Supercongruences involving Apéry-like numbers and binomial coefficients

1.   Introduction

Let $\Bbb Z$ be the set of integers. In 2009, Zagier ^[36] studied the Apéry-like numbers $\{u_n\}$ satisfying

where $a, b, c\in\Bbb Z$, $c\not = 0$ and $u_n\in\Bbb Z$ for $n\ge 1$. Let

where $[x]$ is the greatest integer not exceeding $x$. According to ^[2,36], $\{A'_n\}$, $\{f_n\}$, $\{S_n\}$, $\{a_n\}$, $\{Q_n\}$ and $\{W_n\}$ are Apéry-like sequences with $(a, b, c) = (11, 3, -1), (7, 2, -8),$ $(12, 4, 32), (10, 3, 9), (-17,$ $-6, 72), (-9, -3, 27)$, respectively. The sequence $\{f_n\}$ is called Franel numbers. In ^{[22,23,24,25]}, the author systematically investigated congruences for sums involving $S_n$, $f_n$ and $W_n$. For $\{A'_n\}$, $\{f_n\}$, $\{S_n\}$, $\{a_n\}$, $\{Q_n\}$ and $\{W_n\}$ see A005258, A000172, A081085, A002893, A093388 and A291898 in Sloane's database "The On-Line Encyclopedia of Integer Sequences".

Let $p$ be an odd prime. In ^[29], Z. W. Sun posed many congruences modulo $p^2$ involving Apéry-like numbers. In ^[24,25], the author conjectured many congruences modulo $p^3$ involving Apéry-like numbers. In Section 2, we show that for any odd prime $p$,

and obtain a congruence for $\sum_{n = 0}^{p-1} \binom{2n}n \frac {A_n'}{4^n}\ (\text{ mod}\ p)$. In Section 3, we establish some transformation formulas for congruences involving Apéry-like numbers and obtain some congruences involving $a_n$ and $Q_n$. For example, for any prime $p > 3$ we have

where $(\frac {a}{p})$ is the Legendre symbol. We also pose some conjectures on congruences involving $Q_n$ and $a_n$.

For positive integers $a, b$ and $n$, if $n = ax^2+by^2$ for some integers $x$ and $y$, we briefly write that $n = ax^2+by^2$. Let $p > 3$ be a prime. In 1987, Beukers ^[4] conjectured a congruence equivalent to

This congruence was proved by several authors including Ishikawa ^[8] ($p \equiv 1\ (\text{ mod}\ 4)$), Van Hamme ^[7] ($p \equiv 3\ (\text{ mod}\ 4)$) and Ahlgren ^[1]. Actually, (1.4) follows immediately from the following identity due to Bell (see [5, (6.35)], ^[17]):

In 2003, Rodriguez-Villegas ^[14] posed 22 conjectures on supercongruences modulo $p^2$. In particular, the following congruences are equivalent to conjectures due to Rodriguez-Villegas:

These conjectures have been solved by Mortenson ^[13] and Zhi-Wei Sun ^[28]. Since $\frac { \binom{2k}k}{k+1} = \binom{2k}k- \binom{2k}{k+1},$ from (1.4), (1.6) and ^[28], one may deduce that

Let $p$ be an odd prime, and let $m$ be an integer such that $p\nmid m$. In ^[27,29], Z. W. Sun posed many conjectures for congruences modulo $p^2$ involving the sums

For 13 similar conjectures see ^[16]. Most of these congruences modulo $p$ were proved by the author in ^[17,18,19]. In ^[25,26], the author conjectured many congruences modulo $p^3$ involving the above sums. For instance, for any prime $p\not = 2, 7$,

Let $p$ be an odd prime. In Section 4, we deduce congruences for

In Section 5, based on calculations by Maple, we pose lots of challenging conjectures on congruences modulo $p^3$ for the sums

and congruences modulo $p^2$ for the sums

where $m$ is an integer not divisible by $p$. As two typical conjectures, if $p$ is an odd prime with $p \equiv 1, 2, 4\ (\text{ mod}\ 7)$ and so $p = x^2+7y^2$, then

if $p \equiv 1, 3, 4, 5, 9\ (\text{ mod}\ {11})$ and so $4p = x^2+11y^2$, then

In Section 6, we pose many conjectures on $\sum_{k = 0}^{p-1} \frac { \binom{2k}ku_k}{m^k(2k-1)}$ modulo $p^2$, where $u_n\in\{A'_n, f_n, S_n, a_n,$ $Q_n, W_n\}$.

In addition to the above notation, throughout this paper we use the following notations. For a prime $p$, let $\Bbb Z_p$ be the set of rational numbers whose denominator is not divisible by $p$. For $a\in \Bbb Z_p$, let $q_p(a) = (a^{p-1}-1)/p$ and ${\langle a\rangle_p}$ be determined by ${\langle a\rangle_p}\in\{0, 1, \ldots, p-1\}$ and $a \equiv {\langle a\rangle_p}\ (\text{ mod}\ p)$. Let $H_0 = 0$, $H_n = 1+ \frac 12+\cdots+ \frac 1n$ $(n\ge 1)$ and let $\{E_n\}$ be the Euler numbers given by

2.   Two congruences involving $S_n$ and $A'_n$

Let $\{S_n\}$ be the Apéry-like sequence given by (1.1). In this section, we prove the congruence (1.2). Z. W. Sun stated the identity

which can be easily proved by using WZ method, see ^[34]. Using this identity we see that for any positive integer $p$ and a give sequence $\{c_n\}$,

Thus, for $p\in\{1, 3, 5, \ldots\}$,

Theorem 2.1. Let $p$ be an odd prime, then

Proof. Clearly,

Thus, taking $c_n = n$ in (2.2) and then applying the above gives

Observe that $p\mid \binom{2k}k$ for $\frac p2 < k < p$. By [27, Lemma 2.1],

Now, from the above we deduce that

By [5, (1.79)], $\sum_{k = 0}^n \binom{n+k}k/2^k = 2^n.$ Since $\binom{-x}r = (-1)^r \binom{x-1+r}r$, we see that for $s\ge 1$,

and

It then follows that

Substituting into (2.3) yields

By [12, (25)],

Thus,

and so

which completes the proof.

Theorem 2.2. Let $p$ be a prime with $p\not = 2, 11$, then

Proof. For nonnegative integers $k, m$ and $n$ with $k\le n\le m$, it is known that $\binom mn \binom nk = \binom mk \binom{m-k}{n-k}$. Thus, applying Vandermonde's identity we see that

Note that $p\mid \binom{2k}k$ for $\frac p2 < k < p$ and $\binom{ \frac {p-1}2}k \equiv \binom{- \frac 12}k = \frac { \binom{2k}k}{(-4)^k}\ (\text{ mod}\ p)$ for $k < \frac p2$. Taking $m = \frac {p-1}2$ in the above, we deduce that

Now applying [18, Theorem 4.4], we deduce the result.

Remark 2.1. In ^[29], Z. W. Sun conjectured that for any odd prime $p$,

3.   Congruences involving $a_n$ and $Q_n$

In this section, we establish some transformation formulas for congruences involving Apéry-like numbers, obtain some congruences involving $a_n$ and $Q_n$, and pose ten conjectures on related congruences.

Lemma 3.1. Let $n$ be a nonnegative integer, then

Proof. By [15, (38)], $a_n = \sum_{k = 0}^n \binom nkf_k$. Applying the binomial inversion formula gives $f_n = \sum_{k = 0}^n \binom nk(-1)^{n-k}a_k$. Since $\frac {Q_n}{(-8)^n} = \sum_{k = 0}^n \binom nk(-1)^k \frac {f_k}{8^k}$, applying the binomial inversion formula gives $\frac {f_n}{8^n} = \sum_{k = 0}^n \binom nk \frac {Q_k}{8^k}$. Also,

Applying the binomial inversion formula yields $a_n = \sum_{k = 0}^n \binom nk9^{n-k}Q_k$, which completes the proof.

Theorem 3.1. Let $p > 3$ be a prime, then

Proof. It is well known that $\sum_{n = k}^{p-1} \binom nk = \binom p{k+1}$ and $\binom{p-1}k \equiv (-1)^k\ (\text{ mod}\ p)$ for $0\le k\le p-1$. Since $Q_n = \sum_{k = 0}^n \binom nk(-8)^{n-k}f_k$, we see that

By ^[11], $f_k \equiv (-8)^kf_{p-1-k}\ (\text{ mod}\ p)$. Thus,

By [30, Theorem 1.1 and Lemma 2.5],

Thus,

Using Lemma 3.1,

By ^[11] or [25, Theorem 3.1], $a_r \equiv (\frac {p}{3})9^ra_{p-1-r}\ (\text{ mod}\ p)$. By [32, Lemma 3.2], $a_{p-1} \equiv (\frac {p}{3})(1+2pq_p(3)) \equiv (\frac {p}{3})9^{p-1}\ (\text{ mod}\ {p^2})$. Taking $x = 1$ in [32, (3.6)] gives $\sum_{r = 1}^{p-1} \frac {a_r}r \equiv 0\ (\text{ mod}\ p)$. Now, from the above we deduce that

This completes the proof.

Lemma 3.2. Let $p$ be an odd prime and $m\in\Bbb Z_p$ with $m\not\equiv 0, 1\ (\text{ mod}\ p)$. Suppose that $u_0, u_1, \ldots, u_{p-1}\in\Bbb Z_p$ and $v_n = \sum_{k = 0}^n \binom nku_k$ $(n\ge 0)$. Then

Proof. It is clear that

This proves the lemma.

Theorem 3.2. Suppose that $p$ is an odd prime, $m\in\Bbb Z_p$ and $m\not\equiv 0, 1\ (\text{ mod}\ p)$, then

and

Proof. By Lemma 3.1, $a_n = \sum_{k = 0}^n \binom nkf_k$. From Lemma 3.2, $\sum_{k = 0}^{p-1} \frac {a_k}{m^k} \equiv\sum_{k = 0}^{p-1} \frac {f_k}{(m-1)^k}$ $\ (\text{ mod}\ p)$. Now applying [23, Theorem 2.12 and Lemma 2.4 (with $z = \frac 1{m-1}$)] yields the first result. Since $\frac {Q_n}{(-8)^n} = \sum_{k = 0}^n \binom nk \frac {f_k}{(-8)^k}$, applying Lemma 3.2 gives $\sum_{k = 0}^{p-1} \frac {Q_k}{(-8m)^k} \equiv {\sum_{k = 0}^{p-1}} \frac {f_k}{(-8(m-1))^k}\ (\text{ mod}\ p)$. From [23, Theorem 2.12 and Lemma 2.4 (with $z = \frac 1{-8(m-1)}$)] we deduce the remaining part.

Theorem 3.3. Suppose that $p$ is a prime with $p > 3$, then

Proof. Putting $m = \frac 34, \frac 32$ in Theorem 3.2 yields

Now applying [20, Theorem 3.4] yields the result.

Lemma 3.3. [31, Theorem 2.2] Let $p$ be an odd prime, $u_0, u_1, \ldots, u_{p-1}\in\Bbb Z_p$ and $v_n = \sum_{k = 0}^n \binom nk(-1)^ku_k$ for $n\ge 0$. For $m\in\Bbb Z_p$ with $m\not\equiv 0, 4\ (\text{ mod}\ p)$,

Theorem 3.4. Let $p$ be an odd prime, $m\in\Bbb Z_p$ and $m\not\equiv \pm 2\ (\text{ mod}\ p)$, then

and for $9m-14\not\equiv 0\ (\text{ mod}\ p)$,

Proof. We first note that $p\mid \binom{2k}k$ for $\frac {p+1}2\le k\le p-1$. By Lemma 3.1, taking $u_k = (-1)^kf_k$ and $v_k = a_k$ in Lemma 3.3 gives (3.1). Since $\frac {Q_n}{(-8)^n} = \sum_{k = 0}^n \binom nk \frac {f_k}{(-8)^k}$, taking $u_k = \frac {f_k}{8^k}$ and $v_k = \frac {Q_k}{(-8)^k}$ in Lemma 3.3 gives (3.2). By Lemma 3.1, taking $u_k = \frac {a_k}{9^k}$ and $v_k = \frac {Q_k}{(-9)^k}$ in Lemma 3.3 yields (3.3). Combining (3.3) with (3.1) yields (3.4).

Theorem 3.5. Suppose that $p$ is a prime with $p > 5$, then

Proof. Taking $m = 52$ in (3.1) and then applying [23, Theorem 2.2], we obtain

Taking $m = -4$ in (3.3) gives $\sum_{n = 0}^{p-1} \binom{2n}n \frac {Q_n}{18^n} \equiv \big(\frac {3}{p}\big)\sum_{n = 0}^{p-1} \binom{2n}n \frac {a_n}{54^n}\ (\text{ mod}\ p).$ Taking $m = 2$ in (3.4) and applying the congruence for $\sum_{k = 0}^{p-1} \binom{2k}k \frac {f_k}{(-4)^k}\ (\text{ mod}\ p)$ (see [23, p.124]) yields

Now combining the above proves the theorem.

Theorem 3.6. Suppose that $p$ is a prime such that $p \equiv 1, 19\ (\text{ mod}\ {30})$ and so $p = x^2+15y^2$, then

Proof. Taking $m = 7$ in (3.1) and then applying [23, Theorem 2.5] gives

Putting $m = -47$ in (3.1) and then applying [23, Theorem 2.4] gives

Taking $m = 1, 7$ in (3.3) gives

Now combining the above proves the theorem.

Theorem 3.7. Suppose that $p$ is a prime with $p > 3$, then

Proof. Taking $m = \frac {14}9, - \frac {82}9$ in (3.3) yields

Now applying [23, Theorem 2.14] and [21, Theorem 5.6] yields the first congruence. Taking $m = 0$ in (3.2), $m = 18$ in (3.1) and then applying [23, Theorem 2.10] yields the second congruence. Taking $m = \frac {10}3$ in (3.3) and then applying [21, Theorem 4.3] gives the third congruence.

Based on calculations by Maple, we pose the following conjectures.

Conjecture 3.1. Let $p > 3$ be a prime, then

Conjecture 3.2. Let $p$ be a prime with $p > 3$.

(ⅰ) If $p \equiv 1\ (\text{ mod}\ 3)$ and so $p = x^2+3y^2$ with $3\mid x-1$, then

(ⅱ) If $p \equiv 2\ (\text{ mod}\ 3)$, then

Conjecture 3.3. Let $p$ be a prime with $p > 3$.

(ⅰ) If $p \equiv 1\ (\text{ mod}\ 3)$ and so $p = x^2+3y^2$, then

(ⅱ) If $p \equiv 2\ (\text{ mod}\ 3)$, then

Conjecture 3.4. Let $p > 5$ be a prime, then

(ⅰ) If $p \equiv 1, 17, 19, 23\ (\text{ mod}\ {30})$, then

(ⅱ) If $p \equiv 7, 11, 13, 29\ (\text{ mod}\ {30})$, then

Conjecture 3.5. Let $p > 3$ be a prime, then

Conjecture 3.6. Let $p$ be an odd prime, then

Conjecture 3.7 Let $p$ be a prime with $p\not = 2, 5$, then

Conjecture 3.8. Let $p > 3$ be a prime, then

Conjecture 3.9. Let $p$ be a prime with $p > 3$. If $m\in\{-7, -25, -169, -1519, -70225, 20, 56,650,$ $2450\}$ and $p\nmid m(m-2)$, then

Conjecture 3.10. Let $p$ be a prime with $p > 3$. If $m\in\{-112, -400, -2704, -24304, -1123600\}$ and $p\nmid m(m+4)$, then

4.   Congruences involving $\binom{2k}k^3$

For an odd prime $p$ and $x\in\Bbb Z_p$, the $p$-adic Gamma function $\Gamma_p(x)$ is defined by

Theorem 4.1. [25, Conjecture 4.10] Let $p$ be an odd prime, then

and

Proof. From [10, Theorems 3 and 29],

and

By [35, (9)],

By [24, Theorem 2.8], for $p = x^2+4y^2 \equiv 1\ (\text{ mod}\ 4)$,

Combining (4.1) with (4.2) gives

Now combining all the above proves the theorem.

Theorem 4.2. Suppose that $p$ is an odd prime, then

where $R_1(p) = (2p+2-2^{p-1}) \binom{(p-1)/2}{(p-3)/4}^2$.

Proof. By ^[35], for any positive integer $n$,

From ^[16],

Now, taking $n = \frac {p-1}2$ in (4.4) and then applying (4.5) gives

For $p = x^2+4y^2 \equiv 1\ (\text{ mod}\ 4)$, from [17, Lemma 3.4],

Thus,

For $p \equiv 3\ (\text{ mod}\ 4)$, we see that

and so

Note that $p\mid \binom{2k}k$ for $\frac p2 < k < p$ and

Then we get

Now, combining all the above proves the congruence for $\sum_{k = 0}^{p-1} \frac { \binom{2k}k^3}{64^k(k+1)^2}\ (\text{ mod}\ {p^2})$. By ^[35],

This together with (4.1) and (4.3) yields

For $p \equiv 3\ (\text{ mod}\ 4)$ we see that $\binom{(p-1)/2}{(p-3)/4} = \frac {2(p-1)}{p+1} \binom{(p-3)/2}{(p-3)/4}$ and so

Thus,

It is well known that $\binom{2p-1}{p-1} \equiv 1\ (\text{ mod}\ {p^2})$. Hence,

Since $p\mid \binom{2k}k$ for $\frac p2 < k < p$,

Now combining all the above proves the theorem.

Theorem 4.3. Let $p$ be an odd prime, then

and so

Proof. Using (4.5),

Thus,

and so

By (1.5),

From the above, (1.4) and (4.6),

Hence,

and so

To see the result, we recall that $p\mid \binom{2k}k$ for $\frac p2 < k < p$.

5.   Conjectures on congruences involving binomial coefficients

For $k = 1, 2, 3, \ldots$, it is clear that

where $C_k = \frac 1{k+1} \binom{2k}k$ is the $k$-th Catalan number. For an odd prime $p$, let

Calculations with Maple suggest the following challenging conjectures.

Conjecture 5.1. Let $p > 3$ be a prime, then

Conjecture 5.2. Let $p > 5$ be a prime, then

Remark 5.1. Let $p$ be a prime with $p > 5$. In ^[25], the author conjectured that if $p \equiv 1\ (\text{ mod}\ 3)$ and so $4p = x^2+27y^2$, then

if $p \equiv 2\ (\text{ mod}\ 3)$, then

The congruence for $\sum_{k = 0}^{p-1} \frac { \binom{2k}k^2 \binom{3k}k}{(-192)^k}\ (\text{ mod}\ {p^2})$ was conjectured by Z. W. Sun ^[27] earlier.

Let $p > 3$ be a prime. In ^[27,29], Z. W. Sun conjectured congruences for $\sum_{k = 0}^{p-1} \binom{2k}k^3/m^k$ $\ (\text{ mod}\ {p^2})$ with $m = 1, -8, 16, -64,256, -512, 4096$. Such conjectures were proved by the author in ^[17]. In ^[25], the author conjectured congruences for $\sum_{k = 0}^{p-1} \binom{2k}k^3/m^k$ $\ (\text{ mod}\ {p^3})$ in the cases $m = 1, -8, 16, -64,256, -512, 4096$.

Conjecture 5.3. Let $p$ be an odd prime, then

Conjecture 5.4. Let $p$ be an odd prime, then

Conjecture 5.5. Let $p$ be an odd prime, then

Conjecture 5.6. Let $p > 3$ be a prime, then

Conjecture 5.7. Let $p > 3$ be a prime, then

Conjecture 5.8. Let $p$ be a prime with $p\not = 2, 3, 11$, then

Conjecture 5.9. Let $p > 3$ be a prime, then

Conjecture 5.10. Let $p > 3$ be a prime, then

Moreover,

Conjecture 5.11. Let $p > 3$ be a prime, then

Conjecture 5.12. Let $p > 3$ be a prime, then

Conjecture 5.13. Let $p > 5$ be a prime, then

Conjecture 5.14. Let $p > 3$ be a prime, then

Conjecture 5.15. Let $p$ be an odd prime, then

Conjecture 5.16. Let $p$ be an odd prime, then

Conjecture 5.17. Let $p$ be a prime with $p\not = 2, 7$, then

Conjecture 5.18. Let $p$ be an odd prime, then

Conjecture 5.19. Let $p$ be a prime with $p\not = 2, 5$, then

Remark 5.2. Let $p$ be a prime with $p > 7$ and $p\not = 71$. In ^[25], the author conjectured congruences for $\sum_{k = 0}^{p-1} \frac { \binom{2k}k^2 \binom{4k}{2k}}{256^k}$, $\sum_{k = 0}^{p-1} \frac { \binom{2k}k^2 \binom{4k}{2k}}{28^{4k}}$ and $\sum_{k = 0}^{p-1} \frac { \binom{2k}k \binom{3k}k \binom{6k}{3k}}{20^{3k}}$ modulo $p^3$. The congruence for $\sum_{k = 0}^{p-1} \frac { \binom{2k}k^2 \binom{4k}{2k}}{28^{4k}} \ (\text{ mod}\ {p^2})$ was conjectured by Z. W. Sun ^[27].

For any odd prime $p$, let

Conjecture 5.20. Let $p$ be a prime with $p\not = 2, 7$, then

Conjecture 5.21. Let $p$ be a prime with $p\not = 2, 7$, then

and

Conjecture 5.22. Let $p$ be a prime with $p\not = 2, 3, 7$, then

Conjecture 5.23. Let $p$ be a prime with $p\not = 2, 3, 7$, then

Conjecture 5.24. Let $p$ be a prime with $p > 7$, then

Conjecture 5.25. Let $p > 3$ be a prime, then

Conjecture 5.26. Let $p$ be a prime with $p\not = 2, 5$, and $R_{20}(p) = \binom{ \frac {p-1}2}{[p/20]} \binom{ \frac {p-1}2}{[3p/20]}$, then

Remark 5.3. For any prime $p\not = 2, 5$, the congruence for $\sum_{k = 0}^{p-1} \binom{2k}k^2 \binom{4k}{2k}(-1024)^{-k}$ modulo $p^2$ was first conjectured by Z. W. Sun in ^[27]. Let $p \equiv 1\ (\text{ mod}\ {20})$ be a prime and so $p = x^2+5y^2$. In 1840, Cauchy proved that

see [3, p.291].

Conjecture 5.27. Let $p$ be a prime with $p\not = 2, 5$, then

Moreover, if $(\frac {{-5}}{p}) = 1$, then

if $(\frac {{-5}}{p}) = -1$, then

Conjecture 5.28. Let $p > 3$ be a prime with $(\frac {{-6}}{p}) = -1$ and $R_{24}(p) = \binom{ \frac {p-1}2}{[\frac p{24}]} \binom{ \frac {p-1}2}{[\frac {5p}{24}]}$, then

and

Conjecture 5.29. Let $p > 3$ be a prime, then

Moreover, if $(\frac {{-6}}{p}) = 1$, then

if $(\frac {{-6}}{p}) = -1$, then

Conjecture 5.30. Let $p > 3$ be a prime, then

Moreover, if $(\frac {{-6}}{p}) = 1$, then

if $(\frac {{-6}}{p}) = -1$, then

Remark 5.4. Let $p$ be a prime with $p > 3$. In [25, Conjecture 4.24], the author conjectured the congruences for $\sum_{k = 0}^{p-1} \frac { \binom{2k}k^2 \binom{3k}k}{216^k}$ and $\sum_{k = 0}^{p-1} \frac { \binom{2k}k^2 \binom{4k}{2k}}{48^{2k}}$ modulo $p^3$ in the case $(\frac {{-6}}{p}) = 1$. The corresponding congruences modulo $p^2$ were conjectured by Z. W. Sun in ^[27]. In 2019, Guo and Zudilin ^[6] proved Z. W. Sun's conjecture:

Conjecture 5.31. Let $p > 5$ be a prime.

(ⅰ) If $p \equiv 1, 19\ (\text{ mod}\ {30})$ and so $p = x^2+15y^2$, then

(ⅱ) If $p \equiv 17, 23\ (\text{ mod}\ {30})$ and so $p = 3x^2+5y^2$, then

(ⅲ) If $(\frac {{-15}}{p}) = -1$, then

Moreover,

Conjecture 5.32. Let $p$ be a prime with $p > 5$ and

(ⅰ) If $(\frac {{-10}}{p}) = 1$, then

Moreover,

and

(ⅱ) If $(\frac {{-10}}{p}) = -1$, then

Moreover,

and

Remark 5.5. Let $p > 3$ be a prime. In ^[27], Z. W. Sun conjectured the congruence for $\sum_{k = 0}^{p-1} \frac { \binom{2k}k^2 \binom{4k}{2k}}{12^{4k}}\ (\text{ mod}\ {p^2})$.

Conjecture 5.33. Let $p$ be a prime with $p\not = 2, 11$ and $R_{11}(p) = \binom{[\frac {3p}{11}]}{[\frac p{11}]}^2 \binom{[\frac {6p}{11}]}{[\frac {3p}{11}]}^2/ \binom{[\frac {4p}{11}]}{[\frac {2p}{11}]}^2$.

(ⅰ) If $p \equiv 1, 3, 4, 5, 9\ (\text{ mod}\ {11})$ and so $4p = x^2+11y^2$, then

Moreover,

(ⅱ) If $p \equiv 2, 6, 7, 8, 10\ (\text{ mod}\ {11})$, then

Moreover,

and

Remark 5.6. Let $p$ be a prime with $p\not = 2, 3, 11$. In ^[25], the author conjectured the congruences for $\sum_{k = 0}^{p-1} \frac { \binom{2k}k^2 \binom{3k}k}{64^k}$ and $\sum_{k = 0}^{p-1} \frac { \binom{2k}k \binom{3k}k \binom{6k}{3k}}{(-32)^{3k}}$ modulo $p^3$. The corresponding congruences modulo $p^2$ were conjectured by Z. W. Sun ^[29]. Suppose that $p \equiv 1, 3, 4, 5, 9\ (\text{ mod}\ {11})$ and so $4p = x^2+11y^2$. In ^[9], Lee and Hahn proved that

where $n = [p/11]$. The case $p \equiv 1\ (\text{ mod}\ {11})$ is due to Jacobi.

Conjecture 5.34. Let $p > 3$ be a prime and

(ⅰ) If $(\frac {{p}}{{19}}) = 1$ and so $4p = x^2+19y^2$, then

Moreover,

(ⅱ) If $(\frac {{p}}{{19}}) = -1$, then

Moreover,

and

Conjecture 5.35. Let $p > 5$ be a prime.

(ⅰ) If $(\frac {{p}}{{43}}) = 1$ and so $4p = x^2+43y^2$, then

(ⅱ) If $(\frac {{p}}{{43}}) = -1$, then

Conjecture 5.36. Let $p$ be a prime with $p\not = 2, 3, 5, 11$.

(ⅰ) If $(\frac {{p}}{{67}}) = 1$ and so $4p = x^2+67y^2$, then

(ⅱ) If $(\frac {{p}}{{67}}) = -1$, then

Conjecture 5.37. Let $p$ be a prime with $p\not = 2, 3, 5, 23, 29$. If $(\frac {{p}}{{163}}) = 1$ and so $4p = x^2+163y^2$, then

Remark 5.7. Suppose that $p > 3$ is a prime. In ^[29], Z. W. Sun conjectured the congruences for $\sum_{k = 0}^{p-1} \frac { \binom{2k}k \binom{3k}k \binom{6k}{3k}}{m^{3k}}$ modulo $p^2$ in the cases $m = -32, -96, -960, -5280, -640320$. The corresponding congruences modulo $p$ were proved by the author in ^[19].

Conjecture 5.38. Let $p$ be a prime with $p > 5$.

(ⅰ) If $p \equiv 1, 4\ (\text{ mod}\ {15})$ and so $4p = x^2+75y^2$, then

(ⅱ) If $p \equiv 7, 13\ (\text{ mod}\ {15})$ and so $4p = 3x^2+25y^2$, then

(ⅲ) If $p \equiv 2\ (\text{ mod}\ 3)$, then

Conjecture 5.39. Let $p$ be a prime with $p > 3$.

(ⅰ) If $(\frac {p}{3}) = (\frac {p}{{17}}) = 1$ and so $4p = x^2+51y^2$, then

(ⅱ) If $(\frac {p}{3}) = (\frac {p}{{17}}) = -1$ and so $4p = 3x^2+17y^2$, then

(ⅲ) If $(\frac {{-51}}{p}) = -1$, then

Conjecture 5.40. Let $p$ be a prime with $p > 3$.

(ⅰ) If $(\frac {p}{3}) = (\frac {p}{{41}}) = 1$ and so $4p = x^2+123y^2$, then

(ⅱ) If $(\frac {p}{3}) = (\frac {p}{{41}}) = -1$ and so $4p = 3x^2+41y^2$, then

(ⅲ) If $(\frac {{-123}}{p}) = -1$, then

Conjecture 5.41. Let $p$ be a prime with $p > 3$.

(ⅰ) If $(\frac {p}{3}) = (\frac {p}{{89}}) = 1$ and so $4p = x^2+267y^2$, then

(ⅱ) If $(\frac {p}{3}) = (\frac {p}{{89}}) = -1$ and so $4p = 3x^2+89y^2$, then

(ⅲ) If $(\frac {{-267}}{p}) = -1$, then

Remark 5.8. Let $p > 5$ be a prime. In ^[18], the author proved the congruences modulo $p$ for $\sum_{k = 0}^{p-1} \binom{2k}k^2 \binom{3k}k/m^k$ in the cases $m = -8640, -12^3, -48^3, -300^3$. In ^[29], Z. W. Sun conjectured the corresponding congruences for $\sum_{k = 0}^{p-1} \binom{2k}k^2 \binom{3k}k/m^k\ (\text{ mod}\ {p^2})$ in the cases $m = -12^3, -48^3, -300^3$.

Conjecture 5.42. Let $p > 3$ be a prime.

(ⅰ) If $(\frac {{-1}}{p}) = (\frac {{13}}{p}) = 1$ and so $p = x^2+13y^2$, then

(ⅱ) If $(\frac {{-1}}{p}) = (\frac {{13}}{p}) = -1$ and so $2p = x^2+13y^2$, then

(ⅲ) If $(\frac {{-13}}{p}) = -1$, then

Conjecture 5.43. Let $p$ be a prime with $p\not = 2, 3, 7$.

(ⅰ) If $(\frac {{-1}}{p}) = (\frac {{37}}{p}) = 1$ and so $p = x^2+37y^2$, then

(ⅱ) If $(\frac {{-1}}{p}) = (\frac {{37}}{p}) = -1$ and so $2p = x^2+37y^2$, then

(ⅲ) If $(\frac {{-37}}{p}) = -1$, then

Conjecture 5.44. Let $p$ be a prime with $p\not = 2, 3, 11$.

(ⅰ) If $\big(\frac {{2}}{p}\big) = (\frac {{-11}}{p}) = 1$ and so $p = x^2+22y^2$, then

(ⅱ) If $\big(\frac {{2}}{p}\big) = (\frac {{-11}}{p}) = -1$ and so $p = 2x^2+11y^2$, then

(ⅲ) If $(\frac {{-22}}{p}) = -1$, then

Conjecture 5.45. Let $p$ be a prime with $p\not = 2, 3, 11$.

(ⅰ) If $\big(\frac {{-2}}{p}\big) = (\frac {{29}}{p}) = 1$ and so $p = x^2+58y^2$, then

(ⅱ) If $(\frac {{-2}}{p}) = (\frac {{29}}{p}) = -1$ and so $p = 2x^2+29y^2$, then

(ⅲ) If $(\frac {{-58}}{p}) = -1$, then

Conjecture 5.46. Let $p > 5$ be a prime.

(ⅰ) If $p \equiv 1, 9\ (\text{ mod}\ {20})$ and so $p = x^2+25y^2$, then

(ⅱ) If $p \equiv 13, 17\ (\text{ mod}\ {20})$ and so $2p = x^2+25y^2$, then

(ⅲ) If $p \equiv 3\ (\text{ mod}\ 4)$, then

Remark 5.9. Let $p > 3$ be a prime. The congruences for $\sum_{k = 0}^{p-1} \binom{2k}k^2 \binom{4k}{2k}/m^k\ (\text{ mod}\ {p^2})$ in the cases $m = -82944, -199148544, 1584^2,396^4, -6635520$ were conjectured by Z. W. Sun earlier, see ^[27,29] and arXiv:0911.5665v59.

Conjecture 5.47. Let $p$ be an odd prime.

(ⅰ) If $p \equiv 1\ (\text{ mod}\ 4)$ and so $p = x^2+4y^2$ with $4\mid x-1$, then

(ⅱ) If $p \equiv 3\ (\text{ mod}\ 4)$, then

Conjecture 5.48. Let $p$ be an odd prime, then

and

Conjecture 5.49. Let $p$ be an odd prime, then

Remark 5.10. Let $p$ be an odd prime. In ^[33], Z. W. Sun established the congruences for $\sum_{k = 0}^{p-1} \frac { \binom{2k}k^2}{m^k(2k-1)}\ (\text{ mod}\ {p^2})$ in the cases $m = -16, 8, 32$.

Now we present two general conjectures.

Conjecture 5.50. Let $a, x\in\Bbb Q$ and $d\in\Bbb Z$ with $adx\not = 0$, where $\Bbb Q$ is the set of rational numbers. Suppose that $\sum_{k = 0}^{p-1} \binom ak \binom{-1-a}kx^k \equiv 0\ (\text{ mod}\ p)$ for all odd primes $p$ satisfying $a, x\in\Bbb Z_p$ and $\big(\frac {{d}}{p}\big) = -1$. Then there is a constant $c\in\Bbb Q$ such that for all odd primes $p$ with $c\in\Bbb Z_p$ and $\big(\frac {{d}}{p}\big) = -1$,

Conjecture 5.51. Let $a, m\in\Bbb Q$ and $d\in\Bbb Z$ with $adm\not = 0$. Suppose that

for all odd primes $p$ satisfying $a, m\in\Bbb Z_p$, $p\nmid m$ and $\big(\frac {{d}}{p}\big) = -1$. Then there is a constant $c\in\Bbb Q$ such that for all odd primes $p$ with $c, m\in\Bbb Z_p$, $p\nmid m$ and $\big(\frac {{d}}{p}\big) = -1$,

6.   Conjectures for congruences involving Apéry-like numbers

For an odd prime $p$, let $R_1(p)$–$R_3(p)$ be given by (5.1)–(5.3). With the help of Maple, we discover the following conjectures involving Apéry-like numbers.

Conjecture 6.1. Let $p$ be an odd prime, then

Conjecture 6.2. Let $p > 3$ be a prime, then

Conjecture 6.3. Let $p$ be an odd prime, then

Conjecture 6.4. Let $p > 3$ be a prime, then

and

Conjecture 6.5. Let $p$ be a prime with $p > 3$, then

Conjecture 6.6. Let $p$ be a prime with $p > 3$, then

Conjecture 6.7. Let $p$ be an odd prime, then

Conjecture 6.8. Let $p$ be a prime with $p\not = 2, 3, 7$, then

and

Conjecture 6.9. Let $p$ be a prime with $p\not = 2, 11$, then

Conjecture 6.10. Let $p$ be a prime with $p\not = 2, 3, 19$, then

Conjecture 6.11. Let $p > 3$ be a prime, then

Conjecture 6.12. Let $p > 3$ be a prime, then

Conjecture 6.13. Let $p$ be a prime with $p\not = 2, 5$. If $(\frac {{-5}}{p}) = 1$, then

if $(\frac {{-5}}{p}) = -1$, then

Conjecture 6.14. Let $p > 3$ be a prime. If $(\frac {{-6}}{p}) = 1$, then

if $(\frac {{-6}}{p}) = -1$, then

Conjecture 6.15. Let $p > 3$ be a prime, then

Conjecture 6.16. Let $p > 3$ be a prime, then

Conjecture 6.17. Let $p > 3$ be a prime, then

Conjecture 6.18. Let $p > 3$ be a prime. If $(\frac {{-6}}{p}) = 1$, then

if $(\frac {{-6}}{p}) = -1$, then

Conjecture 6.19. Let $p > 3$ be a prime, then

Conjecture 6.20. Let $p$ be a prime with $p\not = 2, 3, 11$, then

Conjecture 6.21. Let $p$ be a prime with $p\not = 2, 3, 19$. If $(\frac {p}{{19}}) = 1 \; and \;so \; 4p = x^2+19y^2$, the

if $\big(\frac {p}{{19}}\big) = -1$, then

Conjecture 6.22. Let $p$ be a prime with $p > 3$, then

Remark 6.1. In ^[29], Z. W. Sun conjectured that for any prime $p > 3$,

For congruences related to Conjectures 6.1–6.18 see ^{[21,22,23,24,25]}.

7.   Conclusions

In Sections 2–4, we prove some congruences for the sums involving binomial coefficients and Apéry-like numbers modulo $p^r$, where $p$ is an odd prime and $r\in\{1, 2, 3\}$. Based on calculations by Maple, in Sections 3, 5 and 6 we pose 83 challenging conjectures on congruences modulo $p^2$ or $p^3$.

Acknowledgments

The author is supported by the National Natural Science Foundation of China (Grant No. 11771173).

Conflict of interest

The author declares no conflicts of interest regarding the publication of this paper.