. https://doi.org/10.1007/s11214-012-9892-2 doi: 10.1007/s11214-012-9892-2
![]() |
1. | Ji-Cai Liu, On two supercongruences for sums of Apéry-like numbers, 2021, 115, 1578-7303, 10.1007/s13398-021-01092-6 | |
2. | Rong-Hua Wang, Michael X.X. Zhong, q-Rational reduction and q-analogues of series for π, 2023, 116, 07477171, 58, 10.1016/j.jsc.2022.08.020 | |
3. | Qing-hu Hou, Guo-jie Li, Gosper summability of rational multiples of hypergeometric terms, 2021, 27, 1023-6198, 1723, 10.1080/10236198.2021.2007903 | |
4. | Qing-Hu Hou, Ke Liu, Congruences and telescopings of P-recursive sequences, 2021, 27, 1023-6198, 686, 10.1080/10236198.2021.1934462 | |
5. | Ji-Cai Liu, On two congruences involving Franel numbers, 2020, 114, 1578-7303, 10.1007/s13398-020-00935-y | |
6. | Liuquan Wang, Yifan Yang, Ramanujan-type 1/\pi -series from bimodular forms, 2022, 59, 1382-4090, 831, 10.1007/s11139-021-00532-6 | |
7. | Zhi-Wei Sun, On Motzkin numbers and central trinomial coefficients, 2022, 136, 01968858, 102319, 10.1016/j.aam.2021.102319 | |
8. | Ji-Cai Liu, Ramanujan-Type Supercongruences Involving Almkvist–Zudilin Numbers, 2022, 77, 1422-6383, 10.1007/s00025-022-01607-6 | |
9. | Qing-Hu Hou, Zhi-Wei Sun, q-Analogues of Some Series for Powers of \pi , 2021, 25, 0218-0006, 167, 10.1007/s00026-021-00522-x | |
10. | Rong-Hua Wang, Rational Reductions for Holonomic Sequences, 2024, 1009-6124, 10.1007/s11424-024-4034-y | |
11. | Chunli Li, Wenchang Chu, Infinite series about harmonic numbers inspired by Ramanujan–like formulae, 2023, 31, 2688-1594, 4611, 10.3934/era.2023236 | |
12. | Zhi-Wei Sun, 2025, Chapter 21, 978-3-031-65063-5, 413, 10.1007/978-3-031-65064-2_21 | |
13. | Sun Zhi-Wei, Infinite series involving binomial coefficients and harmonic numbers, 2024, 54, 1674-7216, 765, 10.1360/SSM-2024-0007 |