Citation: Habib ur Rehman, Poom Kumam, Kanokwan Sitthithakerngkiet. Viscosity-type method for solving pseudomonotone equilibrium problems in a real Hilbert space with applications[J]. AIMS Mathematics, 2021, 6(2): 1538-1560. doi: 10.3934/math.2021093
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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
limn→∞n√|Tn(b,c)|=√b2−4c. |
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
∞∑k=03990k+1147(−288)3kTk(62,952)3=43295π(94√2+195√14) |
as well as its
p−1∑k=03990k+1147(−288)3kTk(62,952)3≡p19(4230(−2p)+17563(−14p)) (mod p2), |
where
In 1905, J. W. L. Glaisher [15] proved that
∞∑k=0(4k−1)(2kk)4(2k−1)4256k=−8π2. |
This actually follows from the following finite identity observed by the author [38]:
n∑k=0(4k−1)(2kk)4(2k−1)4256k=−(8n2+4n+1)(2nn)4256n for all n∈N. |
Motivated by Glaisher's identity and Ramanujan-type series for
Theorem 1.1. We have the following identities:
∞∑k=0k(4k−1)(2kk)3(2k−1)2(−64)k=−1π, | (1.1) |
∞∑k=0(4k−1)(2kk)3(2k−1)3(−64)k=2π, | (1.2) |
∞∑k=0(12k2−1)(2kk)3(2k−1)2256k=−2π, | (1.3) |
∞∑k=0k(6k−1)(2kk)3(2k−1)3256k=12π, | (1.4) |
∞∑k=0(28k2−4k−1)(2kk)3(2k−1)2(−512)k=−3√2π, | (1.5) |
∞∑k=0(30k2+3k−2)(2kk)3(2k−1)3(−512)k=27√28π, | (1.6) |
∞∑k=0(28k2−4k−1)(2kk)3(2k−1)24096k=−3π, | (1.7) |
∞∑k=0(42k2−3k−1)(2kk)3(2k−1)34096k=278π, | (1.8) |
∞∑k=0(34k2−3k−1)(2kk)2(3kk)(2k−1)(3k−1)(−192)k=−10√3π, | (1.9) |
∞∑k=0(64k2−11k−7)(2kk)2(3kk)(k+1)(2k−1)(3k−1)(−192)k=−125√39π, | (1.10) |
∞∑k=0(14k2+k−1)(2kk)2(3kk)(2k−1)(3k−1)216k=−√3π, | (1.11) |
∞∑k=0(90k2+7k+1)(2kk)2(3kk)(k+1)(2k−1)(3k−1)216k=9√32π, | (1.12) |
∞∑k=0(34k2−3k−1)(2kk)2(3kk)(2k−1)(3k−1)(−12)3k=−2√3π, | (1.13) |
∞∑k=0(17k+5)(2kk)2(3kk)(k+1)(2k−1)(3k−1)(−12)3k=9√3π, | (1.14) |
∞∑k=0(111k2−7k−4)(2kk)2(3kk)(2k−1)(3k−1)1458k=−454π, | (1.15) |
∞∑k=0(1524k2+899k+263)(2kk)2(3kk)(k+1)(2k−1)(3k−1)1458k=33754π, | (1.16) |
∞∑k=0(522k2−55k−13)(2kk)2(3kk)(2k−1)(3k−1)(−8640)k=−54√155π, | (1.17) |
∞∑k=0(1836k2+2725k+541)(2kk)2(3kk)(k+1)(2k−1)(3k−1)(−8640)k=2187√155π, | (1.18) |
∞∑k=0(529k2−45k−16)(2kk)2(3kk)(2k−1)(3k−1)153k=−55√32π, | (1.19) |
∞∑k=0(77571k2+68545k+16366)(2kk)2(3kk)(k+1)(2k−1)(3k−1)153k=59895√32π, | (1.20) |
∞∑k=0(574k2−73k−11)(2kk)2(3kk)(2k−1)(3k−1)(−48)3k=−20√3π, | (1.21) |
∞∑k=0(8118k2+9443k+1241)(2kk)2(3kk)(k+1)(2k−1)(3k−1)(−48)3k=2250√3π, | (1.22) |
∞∑k=0(978k2−131k−17)(2kk)2(3kk)(2k−1)(3k−1)(−326592)k=−990√749π, | (1.23) |
∞∑k=0(592212k2+671387k2+77219)(2kk)2(3kk)(k+1)(2k−1)(3k−1)(−326592)k=4492125√749π, | (1.24) |
∞∑k=0(116234k2−17695k−1461)(2kk)2(3kk)(2k−1)(3k−1)(−300)3k=−2650√3π, | (1.25) |
∞∑k=0(223664832k2+242140765k+18468097)(2kk)2(3kk)(k+1)(2k−1)(3k−1)(−300)3k=33497325√3π, | (1.26) |
∞∑k=0(122k2+3k−5)(2kk)2(4k2k)(2k−1)(4k−1)648k=−212π, | (1.27) |
∞∑k=0(1903k2+114k+41)(2kk)2(4k2k)(k+1)(2k−1)(4k−1)648k=3432π, | (1.28) |
∞∑k=0(40k2−2k−1)(2kk)2(4k2k)(2k−1)(4k−1)(−1024)k=−4π, | (1.29) |
∞∑k=0(8k2−2k−1)(2kk)2(4k2k)(k+1)(2k−1)(4k−1)(−1024)k=−165π, | (1.30) |
∞∑k=0(176k2−6k−5)(2kk)2(4k2k)(2k−1)(4k−1)482k=−8√3π, | (1.31) |
∞∑k=0(208k2+66k+23)(2kk)2(4k2k)(k+1)(2k−1)(4k−1)482k=128√3π, | (1.32) |
∞∑k=0(6722k2−411k−152)(2kk)2(4k2k)(2k−1)(4k−1)(−632)k=−195√7π, | (1.33) |
∞∑k=0(281591k2−757041k−231992)(2kk)2(4k2k)(k+1)(2k−1)(4k−1)(−632)k=−274625√7π, | (1.34) |
∞∑k=0(560k2−42k−11)(2kk)2(4k2k)(2k−1)(4k−1)124k=−24√2π, | (1.35) |
∞∑k=0(112k2+114k+23)(2kk)2(4k2k)(k+1)(2k−1)(4k−1)124k=256√25π, | (1.36) |
∞∑k=0(248k2−18k−5)(2kk)2(4k2k)(2k−1)(4k−1)(−3×212)k=−28√3π, | (1.37) |
∞∑k=0(680k2+1482k+337)(2kk)2(4k2k)(k+1)(2k−1)(4k−1)(−3×212)k=5488√39π, | (1.38) |
∞∑k=0(1144k2−102k−19)(2kk)2(4k2k)(2k−1)(4k−1)(−21034)k=−60π, | (1.39) |
∞∑k=0(3224k2+4026k+637)(2kk)2(4k2k)(k+1)(2k−1)(4k−1)(−21034)k=2000π, | (1.40) |
∞∑k=0(7408k2−754k−103)(2kk)2(4k2k)(2k−1)(4k−1)284k=−560√33π, | (1.41) |
∞∑k=0(3641424k2+4114526k+493937)(2kk)2(4k2k)(k+1)(2k−1)(4k−1)284k=896000√3π, | (1.42) |
∞∑k=0(4744k2−534k−55)(2kk)2(4k2k)(2k−1)(4k−1)(−214345)k=−1932√525π, | (1.43) |
∞∑k=0(18446264k2+20356230k+1901071)(2kk)2(4k2k)(k+1)(2k−1)(4k−1)(−214345)k=66772496√525π, | (1.44) |
∞∑k=0(413512k2−50826k−3877)(2kk)2(4k2k)(2k−1)(4k−1)(−210214)k=−12180π, | (1.45) |
∞∑k=0(1424799848k2+1533506502k+108685699)(2kk)2(4k2k)(k+1)(2k−1)(4k−1)(−210214)k=341446000π, | (1.46) |
∞∑k=0(71312k2−7746k−887)(2kk)2(4k2k)(2k−1)(4k−1)15842k=−840√11π, | (1.47) |
∞∑k=0(50678512k2+56405238k+5793581)(2kk)2(4k2k)(k+1)(2k−1)(4k−1)15842k=5488000√11π, | (1.48) |
∞∑k=0(7329808k2−969294k−54073)(2kk)2(4k2k)(2k−1)(4k−1)3964k=−120120√2π, | (1.49) |
∞∑k=0(2140459883152k2+2259867244398k+119407598201)(2kk)2(4k2k)(k+1)(2k−1)(4k−1)3964k=44×18203√2π, | (1.50) |
∞∑k=0(164k2−k−3)(2kk)(3kk)(6k3k)(2k−1)(6k−1)203k=−7√52π, | (1.51) |
∞∑k=0(2696k2+206k+93)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)203k=686√5π, | (1.52) |
∞∑k=0(220k2−8k−3)(2kk)(3kk)(6k3k)(2k−1)(6k−1)(−215)k=−7√2π, | (1.53) |
∞∑k=0(836k2−1048k−309)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)(−215)k=−686√2π, | (1.54) |
∞∑k=0(504k2−11k−8)(2kk)(3kk)(6k3k)(2k−1)(6k−1)(−15)3k=−9√15π, | (1.55) |
∞∑k=0(189k2−11k−8)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)(−15)3k=−243√1535π, | (1.56) |
∞∑k=0(516k2−19k−7)(2kk)(3kk)(6k3k)(2k−1)(6k−1)(2×303)k=−11√152π, | (1.57) |
∞∑k=0(3237k2+1922k+491)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)(2×303)k=3993√1510π, | (1.58) |
∞∑k=0(684k2−40k−7)(2kk)(3kk)(6k3k)(2k−1)(6k−1)(−96)3k=−9√6π, | (1.59) |
∞∑k=0(2052k2+2536k+379)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)(−96)3k=486√6π, | (1.60) |
∞∑k=0(2556k2−131k−29)(2kk)(3kk)(6k3k)(2k−1)(6k−1)663k=−63√334π, | (1.61) |
∞∑k=0(203985k2+212248k+38083)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)663k=83349√334π, | (1.62) |
∞∑k=0(5812k2−408k−49)(2kk)(3kk)(6k3k)(2k−1)(6k−1)(−3×1603)k=−253√309π, | (1.63) |
∞∑k=0(3471628k2+3900088k+418289)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)(−3×1603)k=32388554√30135π, | (1.64) |
∞∑k=0(35604k2−2936k−233)(2kk)(3kk)(6k3k)(2k−1)(6k−1)(−960)3k=−189√15π, | (1.65) |
∞∑k=0(13983084k2+15093304k+1109737)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)(−960)3k=4500846√155π, | (1.66) |
∞∑k=0(157752k2−11243k−1304)(2kk)(3kk)(6k3k)(2k−1)(6k−1)2553k=−513√2552π, | (1.67) |
∞∑k=0(28240947k2+31448587k+3267736)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)2553k=45001899√25570π, | (1.68) |
∞∑k=0(2187684k2−200056k−11293)(2kk)(3kk)(6k3k)(2k−1)(6k−1)(−5280)3k=−1953√330π, | (1.69) |
∞∑k=0(101740699836k2+107483900696k+5743181813)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)(−5280)3k=4966100118√3305π, | (1.70) |
∞∑k=0(16444841148k2−1709536232k−53241371)(2kk)(3kk)(6k3k)(2k−1)(6k−1)(−640320)3k=−1672209√10005π, | (1.71) |
and
∞∑k=0P(k)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)(−640320)3k=18×5574033√100055π, | (1.72) |
where
P(k):=637379600041024803108k2+657229991696087780968k+19850391655004126179. |
Recall that the Catalan numbers are given by
Cn:=(2nn)n+1=(2nn)−(2nn+1) (n∈N). |
For
(2kk)2k−1={−1if k=0,2Ck−1if k>0. |
Thus, for any
∞∑k=0(ak2+bk+c)(2kk)3(2k−1)3mk=−c+∞∑k=1(ak2+bk+c)(2Ck−1)3mk=−c+8m∞∑k=0a(k+1)2+b(k+1)+cmkC3k. |
For example, (1.2) has the equivalent form
∞∑k=04k+3(−64)kC3k=8−16π.(1.2′) |
For any odd prime
(p+1)/2∑k=0(4k−1)(2kk)3(2k−1)3(−64)k≡p(−1p)+p3(Ep−3−2) (mod p4) |
(where
(p−1)/2∑k=04k+3(−64)kC3k≡8(1−p(−1p)−p3(Ep−3−2)) (mod p4). |
Recently, C. Wang [50] proved that for any prime
(p+1)/2∑k=0(3k−1)(2kk)3(2k−1)216k≡p+2p3(−1p)(Ep−3−3) (mod p4) |
and
p−1∑k=0(3k−1)(2kk)3(2k−1)216k≡p−2p3 (mod p4). |
(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
∞∑k=0(56k2+118k+61)(2kk)3(k+1)24096k=192π |
and
∞∑k=0(420k2+992k+551)(2kk)3(k+1)2(2k−1)4096k=−1728π. |
Now we state our second theorem.
Theorem 1.2. We have the identities
∞∑k=128k2+31k+8(2k+1)2k3(2kk)3=π2−82, | (1.73) |
∞∑k=142k2+39k+8(2k+1)3k3(2kk)3=9π2−882, | (1.74) |
∞∑k=1(8k2+5k+1)(−8)k(2k+1)2k3(2kk)3=4−6G, | (1.75) |
∞∑k=1(30k2+33k+7)(−8)k(2k+1)3k3(2kk)3=54G−52, | (1.76) |
∞∑k=1(3k+1)16k(2k+1)2k3(2kk)3=π2−82, | (1.77) |
∞∑k=1(4k+1)(−64)k(2k+1)2k2(2kk)3=4−8G, | (1.78) |
∞∑k=1(4k+1)(−64)k(2k+1)3k3(2kk)3=16G−16, | (1.79) |
∞∑k=1(2k2−11k−3)8k(2k+1)(3k+1)k3(2kk)2(3kk)=48−5π22, | (1.80) |
∞∑k=2(178k2−103k−39)8k(k−1)(2k+1)(3k+1)k3(2kk)2(3kk)=1125π2−1109636, | (1.81) |
∞∑k=1(5k+1)(−27)k(2k+1)(3k+1)k2(2kk)2(3kk)=6−9K, | (1.82) |
∞∑k=2(45k2+5k−2)(−27)k−1(k−1)(2k+1)(3k+1)k3(2kk)2(3kk)=37−48K16, | (1.83) |
∞∑k=1(98k2−21k−8)81k(2k+1)(4k+1)k3(2kk)2(4k2k)=216−20π2, | (1.84) |
∞∑k=2(1967k2−183k−104)81k(k−1)(2k+1)(4k+1)k3(2kk)2(4k2k)=20000π2−190269120, | (1.85) |
∞∑k=1(46k2+3k−1)(−144)k(2k+1)(4k+1)k3(2kk)2(4k2k)=72−2252K, | (1.86) |
∞∑k=2(343k2+18k−16)(−144)k(k−1)(2k+1)(4k+1)k3(2kk)2(4k2k)=9375K−704810, | (1.87) |
where
G:=∞∑k=0(−1)k(2k+1)2 and K:=∞∑k=0(k3)k2. |
For
(k−1)k(2kk)=2(2j+1)j(2jj). |
Thus, for any
∞∑j=1(aj2+bj+c)mj(2j+1)3j3(2jj)3=8m∞∑k=2(a(k−1)2+b(k−1)+c)mk(k−1)3k3(2kk)3. |
For example, (1.77) has the following equivalent form
∞∑k=2(2k−1)(3k−2)16k(k−1)3k3(2kk)3=π2−8.(1.77′) |
In contrast with the Domb numbers, we introduce a new kind of numbers
Sn:=n∑k=0(nk)2TkTn−k (n=0,1,2,…). |
The values of
1,2,10,68,586,5252,49204,475400,4723786,47937812,494786260 |
respectively. We may extend the numbers
Sn(b,c):=n∑k=0(nk)2Tk(b,c)Tn−k(b,c) (n=0,1,2,…). |
Note that
Now we state our third theorem.
Theorem 1.3. We have
∞∑k=07k+324kSk(1,−6)=15√2π, | (1.88) |
∞∑k=012k+5(−28)kSk(1,7)=6√7π, | (1.89) |
∞∑k=084k+2980kSk(1,−20)=24√15π, | (1.90) |
∞∑k=03k+1(−100)kSk(1,25)=258π, | (1.91) |
∞∑k=0228k+67224kSk(1,−56)=80√7π, | (1.92) |
∞∑k=0399k+101(−676)kSk(1,169)=25358π, | (1.93) |
∞∑k=02604k+5632600kSk(1,−650)=850√393π, | (1.94) |
∞∑k=039468k+7817(−6076)kSk(1,1519)=4410√31π, | (1.95) |
∞∑k=041667k+78799800kSk(1,−2450)=40425√64π, | (1.96) |
∞∑k=074613k+10711(−5302)kSk(1,2652)=161517548π. | (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
∞∑k=0bk+cmkak=r∑i=1λi√diπ | (1.98) |
for some nonzero rational numbers
p−1∑k=0bk+cmkak≡p(r∑i=1ci(εidip)+∑r<j≤3cj(djp)) (mod p2), | (1.99) |
where
For a Ramanujan-type series of the form (1.98), we call
Conjecture 1.2. Let
p−1∑k=0bk+cmkak≡p(c1(d1p)+c2(d2p)+c3(d3p)) (mod p2) | (1.100) |
for all primes
1(pn)2(pn−1∑k=0bk+cmkak−pδn−1∑k=0bk+cmkak)∈Zp for all n∈Z+. |
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
p−1∑k=0(8k+5)T2k≡3p(−3p) (mod p2) |
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
ak≡(dp)Dkap−1−k (mod p) | (1.101) |
for any prime
∞∑k=0bk+cmkak=λ1√d1+λ2√d2+λ3√d3π |
for some
p−1∑k=0akmk≡(dp)p−1∑k=0ak(D/m)k (mod p2) | (1.102) |
for any prime
Remark 1.3 (ⅰ) For any prime
(ⅱ) For any
Tk(b,c)≡(b2−4cp)(b2−4c)kTp−1−k(b,c) (mod p) | (1.103) |
for all
For a series
In Section 10, we pose two curious conjectural series for
Lemma 2.1. Let
n∑k=0((64−m)k3−32k2−16k+8)(2kk)3(2k−1)2mk=8(2n+1)mn(2nn)3, | (2.1) |
n∑k=0((64−m)k3−96k2+48k−8)(2kk)3(2k−1)3mk=8mn(2nn)3, | (2.2) |
n∑k=0((108−m)k3−54k2−12k+6)(2kk)2(3kk)(2k−1)(3k−1)mk=6(3n+1)mn(2nn)2(3nn), | (2.3) |
n∑k=0((108−m)k3−(54+m)k2−12k+6)(2kk)2(3kk)(k+1)(2k−1)(3k−1)mk=6(3n+1)(n+1)mn(2nn)2(3nn), | (2.4) |
n∑k=0((256−m)k3−128k2−16k+8)(2kk)2(4k2k)(2k−1)(4k−1)mk=8(4n+1)mn(2nn)2(4n2n), | (2.5) |
n∑k=0((256−m)k3−(128+m)k2−16k+8)(2kk)2(4k2k)(k+1)(2k−1)(4k−1)mk=8(4n+1)(n+1)mn(2nn)2(4n2n), | (2.6) |
n∑k=0((1728−m)k3−864k2−48k+24)(2kk)(3kk)(6k3k)(2k−1)(6k−1)mk=24(6n+1)mn(2nn)(3nn)(6n3n), | (2.7) |
n∑k=0((1728−m)k3−(864+m)k2−48k+24)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)mk=24(6n+1)(n+1)mn(2nn)(3nn)(6n3n). | (2.8) |
Remark 2.1. The eight identities in Lemma 2.1 can be easily proved by induction on
(2nn)∼4n√nπ, (2nn)(3nn)∼√327n2nπ, | (2.9) |
(2nn)(4n2n)∼64n√2nπ, (3nn)(6nn)∼432n2nπ. | (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
∞∑k=0(16k3−4k2−2k+1)(2kk)3(2k−1)2(−64)k=limn→+∞2n+1(−64)n(2nn)3=0. |
Note that
16k3−4k2−2k+1=(4k+1)(2k−1)2+2k(4k−1) |
and recall Bauer's series
∞∑k=0(4k+1)(2kk)3(−64)k=2π. |
So, we get
∞∑k=0k(4k−1)(2kk)3(2k−1)2(−64)k=−12∞∑k=0(4k+1)(2kk)3(−64)k=−1π. |
This proves (1.1). By (2.2) with
n∑k=0(4k−1)(4k2−2k+1)(2kk)3(2k−1)3(−64)k=(2nn)3(−64)n |
and hence
∞∑k=0(2k(2k−1)(4k−1)+4k−1)(2kk)3(2k−1)3(−64)k=limn→+∞(2nn)3(−64)n=0. |
Combining this with
In view of (2.7) with
n∑k=0(10939058860032072k3−36k2−2k+1)(2kk)(3kk)(6k3k)(2k−1)(6k−1)(−640320)3k=6n+1(−640320)3n(2nn)(3nn)(6n3n). |
and hence
∞∑k=0(10939058860032072k3−36k2−2k+1)(2kk)(3kk)(6k3k)(2k−1)(6k−1)(−640320)3k=0. |
In 1987, D. V. Chudnovsky and G. V. Chudnovsky [8] got the formula
∞∑k=0545140134k+13591409(−640320)3k(2kk)(3kk)(6k3k)=3×5336022π√10005, |
which enabled them to hold the world record for the calculation of
10939058860032072k3−36k2−2k+1=1672209(2k−1)(6k−1)(545140134k+13591409)+426880(16444841148k2−1709536232k−53241371) |
and hence
∞∑k=0(16444841148k2−1709536232k−53241371)(2kk)(3kk)(6k3k)(2k−1)(6k−1)(−640320)3k=−1672209426880×3×5336022π√10005=−1672209√10005π. |
This proves
By (2.8) with
n∑k=0(10939058860032072k3+10939058860031964k2−2k+1)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)(−640320)3k=6n+1(n+1)(−640320)3n(2nn)(3nn)(6n3n) |
and hence
∞∑k=0(10939058860032072k3+10939058860031964k2−2k+1)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)(−640320)3k=0. |
Note that
2802461(10939058860032072k3+10939058860031964k2−2k+1)=1864188626454(k+1)(16444841148k2−1709536232k−53241371)+5P(k). |
Therefore, with the help of
∞∑k=0P(k)(2kk)(3kk)(6k3k)(k+1)(2k−1)(6k−1)(−640320)3k=−18641886264545×(−1672209)√10005π=18×5574033√100055π. |
This proves
The identities (1.3)–(1.70) can be proved similarly.
Lemma 2.2. Let
n∑k=1mk((m−64)k3−32k2+16k+8)(2k+1)2k3(2kk)3=mn+1(2n+1)2(2nn)3−m, | (2.11) |
n∑k=1mk((m−64)k3−96k2−48k−8)(2k+1)3k3(2kk)3=mn+1(2n+1)3(2nn)3−m, | (2.12) |
n∑k=1mk((m−108)k3−54k2+12k+6)(2k+1)(3k+1)k3(2kk)2(3kk)=mn+1(2n+1)(3n+1)(2nn)2(3nn)−m, | (2.13) |
∑1<k≤nmk((m−108)k3−(54+m)k2+12k+6)(k−1)(2k+1)(3k+1)k3(2kk)2(3kk)=mn+1n(2n+1)(3n+1)(2nn)2(3nn)−m2144, | (2.14) |
n∑k=1mk((m−256)k3−128k2+16k+8)(2k+1)(4k+1)k3(2kk)2(4k2k)=mn+1(2n+1)(4n+1)(2nn)2(4n2n)−m, | (2.15) |
∑1<k≤nmk((m−256)k3−(128+m)k2+16k+8)(k−1)(2k+1)(4k+1)k3(2kk)2(4k2k)=mn+1n(2n+1)(4n+1)(2nn)2(4n2n)−m2360. | (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:
∞∑k=121k−8k3(2kk)3=π26, ∞∑k=1(4k−1)(−64)kk3(2kk)3=−16G,∞∑k=1(3k−1)(−8)kk3(2kk)3=−2G, ∞∑k=1(3k−1)16kk3(2kk)3=π22,∞∑k=1(15k−4)(−27)k−1k3(2kk)2(3kk)=K, ∞∑k=1(5k−1)(−144)kk3(2kk)2(4k2k)=−452K,∞∑k=1(11k−3)64kk2(2kk)2(3kk)=8π2, ∞∑k=1(10k−3)8kk3(2kk)2(3kk)=π22, ∞∑k=1(35k−8)81kk3(2kk)2(4k2k)=12π2. |
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
n∑k=116k(−48k3−32k2+16k+8)(2k+1)2k3(2kk)3=16n+1(2n+1)2(2nn)3−16 |
for all
∞∑k=116k(6k3+4k2−2k−1)(2k+1)2k3(2kk)3=limn→+∞(−2×16n(2n+1)2(2nn)3+2)=2. |
Notice that
2(6k3+4k2−2k−1)=(2k+1)2(3k−1)−(3k+1). |
So we have
−∞∑k=1(3k+1)16k(2k+1)2k3(2kk)3=2×2−∞∑k=1(3k−1)16kk3(2kk)3=4−π22 |
and hence (1.77) holds.
By (2.11) with
n∑k=1(−64)k(−128k3−32k2+16k+8)(2k+1)2k3(2kk)3=(−64)n+1(2n+1)2(2nn)3+64 |
for all
∞∑k=1(−64)k(16k3+4k2−2k−1)(2k+1)2k3(2kk)3=−8+limn→+∞8(−64)n(2n+1)2(2nn)3=−8. |
Since
∞∑k=1(4k−1)(−64)kk3(2kk)3=−16G, |
we see that
−16G−2∞∑k=1(4k+1)(−64)k(2k+1)2k2(2kk)3=−8 |
and hence (1.78) holds. In light of (2.12) with
n∑k=1(−64)k(−128k3−96k2−48k−8)(2k+1)3k3(2kk)3=(−64)n+1(2n+1)3(2nn)3+64 |
for all
∞∑k=1(−64)k(16k3+12k2+6k+1)(2k+1)3k3(2kk)3=−8+limn→+∞8(−64)n(2n+1)3(2nn)3=−8. |
Since
∞∑k=1(4k+1)(−64)k(2k+1)3k3(2kk)3=−8−2(4−8G)=16G−16. |
This proves (1.79).
By (2.13) with
∞∑k=1(45k3+18k2−4k−2)(−27)k(2k+1)(3k+1)k3(2kk)2(3kk)=−9. |
As
2(45k3+18k2−4k−2)=(15k−4)(2k+1)(3k+1)−3k(5k+1) |
and
∞∑k=1(15k−4)(−27)kk3(2kk)2(3kk)=−27K, |
we see that (1.82) follows. By (2.14) with
−3∞∑k=2(−27)k(45k3+9k2−4k−2)(k−1)(2k+1)(3k+1)k3(2kk)2(3kk)=−(−27)2144 |
and hence
∞∑k=2(−27)k(45k3+9k2−4k−2)(k−1)(2k+1)(3k+1)k3(2kk)2(3kk)=2716. |
As
45k3+9k2−4k−2=9(k−1)k(5k+1)+(45k2+5k−2), |
with the aid of (1.82) we get
∞∑k=2(−27)k(45k2+5k−2)(k−1)(2k+1)(3k+1)k3(2kk)2(3kk)=2716−9(6−9K−6(−27)122)=2716(48K−37) |
and hence
Other identities in Theorem 1.2 can be proved similarly.
For integers
sn,k:=1(nk)k∑i=0(n2i)(n2(k−i))(2ii)(2(k−i)k−i). | (2.17) |
For
tn:=∑0<k≤n(n−1k−1)(−1)k4n−ksn+k,k. | (2.18) |
Lemma 2.3. For any
n∑k=0(nk)(−1)k4n−ksn+k,k=fn | (2.19) |
and
(2n+1)tn+1+8ntn=(2n+1)fn+1−4(n+1)fn, | (2.20) |
where
Proof. For
F(n,i,k)=(nk)(−1)k4n−k(n+kk)(n+k2i)(2ii)(n+k2(k−i))(2(k−i)k−i). |
By the telescoping method for double summation [7], for
F(n,i,k):=F(n,i,k)+7n2+21n+168(n+1)2F(n+1,i,k)−(n+2)28(n+1)2F(n+2,i,k) |
with
F(n,i,k)=(G1(n,i+1,k)−G1(n,i,k))+(G2(n,i,k+1)−G2(n,i,k)), |
where
G1(n,i,k):=i2(−k+i−1)(−1)k+14n−kn!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
G2(n,i,k):=2(k−i)(−1)k4n−kn!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
−10n4+(i−10k−68)n3+(−24i2+(32k+31)i+2k2−67k−172)n2+(36i3+(−68k−124)i2+(39k2+149k+104)i+2k3−8k2−145k−192)n+60i3+(−114k−140)i2+(66k2+160k+92)i+3k3−19k2−102k−80 |
and
10(i−k)n4+(−20i2+(46k+47)i−6k2−47k)n3+(72i3+(−60k−38)i2+(22k2+145k+90)i+4k3−11k2−90k)n2+(72k+156)i3n+(−72k2−60k−10)i2n+(18k3+4k2+165k+85)in+(22k3−5k2−85k)n+(120k+60)i3+(−120k2+68k−4)i2+(30k3−56k2+86k+32)i+26k3−6k2−32k |
respectively. Therefore
n+2∑k=0k∑i=0F(n,i,k)=n+2∑k=0(G1(n,k+1,k)−G1(n,0,k))+n+2∑i=0(G2(n,i,n+3)−G2(n,i,i))=n+2∑k=0(0−0)+n+2∑i=0(0−0)=0, |
and hence
u(n):=n∑k=0(nk)(−1)k4n−ksn+k,k |
satisfies the recurrence relation
8(n+1)2u(n)+(7n2+21n+16)u(n+1)−(n+2)2u(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
8(n+1)(n+2)(18n3+117n2+249n+172)v(n)+(126n5+1197n4+4452n3+8131n2+7350n+2656)v(n+1)=(n+3)2(18n3+63n2+69n+22)v(n+2). |
In view of the above, we have completed the proof of Lemma 2.3.
Lemma 2.4. For any
Sn(4,c)=⌊n/2⌋∑k=0(n−kk)(2(n−k)n−k)ck4n−2ksn,k. | (2.21) |
Proof. For each
Tk(4,c)Tn−k(4,c)=⌊k/2⌋∑i=0(k2i)(2ii)4k−2ici⌊(n−k)/2⌋∑j=0(n−k2j)(2jj)4n−k−2jcj=⌊n/2⌋∑r=0cr4n−2r∑i,j∈Ni+j=r(k2i)(n−k2j)(2ii)(2jj). |
If
n∑k=0(nk)2(k2i)(n−k2j)=(n2i)(n2j)n−2j∑k=2i(n−2ik−2i)(n−2jn−k−2j)=(n2i)(n2j)(2n−2(i+j)n−2(i+j))=(2n−2rn)(n2i)(n2j) |
with the aid of the Chu-Vandermonde identity. Therefore
Sn(4,c)=⌊n/2⌋∑k=0ck4n−2k(2n−2kn)(nk)sn,k=⌊n/2⌋∑k=0ck4n−2k(2n−2kn−k)(n−kk)sn,k. |
This proves (2.21).
Lemma 2.5. For
sk+l,k≤(2k+1)4kl(k+ll). | (2.22) |
Proof. Let
(nk)sn,k≤∑i,j∈Ni+j=k(n2i)(n2j)∑i,j∈Ni+j=k(2ii)(2jj)≤∑s,t∈Ns+t=2k(ns)(nt)∑i,j∈Ni+j=k4i4j=(2n2k)(k+1)4k |
and
(2n2k)(nk)=(2n2l)(nl)=l−1∏j=02(j+k)+12j+1≤(2k+1)∏0<j<l2(j+k)2j=(2k+1)(k+l−1l−1). |
Hence
sk+l,k≤(k+1)4k(2k+1)lk+l(k+ll)≤(2k+1)4kl(k+ll). |
This proves (2.22).
To prove Theorem 1.3, we need an auxiliary theorem.
Theorem 2.6. Let
∞∑n=0(an+b)Sn(4,−m)mn=1m+16∞∑n=0(2a(m+4)n−8a+b(m+16))(2nn)fnmn. | (2.23) |
Proof. Let
N∑n=0Sn(4,−m)mn=N∑n=01mn⌊n/2⌋∑k=0(−m)k4n−2k(2n−2kn−k)(n−kk)sn,k |
=N∑l=0(2ll)mlN−l∑k=0(lk)(−1)k4l−ksl+k,k=⌊N/2⌋∑l=0(2ll)mll∑k=0(lk)(−1)k4l−ksl+k,k+∑N/2<l≤N(2ll)mlN−l∑k=0(lk)(−1)k4l−ksl+k,k |
and similarly
N∑n=0nSn(4,−m)mn=N∑l=0(2ll)mlN−l∑k=0(lk)(−1)k4l−k(k+l)sl+k,k=N∑l=0l(2ll)mlN−l∑k=0((lk)+(l−1k−1))(−1)k4l−ksl+k,k=⌊N/2⌋∑l=0l(2ll)mll∑k=0((lk)+(l−1k−1))(−1)k4l−ksl+k,k+∑N/2<l≤Nl(2ll)mlN−l∑k=0((lk)+(l−1k−1))(−1)k4l−ksl+k,k, |
where we consider
If
|N−l∑k=0(lk)(−1)k4l−ksl+k,k|≤l∑k=0(lk)4l−ksl+k,k≤l∑k=0(lk)4l−k(2k+1)4kl(k+ll)≤l(2l+1)4ll∑k=0(lk)(l+kk)=l(2l+1)4lPl(3), |
where
|∑N/2<l≤N(2ll)mlN−l∑k=0(lk)(−1)k4l−ksl+k,k|≤∑N/2<l≤Nl(2l+1)(16m)lPl(3)≤∑l>N/2l(2l+1)Pl(3)(16m)l |
and
|∑N/2<l≤Nl(2ll)mlN−l∑k=0((lk)+(l−1k−1))(−1)k4l−ksl+k,k|≤∑N/2<l≤Nl4lmll∑k=02(lk)4l−ksl+k,k |
≤∑N/2<l≤N2l2(2l+1)(16m)lPl(3)≤2∑l>N/2l2(2l+1)Pl(3)(16m)l. |
Recall that
Pl(3)=Tl(3,2)∼(3+2√2)l+1/224√2√lπ as l→+∞. |
As
∞∑l=0l2(2l+1)Pl(3)(16m)l |
converges. Thus
limN→+∞∑l>N/2l(2l+1)Pl(3)(16m)l=0=limN→+∞∑l>N/2l2(2l+1)Pl(3)(16m)l |
and hence by the above we have
∞∑n=0Sn(4,−m)mn=∞∑l=0(2ll)mll∑k=0(lk)(−1)k4l−ksl+k,k |
and
∞∑n=0nSn(4,−m)mn=∞∑l=0l(2ll)mll∑k=0((lk)+(l−1k−1))(−1)k4l−ksl+k,k. |
Therefore, with the aid of (2.19), we obtain
∞∑n=0Sn(4,−m)mn=∞∑n=0(2nn)mnfn | (2.24) |
and
∞∑n=0nSn(4,−m)mn=∞∑n=0n(2nn)mn(fn+tn). | (2.25) |
In view of (2.25) and (2.20),
(m+16)∞∑n=0nSn(4,−m)mn=∞∑n=1n(2nn)mn−1(fn+tn)+16∞∑n=0n(2nn)mn(fn+tn)=∞∑n=0(n+1)(2n+2n+1)(fn+1+tn+1)+16n(2nn)(fn+tn)mn=2∞∑n=0(2nn)mn((2n+1)(fn+1+tn+1)+8n(fn+tn)) |
=2∞∑n=0(2nn)mn(2(2n+1)fn+1+4(n−1)fn)=2∞∑n=0(n+1)(2n+2n+1)fn+1mn+8∞∑n=0(n−1)(2nn)fnmn=2∞∑n=0n(2nn)fnmn−1+8∞∑n=0(n−1)(2nn)fnmn=2∞∑n=0((m+4)n−4)(2nn)fnmn. |
Combining this with (2.24), we immediately obtain the desired (2.23).
Proof of Theorem 1.3. Let
4nTn(1,m)=⌊n/2⌋∑k=0(n2k)(2kk)4n−2k(16m)k=Tn(4,16m) |
for any
∞∑n=0(an+b)Sn(1,m)(−4m)n=∞∑n=0(an+b)Sn(4,16m)(−16m)n=116−16m∞∑n=0(2a(4−16m)n−8a+(16−16m)b)(2nn)fn(−16m)n=12(m−1)∞∑n=0(a(4m−1)n+a+2b(m−1))(2nn)fn(−16m)n. |
Therefore
∞∑k=07k+324kSk(1,−6)=52∞∑k=05k+196k(2kk)fk,∞∑k=012k+5(−28)kSk(1,7)=3∞∑k=09k+2(−112)k(2kk)fk,∞∑k=084k+2980kSk(1,−20)=27∞∑k=06k+1320k(2kk)fk,∞∑k=03k+1(−100)kSk(1,25)=116∞∑k=099k+17(−400)k(2kk)fk,∞∑k=0228k+67224kSk(1,−56)=5∞∑k=090k+13896k(2kk)fk,∞∑k=0399k+101(−676)kSk(1,169)=1516∞∑k=0855k+109(−2704)k(2kk)fk,∞∑k=02604k+5632600kSk(1,−650)=51∞∑k=0102k+1110400k(2kk)fk,∞∑k=039468k+7817(−6076)kSk(1,1519)=135∞∑k=0585k+58(−24304)k(2kk)fk,∞∑k=041667k+78799800kSk(1,−2450)=2972∞∑k=0561k+5339200k(2kk)fk,∞∑k=074613k+10711(−5302)kSk(1,2652)=2332∞∑k=0207621k+14903(−10602)k(2kk)fk. |
∞∑k=05k+196k(2kk)fk=3√2π, ∞∑k=09k+2(−112)k(2kk)fk=2√7π,∞∑k=06k+1320k(2kk)fk=8√159π, ∞∑k=099k+17(−400)k(2kk)fk=50π,∞∑k=090k+13896k(2kk)fk=16√7π, ∞∑k=0855k+109(−2704)k(2kk)fk=338π,∞∑k=0102k+1110400k(2kk)fk=50√399π, ∞∑k=0585k+58(−24304)k(2kk)fk=98√313π,∞∑k=0561k+5339200k(2kk)fk=1225√618π, ∞∑k=0207621k+14903(−10602)k(2kk)fk=1404503π. |
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:
∞∑k=050k+1(−256)k(2kk)(2kk+1)Tk(1,16)=83π,(Ⅰ1′) |
∞∑k=0(100k2−4k−7)(2kk)2Tk(1,16)(2k−1)2(−256)k=−24π,(Ⅰ1″ |
\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.
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