Jacobi Polynomials and Congruences Involving Some Higher-Order Catalan Numbers and Binomial Coefficients

Khodabakhsh Hessami Pilehrood and Tatiana Hessami Pilehrood

Fields Institute for Research in Mathematical Sciences 222 College St. Toronto, Ontario M5T 3J1 Canada

                  hessamik@gmail.com
                  hessamit@gmail.com

Abstract

In this paper, we study congruences on sums of products of binomial coefficients that can be proved by using properties of the Jacobi polynomials. We give special attention to polynomial congruences containing Catalan numbers, second-order Catalan numbers, the sequence S_{n} = \frac{\binom{6n}{3n}\binom{3n}{2n}}{2\binom{2n}{n}(2n+1)}, and the binomial coefficients \binom{3n}{n} and \binom{4n}{2n}. As an application, we address several conjectures of Z. W. Sun on congruences of sums involving S_{n} and we prove a cubic residuacity criterion in terms of sums of the binomial coefficients \binom{3n}{n} conjectured by Z. H. Sun.

1 Introduction

In this paper, building on our previous work with Tauraso [3], we continue to apply properties of the Jacobi polynomials P_{n}^{(±1/2,∓1/2)}(x) for proving polynomial and numerical congruences containing sums of binomial coefficients. In particular, we derive polynomial congruences for sums involving binomial coefficients \binom{3n}{n}, \binom{4n}{2n}, Catalan numbers (\underline{A000108})

C_{n} = \frac{1}{n + 1}\binom{2n}{n} = \binom{2n}{n} − \binom{2n}{n − 1}, n = 0, 1, 2, …,

second-order Catalan numbers (\underline{A001764})

C_{n}^{(2)} = \frac{1}{2n + 1}\binom{3n}{n} = \binom{3n}{n} − 2\binom{3n}{n − 1}, n = 0, 1, 2, …,

and the sequence (\underline{A176898})

S_{n} = \frac{\binom{6n}{3n}\binom{3n}{n}}{2\binom{2n}{n}(2n + 1)},  n = 0, 1, 2, …,  (1)

arithmetical properties of which have been studied very recently by Sun [13] and Guo [2]. Recall that the Jacobi polynomials P_{n}^{(α,β)}(x) are defined by

P_{n}^{(α,β)}(x) = \frac{(α + 1)_{n}}{n!} F (−n, n + α + β + 1; α + 1; (1 − x)/2), α, β > −1,  (2)

where

F (a, b; c; z) = ∑_{k=0}^{∞} \frac{(a)_{k}(b)_{k}}{k!(c)_{k}} z^{k},

is the Gauss hypergeometric function and (a)_{0} = 1, (a)_{k} = a(a + 1) ⋯ (a + k − 1), k ≥ 1, is the Pochhammer symbol.

The polynomials P_{n}^{(α,β)}(x) satisfy the three-term recurrence relation [14, Sect. 4.5]

\begin{matrix}2(n + 1) & (n + α + β + 1)(2n + α + β)P_{n+1}^{(α,β)}(x) \\ = ((2n + α + β + 1)(α^{2} − β^{2}) + (2n + α + β)_{3}x) P_{n}^{(α,β)}(x) \\ − 2(n + α)(n + β)(2n + α + β + 2)P_{n−1}^{(α,β)}(x) \end{matrix}  (3)

with the initial conditions P_{0}^{(α,β)}(x) = 1, P_{1}^{(α,β)}(x) = (x(α + β + 2) + α − β)/2. While in [3] we studied binomial sums arising from the truncation of the series

arcsin(z) = ∑_{k=0}^{∞} \frac{\binom{2k}{k}z^{2k+1}}{4^{k}(2k + 1)},  |z| ≤ 1,  (4)

the purpose of the present paper is to consider a quadratic transformation of the Gauss hypergeometric function given by [6, p. 210]

\frac{sin(a arcsin(z))}{a} = zF (\frac{1 + a}{2}, \frac{1 − a}{2}; \frac{3}{2}; z^{2}) ,  |z| ≤ 1,  (5)

which essentially can be regarded as a generalization of series (4). Note that letting a approach zero in (5) yields (4). On the other side, identity (5) serves as a source of generating functions for some special sequences of numbers including those mentioned above. Namely, for a = 1/2, 1/3, 2/3, we have

sin (\frac{arcsin(z)}{2}) = 2 ∑_{k=0}^{∞} C_{2k} (\frac{z}{4})^{2k+1} ,  |z| ≤ 1,  (6)

sin (\frac{arcsin(z)}{3}) = \frac{z}{3} ∑_{k=0}^{∞} C_{k}^{(2)} (\frac{4z^{2}}{27} )^{k} ,  |z| ≤ 1,  (7)

sin (\frac{2}{3} arcsin(z)) = \frac{4z}{3} ∑_{k=0}^{∞} S_{k} (\frac{z^{2}}{108})^{k} ,  |z| ≤ 1.  (8)

In this paper, we develop a unified approach for the calculation of polynomial congruences modulo a prime p arising from the truncation of the series (6)–(8) and polynomial congruences involving binomial coefficients \binom{3k}{k}, \binom{4k}{2k} and also the sequence (2k+1)S_{k} within various ranges of summation depending on a prime p.

Note that the congruences involving binomial coefficients \binom{3k}{k}, \binom{4k}{2k} have been studied extensively from different points of view [9, 10, 11, 12, 16]. Z. H. Sun [10, 11] studied congruences for the sums ∑_{k=1}^{⌊p/3⌋} \binom{3k}{k}t^{k} and ∑_{k=1}^{⌊p/4⌋} \binom{4k}{2k}t^{k} using congruences for Lucas sequences and properties of the cubic and quartic residues. Sun [9] also investigated interesting connections between values of ∑_{k=1}^{⌊p/3⌋} \binom{3k}{k}t^{k} (mod p), solubility of cubic congruences, and cubic residuacity criteria. Zhao, Pan, and Sun [16] obtained first congruences for the sums ∑_{k=1}^{p−1} \binom{3k}{k}t^{k} and ∑_{k=1}^{p−1} C_{k}^{(2)}t^{k} at t = 2 with the help of some combinatorial identity. Later Z. W. Sun [12] gave explicit congruences for t = −4, \frac{1}{6}, \frac{1}{7}, \frac{1}{8}, \frac{1}{9}, \frac{1}{13}, \frac{3}{8}, \frac{4}{27} by applying properties of third-order recurrences and cubic residues.

Our approach is based on reducing values of the finite sums discussed above modulo a prime p to values of the Jacobi polynomials P^{(±1/2,∓1/2)}(x), which is done in Section 2, and then investigating congruences for the Jacobi polynomials in subsequent sections. In Section 3, we deal with polynomial congruences involving binomial coefficients \binom{4k}{2k} and evenindexed Catalan numbers C_{2k}. In Section 4, we study polynomial congruences containing binomial coefficients \binom{3k}{k} and second-order Catalan numbers C_{k}^{(2)}. In Sections 5 and 6, we apply the theory of cubic residues developed in [8] to study congruences for polynomials of the form

∑_{k=(p+1)/2}^{⌊2p/3⌋} \binom{3k}{k}t^{k}, ∑_{k=1}^{p−1} \binom{3k}{k}t^{k}, ∑_{k=1}^{p−1} C_{k}^{(2)}t^{k}, ∑_{k=0}^{p−1} S_{k}t^{k}, ∑_{k=0}^{p−1}(2k + 1)S_{k}t^{k}.

As a result, we prove several cubic residuacity criteria in terms of these sums, one of which, in terms of ∑_{k=(p+1)/2}^{⌊2p/3⌋} \binom{3k}{k}t^{k}, confirms a question posed by Z. H. Sun [9, Conj. 2.1].

In Section 6, we derive polynomial congruences for the sums ∑_{k=0}^{⌊p/6⌋} S_{k}t^{k}, ∑_{k=0}^{p−1} S_{k}t^{k}, ∑_{k=0}^{⌊p/6⌋}(2k + 1)S_{k}t^{k}, ∑_{k=0}^{p−1}(2k + 1)S_{k}t^{k} and also give many numerical congruences which are new and have not appeared in the literature before. In particular, we show that

∑_{k=0}^{p−1} \frac{S_{k}}{108^{k}} ≡ \frac{1}{2} (\frac{3}{p}) (mod p)

confirming a conjecture of Z. W. Sun [13, Conj. 2]. Finally, in Section 7, we prove a closed form formula for a companion sequence of S_{n} answering another question of Sun [13, Conj. 4].

2 Main theorem

For a non-negative integer n, we consider the sequence w_{n}(x) defined [3, Sect. 3] by

w_{n}(x) := (2n + 1)F (−n, n + 1; 3/2; (1 − x)/2) = \frac{n!}{(1/2)_{n}} P_{n}^{(1/2,−1/2)}(x).  (9)

From (3) it follows that w_{n}(x) satisfies a second-order linear recurrence with constant coefficients

w_{n+1}(x) = 2xw_{n}(x) − w_{n−1}(x)

and initial conditions w_{0}(x) = 1, w_{1}(x) = 1 + 2x. This yields the following formulae:

w_{n}(x) = \begin{cases}\frac{(α + 1)α^{n} − (α^{−1} + 1)α^{−n}}{α − α^{−1}} , & if x ≠ ±1; \\ 2n + 1, & if x = 1; \\ (−1)^{n}, & if x = −1, \end{cases}  (10)

where α = x + \sqrt{x^{2} − 1}. Note that for x ∈ (−1, 1) we also have an alternative representation

w_{n}(x) = cos(n arccos x) + \frac{x + 1}{\sqrt{1 − x^{2}}} sin(n arccos x).  (11)

By the well-known symmetry property of the Jacobi polynomials

P_{n}^{(α,β)}(x) = (−1)^{n}P_{n}^{(β,α)}(−x)

and formula (2), we get one more expression of w_{n}(x) in terms of the Gauss hypergeometric function

w_{n}(x) = (−1)^{n}F (−n, n + 1; 1/2; (1 + x)/2).  (12)

For a given prime p, let D_{p} denote the set of those rational numbers whose denominator is not divisible by p. Let φ(m) be the Euler totient function and let (\frac{a}{p}) be the Legendre symbol. We put (\frac{a}{p}) = 0 if p|a. For c = a/b ∈ D_{p} written in its lowest terms, we define (\frac{c}{p}) = (\frac{ab}{p}) in view that the congruences x^{2} ≡ c (mod p) and (bx)^{2} ≡ ab (mod p) are equivalent. It is clear that (\frac{c}{p}) has all the formal properties of the ordinary Legendre symbol. For any rational number x, let v_{p}(x) denote the p-adic order of x.

Theorem 1. Let m be a positive integer with φ(m) = 2, i.e., m ∈ {3, 4, 6}, and let p be a prime greater than 3. Then for any t ∈ D_{p}, we have

∑_{k=0}^{⌊p/m⌋} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k + 1)!} t^{k} ≡ \frac{1}{1 + 2⌊p/m⌋} w_{⌊ \frac{p}{m} ⌋}(1 − t/2) (mod p),  (13)

∑_{k=0}^{⌊p/m⌋} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k)!} t^{k} ≡ (−1)^{⌊p/m⌋} w_{⌊ \frac{p}{m} ⌋}(t/2 − 1) (mod p),  (14)

∑_{k=(p−1)/2}^{⌊(m−1)p/m⌋} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k + 1)!} t^{k} ≡ \frac{−1}{m(1 + 2⌊p/m⌋)}(w_{⌊ \frac{(m−1)p}{m} ⌋}(1 − t/2)+ w_{⌊ \frac{p}{m} ⌋}(1 − t/2)) (mod p), (15)

∑_{k=(p+1)/2}^{⌊(m−1)p/m⌋} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k)!} t^{k} ≡ \frac{(−1)^{⌊p/m⌋}}{m} (w_{⌊ \frac{(m−1)p}{m} ⌋}(t/2 − 1) − w_{⌊ \frac{p}{m} ⌋}(t/2 − 1)) (mod p).

Proof. Let m ∈ {3, 4, 6}, i.e., φ(m) = 2. Suppose p is an odd prime greater than 3 and p ≡ r (mod m), where r ∈ \{1, m − 1\}. We put n = \frac{p−r}{m}. Then p = mn + r and from (9) we have

w_{n}(x) = (2n + 1)F (−n, n + 1; \frac{3}{2}; \frac{1 − x}{2}) = \frac{2p − 2r + m}{m} ∑_{k=0}^{n} \frac{(\frac{r−p}{m})k (\frac{m−r+p}{m})k}{(\frac{3}{2})_{k} k!} (\frac{1 − x}{2})^{k} .

Since (\frac{3}{2})_{k} = \frac{3·5·...·(2k+1)}{2^{k}} and 2k + 1 ≤ 2n + 1 < p, the denominators of the summands are coprime to p and we have

w_{n}(x)≡ \frac{m − 2r}{m} ∑_{k=0}^{⌊p/m⌋} \frac{(\frac{r}{m})k (\frac{m−r}{m})k}{(\frac{3}{2})_{k} k!} (\frac{1 − x}{2})^{k}= \frac{m − 2r}{m} ∑_{k=0}^{⌊p/m⌋} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(\frac{3}{2})_{k} k!} (\frac{1 − x}{2})^{k}(mod p)

or

w_{n}(x) ≡ \frac{m − 2r}{m} ∑_{k=0}^{⌊p/m⌋} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k + 1)!} (2(1 − x))^{k} (mod p).

Replacing x by 1 − t/2, we get (13).

Applying formula (12) to w_{n}(x), similarly as before, we get

w_{n}(x) = (−1)^{n}F (−n, n + 1; 1/2; (1 + x)/2) = (−1)^{n} ∑_{k=0}^{n} \frac{(\frac{r−p}{m})k (\frac{p−r+m}{m})k}{(\frac{1}{2})_{k} k!} (\frac{1 + x}{2})^{k}

or

w_{n}(x) ≡ (−1)^{n}∑_{k=0}^{n} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(\frac{1}{2})_{k} k!} (\frac{1 + x}{2})^{k} = (−1)^{n} ∑_{k=0}^{n} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k)!} (2(1 + x))^{k} (mod p).

Substituting t = 2(1 + x), we obtain (14).

To prove the other two congruences, we consider (m − 1)p modulo m. It is clear that (m−1)p ≡ r (mod m), where r ∈ \{1, m−1\}. We put n = \frac{(m−1)p−r}{m}. Then (m−1)p = mn+r and from (9) we have

w_{n}(x) = \frac{2(m − 1)p − 2r + m}{m} ∑_{k=0}^{n} \frac{(\frac{r−(m−1)p}{m})k (\frac{(m−1)p+m−r}{m})k}{(\frac{3}{2})_{k} k!} (\frac{1 − x}{2})^{k} .  (16)

Note that p divides (\frac{3}{2})_{k} if and only if k ≥ (p − 1)/2. Moreover, p^{2} does not divide (\frac{3}{2})_{k} for any k from the range of summation. Similarly, we have

(\frac{r − (m − 1)p}{m})_{k} = ∏_{l=0}^{k−1} \frac{r + ml − (m − 1)p}{m}.

All possible multiples of p among the numbers r + ml, where 0 ≤ l ≤ k − 1 ≤ \frac{(m−1)p−r−m}{m}, could be only of the form r + ml = jp with 1 ≤ j ≤ m − 2. This implies that jp ≡ r ≡ −p (mod m) or (j + 1)p ≡ 0 (mod m), which is impossible, since gcd(p, m) = 1 and j + 1 < m. So p does not divide (\frac{r−(m−1)p}{m})_{k}. Considering

(\frac{(m − 1)p + m − r}{m})_{k} = ∏_{l=1}^{k} \frac{(m − 1)p + ml − r}{m},

we see that p divides (\frac{(m−1)p+m−r}{m})_{k} if and only if k ≥ \frac{p+r}{m}. Moreover, p^{2} does not divide (\frac{(m−1)p+m−r}{4})_{k} for any k from the range of summation. Indeed, if we had ml − r = jp for some 1 < j ≤ m − 1, then p ≡ −r ≡ jp (mod m) and therefore p(j − 1) ≡ 0 (mod m), which is impossible. From the divisibility properties of the Pochhammer’s symbols above and (16) we easily conclude that

w_{⌊ \frac{(m−1)p}{m} ⌋}(x) ≡ \frac{m − 2r}{m} ( ∑_{k=0}^{⌊p/m⌋} + ∑_{k=(p−1)/2}^{⌊(m−1)p/m⌋})\frac{(\frac{r−(m−1)p}{m})k (\frac{(m−1)p+m−r}{m})k}{(\frac{3}{2})_{k} k!} (\frac{1 − x}{2})^{k} (mod p)

and therefore,

w_{⌊ \frac{(m−1)p}{m} ⌋}(x) ≡ \frac{m − 2r}{m} ( ∑_{k=0}^{⌊p/m⌋} + m ∑_{k=(p−1)/2}^{⌊(m−1)p/m⌋}) \frac{(\frac{r}{m})k (\frac{m−r}{m})k}{(\frac{3}{2})_{k} k!} (\frac{1 − x}{2})^{k} (mod p),

where for the second sum, we employed the congruence

\begin{aligned}\begin{matrix}(\frac{(m−1)p+m−r}{m}) & k \\ (\frac{3}{2}) & k \end{matrix} & = \frac{\frac{(m−1)p+m−r}{m} · \frac{(m−1)p+2m−r}{m} ⋯ \frac{(m−1)p+m· \frac{p+r}{m} −r}{m} ⋯ \frac{(m−1)p+mk−r}{m}}{\frac{3}{2} · \frac{5}{2} ⋯ \frac{p}{2} ⋯ \frac{2k+1}{2}} \\ & ≡ m\begin{matrix}(\frac{m−r}{m}) & k \\ (\frac{3}{2}) & k \end{matrix} (mod p) \end{aligned}

valid for (p − 1)/2 ≤ k ≤ ⌊(m − 1)p/m⌋. Now by (13), we obtain

\frac{m}{m − 2r} w_{⌊ \frac{(m−1)p}{m} ⌋}(x) ≡ \frac{1}{1 + 2⌊p/m⌋} w_{⌊ \frac{p}{m} ⌋}(x)+m ∑_{k=(p−1)/2}^{⌊(m−1)p/m⌋} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k + 1)!} (2(1 − x))^{k} (mod p).

Taking into account that ⌊\frac{(m−1)p}{m}⌋ = p − 1 − ⌊\frac{p}{m}⌋ and replacing x by 1 − t/2, we get the desired congruence (15).

Finally, applying formula (12) and following the same line of arguments as for proving (15), we have

w_{⌊ \frac{(m−1)p}{m} ⌋}(x) = (−1)^{n}F (−n, n + 1; 1/2; (1 + x)/2)

= (−1)^{⌊ \frac{(m−1)p}{m} ⌋} ∑_{k=0}^{⌊(m−1)p/m⌋} \frac{(\frac{r−(m−1)p}{m})k (\frac{(m−1)p+m−r}{m})k}{(\frac{1}{2})_{k} k!} (\frac{1 + x}{2})^{k}

≡ (−1)^{⌊ \frac{(m−1)p}{m} ⌋} ( ∑_{k=0}^{⌊p/m⌋} + ∑_{k=(p+1)/2}^{⌊(m−1)p/m⌋}) \frac{(\frac{r−(m−1)p}{m})k (\frac{(m−1)p+m−r}{m})k}{(\frac{1}{2})_{k} k!} (\frac{1 + x}{2})^{k}

≡ (−1)^{⌊ \frac{(m−1)p}{m} ⌋} ( ∑_{k=0}^{⌊p/m⌋} + m ∑_{k=(p+1)/2}^{⌊(m−1)p/m⌋}) \frac{(\frac{1}{m})k (\frac{(m−1)}{m})k}{(2k)!} (2(1 + x))^{k} (mod p).

Now by (15), we obtain

(−1)^{⌊ \frac{(m−1)p}{m} ⌋}w_{⌊ \frac{(m−1)p}{m} ⌋}(x) ≡ (−1)^{⌊ \frac{p}{m} ⌋}w_{⌊ \frac{p}{m} ⌋}(x)+m ∑_{k=(p+1)/2}^{⌊(m−1)p/m⌋} \frac{(\frac{1}{m})k (\frac{(m−1)}{m})k}{(2k)!} (2(1 + x))^{k} (mod p)

and after the substitution x = t/2 − 1, we derive the last congruence of the theorem.

Corollary 2. Let m be a positive integer with φ(m) = 2, i.e., m ∈ {3, 4, 6}, and let p be a prime greater than 3. Then for any t ∈ D_{p}, we have

∑_{k=0}^{p−1} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k + 1)!} t^{k} ≡ \frac{1}{m(1 + 2⌊p/m⌋)}((m − 1)w_{⌊ \frac{p}{m} ⌋}(1 − t/2)−w_{⌊ \frac{(m−1)p}{m} ⌋}(1 − t/2)) (mod p),

∑_{k=0}^{p−1} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k)!} t^{k} ≡ \frac{(−1)^{⌊p/m⌋}}{m} ((m − 1)w_{⌊ \frac{p}{m} ⌋}(t/2 − 1) + w_{⌊ \frac{(m−1)p}{m} ⌋}(t/2 − 1)) (mod p).

Proof. Let p = mn + r, where r ∈ \{1, m − 1\}. If n + 1 = ⌊\frac{p}{m}⌋ + 1 ≤ k ≤ \frac{p−3}{2}, then v_{p}((2k + 1)!) = 0 and v_{p}((\frac{r}{m})_{k}) ≥ 1, since the product ∏_{l=0}^{k−1}(r + lm) is divisible by p.

If (m − 1)n + r = ⌊\frac{(m−1)p}{m}⌋ + 1 ≤ k ≤ p − 1, then it is easy to see that v_{p}((2k + 1)!) = v_{p}((2k)!) = 1, v_{p}((\frac{r}{m})_{k}) ≥ 1, and the product ∏_{l=1}^{k}(lm − r) contains the factor (m − 1)p. This implies that v_{p}((\frac{m−r}{m})_{k}) ≥ 1, and therefore we have

∑_{k=0}^{p−1} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k + 1)!} t^{k} ≡ ∑_{k=0}^{⌊p/m⌋} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k + 1)!} t^{k} + ∑_{k=(p−1)/2}^{⌊(m−1)p/m⌋} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k + 1)!} t^{k} (mod p),

∑_{k=0}^{p−1} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k)!} t^{k} ≡ ∑_{k=0}^{⌊p/m⌋} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k)!} t^{k} + ∑_{k=(p+1)/2}^{⌊(m−1)p/m⌋} \frac{(\frac{1}{m})k (\frac{m−1}{m})k}{(2k)!} t^{k} (mod p).

Finally, applying Theorem 1, we conclude the proof of the corollary.

3 Polynomial congruences involving Catalan numbers

In this section, we consider applications of Theorem 1 when m = 4. In this case, we get polynomial congruences involving even-indexed Catalan numbers C_{2n} (sequence \underline{A048990} in the OEIS [7]) and binomial coefficients \binom{4n}{2n} (sequence \underline{A001448}).

Theorem 3. Let p be an odd prime and let t ∈ D_{p}. Then

∑_{k=0}^{⌊p/4⌋} C_{2k}t^{k} ≡ 2(−1)^{\frac{p−1}{2}} w_{⌊ \frac{p}{4} ⌋}(1 − 32t) (mod p),

∑_{k=(p−1)/2}^{⌊3p/4⌋} C_{2k}t^{k} ≡ \frac{(−1)^{\frac{p+1}{2}}}{2} (w_{⌊ \frac{3p}{4} ⌋}(1 − 32t) + w_{⌊ \frac{p}{4} ⌋}(1 − 32t)) (mod p),

∑_{k=0}^{⌊p/4⌋} \binom{4k}{2k}t^{k} ≡ (\frac{−2}{p}) w_{⌊ \frac{p}{4} ⌋}(32t − 1) (mod p),

∑_{k=(p+1)/2}^{⌊3p/4⌋} \binom{4k}{2k}t^{k} ≡ \frac{1}{4} (\frac{−2}{p}) (w_{⌊ \frac{3p}{4} ⌋}(32t − 1) − w_{⌊ \frac{p}{4} ⌋}(32t − 1)) (mod p).

Proof. We put m = 4 in Theorem 1. Then for any odd prime p, we have p = 4l + r, where l is non-negative integer and r ∈ {1, 3}. Hence,

\frac{1}{1 + 2⌊p/4⌋} = \frac{1}{1 + 2l} = \frac{2}{p + 2 − r} ≡ \frac{2}{2 − r} = 2(−1)^{(p−1)/2} (mod p).

Moreover, (−1)^{⌊p/4⌋} = (−1)^{l} = (\frac{−2}{p}). Now noticing that

C_{2k} = \frac{1}{2k + 1}\binom{4k}{2k} = \frac{(4k)!}{(2k)!(2k + 1)!} = \frac{(\frac{3}{4})_{k}(\frac{1}{4})_{k}}{(2k + 1)!} (64)^{k}

and replacing t by 64t in Theorem 1, we get the desired congruences.

Corollary 4. Let p be an odd prime and let t ∈ D_{p}. Then

∑_{k=0}^{p−1} C_{2k}t^{k} ≡ \frac{1}{2} (\frac{−1}{p}) (3w_{⌊ \frac{p}{4} ⌋}(1 − 32t) − w_{⌊ \frac{3p}{4} ⌋}(1 − 32t)) (mod p),

∑_{k=0}^{p−1} \binom{4k}{2k}t^{k} ≡ \frac{1}{4} (\frac{−2}{p}) (w_{⌊ \frac{3p}{4} ⌋}(32t − 1) + 3w_{⌊ \frac{p}{4} ⌋}(32t − 1)) (mod p).

Evaluating values of the sequences w_{⌊ \frac{p}{4} ⌋}(x) and w_{⌊ \frac{3p}{4} ⌋}(x) modulo p, we get numerical congruences for the above sums. Here are some typical examples.

Corollary 5. Let p be a prime greater than 3. Then

∑_{k=0}^{p−1} \frac{C_{2k}}{16^{k}} ≡ (\frac{2}{p}) ,  ∑_{k=0}^{⌊p/4⌋} \frac{C_{2k}}{16^{k}} ≡ 2 (\frac{2}{p}) (mod p),

∑_{k=0}^{p−1} \frac{\binom{4k}{2k}}{16^{k}} ≡ \frac{1}{4} (\frac{2}{p}) ,  ∑_{k= \frac{p+1}{2}}^{⌊3p/4⌋} \frac{\binom{4k}{2k}}{16^{k}} ≡ −\frac{1}{4} (\frac{2}{p}) (mod p),

∑_{k=0}^{⌊p/4⌋} \frac{C_{2k}}{32^{k}} ≡ 2 (\frac{−1}{p}) (−1)^{⌊p/8⌋},  ∑_{k=0}^{⌊p/4⌋} \frac{\binom{4k}{2k}}{32^{k}} ≡ (\frac{−2}{p}) (−1)^{⌊p/8⌋} (mod p),

∑_{k=0}^{p−1} \frac{C_{2k}}{32^{k}} ≡ \begin{cases}(−1)^{(p−1)/2+⌊p/8⌋} (mod p), & if p ≡ ±1 (mod 8); \\ 2(−1)^{(p−1)/2+⌊p/8⌋} (mod p), & if p ≡ ±3 (mod 8), \end{cases}

∑_{k=0}^{p−1} \frac{\binom{4k}{2k}}{32^{k}} ≡ \begin{cases}(\frac{−2}{p})(−1)^{⌊p/8⌋} (mod p), & if p ≡ ±1 (mod 8); \\ \frac{1}{2}(\frac{−2}{p})(−1)^{⌊p/8⌋} (mod p), & if p ≡ ±3 (mod 8). \end{cases}

Proof. The proof easily follows from the fact that

w_{n}(−1) = (−1)^{n}, w_{n}(0) = (−1)^{⌊n/2⌋}, and w_{n}(1) = 2n + 1.  (17)

Corollary 6. Let p be a prime greater than 3. Then

(\frac{2}{p}) ∑_{k=0}^{p−1} \frac{C_{2k}}{64^{k}} ≡ \begin{cases}1 (mod p), & if p ≡ ±1 (mod 12); \\ −7/2 (mod p), & if p ≡ ±5 (mod 12), \end{cases}

(\frac{2}{p}) ∑_{k=0}^{p−1} \frac{\binom{4k}{2k}}{64^{k}} ≡ \begin{cases}1 (mod p), & if p ≡ ±1 (mod 12); \\ 1/4 (mod p), & if p ≡ ±5 (mod 12), \end{cases}

∑_{k=0}^{p−1} C_{2k} (\frac{3}{64})^{k} ≡ \begin{cases}1 (mod p), & if p ≡ ±1 (mod 12); \\ 1/2 (mod p), & if p ≡ ±5 (mod 12), \end{cases}

∑_{k=0}^{p−1} \binom{4k}{2k} (\frac{3}{64})^{k} ≡ \begin{cases}1 (mod p), & if p ≡ ±1 (mod 12); \\ −5/4 (mod p), & if p ≡ ±5 (mod 12). \end{cases}

Proof. We can easily evaluate by (11),

w_{n}(1/2) = 2 cos(\frac{π(n − 1)}{3}) and w_{n}(−1/2) = \frac{2}{\sqrt{3}} sin(\frac{π(2n + 1)}{3}).  (18)

Hence we obtain

w_{⌊ \frac{p}{4} ⌋}(1/2) ≡ \begin{cases}(−1)^{⌊p/4⌋} (mod p), & if p ≡ ±1 (mod 12); \\ −2(−1)^{⌊p/4⌋} (mod p), & if p ≡ ±5 (mod 12), \end{cases}

w_{⌊ \frac{p}{4} ⌋}(−1/2) ≡ \begin{cases}(−1)^{(p−1)/2} (mod p), & if p ≡ ±1 (mod 12); \\ 0 (mod p), & if p ≡ ±5 (mod 12) \end{cases}

and w_{⌊3p/4⌋}(1/2) ≡ (−1)^{⌊p/4⌋}, w_{⌊3p/4⌋}(−1/2) ≡ (−1)^{(p−1)/2} (mod p). Applying Corollary 4 and the equality (−1)^{(p−1)/2+⌊p/4⌋} = (\frac{2}{p}), we get the desired congruences.

Lemma 7. For any x ≠ ±1, we have

w_{n}(2x^{2} − 1) = \frac{α^{2n+1} − α^{−2n−1}}{α − α^{−1}} , where α = x + \sqrt{x^{2} − 1}.

Proof. By (10), we obtain

w_{2n}(x) = \frac{(α + 1)α^{2n} − (α^{−1} + 1)α^{−2n}}{α − α^{−1}}

= \frac{(α^{2} + 1)α^{2n} − (α^{−2} + 1)α^{−2n} + (α + α^{−1})(α^{2n} − α^{−2n})}{α^{2} − α^{−2}}

= w_{n}(2x^{2} − 1) + \frac{α^{2n} − α^{−2n}}{α − α^{−1}} .

This implies

w_{n}(2x^{2} − 1) = w_{2n}(x) − \frac{α^{2n} − α^{−2n}}{α − α^{−1}} = \frac{α^{2n+1} − α^{−2n−1}}{α − α^{−1}} ,

and the lemma follows. Lemma 8. Let p be a prime, p > 3, and let x ∈ D_{p}. Then

w_{⌊ \frac{p}{4} ⌋}(2x^{2} − 1) ≡ \frac{1}{2} (\frac{−2}{p}) ((\frac{1 − x}{p}) + (\frac{1 + x}{p})) (mod p),
w_{⌊ \frac{3p}{4} ⌋}(2x^{2} − 1) ≡ \frac{1}{2} (\frac{−2}{p}) ((\frac{1 − x}{p}) (1 + 2x) + (\frac{1 + x}{p}) (1 − 2x)) (mod p).

Proof. First, we suppose that x ≠ ±1. Then, by Lemma 7, if p ≡ 1 (mod 4), we have

w_{⌊ \frac{p}{4} ⌋}(2x^{2} − 1) = w_{\frac{p−1}{4}} (2x^{2} − 1) = \frac{α^{\frac{p+1}{2}} − α^{− \frac{p+1}{2}}}{α − α^{−1}}

= \frac{(\sqrt{\frac{x+1}{2}} + \sqrt{\frac{x−1}{2}})^{p+1} − (\sqrt{\frac{x+1}{2}} − \sqrt{\frac{x−1}{2}})^{p+1}}{2\sqrt{x^{2} − 1}}

= \frac{1}{\sqrt{x^{2} − 1}}_{∑_{k=1}^{p} ; k is odd}\binom{p}{k} (\frac{x − 1}{2})^{\frac{k}{2}} (\frac{x + 1}{2})^{\frac{p+1−k}{2}}

+ \frac{1}{\sqrt{x^{2} − 1}}_{∑_{k=0}^{p−1} ; k is even}\binom{p}{k} (\frac{x − 1}{2})^{\frac{k+1}{2}} (\frac{x + 1}{2})^{\frac{p−k}{2}}

≡ \frac{(x − 1)^{\frac{p−1}{2}}}{2^{\frac{p+1}{2}}} + \frac{(x + 1)^{\frac{p−1}{2}}}{2^{\frac{p+1}{2}}} ≡ \frac{1}{2} (\frac{2}{p}) ((\frac{x − 1}{p}) + (\frac{x + 1}{p})) (mod p).

If p ≡ 3 (mod 4), then, by Lemma 7, we have

w_{⌊ \frac{p}{4} ⌋}(2x^{2} − 1) = w_{\frac{p−3}{4}} (2x^{2} − 1) = \frac{α^{\frac{p−1}{2}} − α^{− \frac{p−1}{2}}}{α − α^{−1}}

= \frac{(\sqrt{\frac{x+1}{2}} + \sqrt{\frac{x−1}{2}})^{p−1} − (\sqrt{\frac{x+1}{2}} − \sqrt{\frac{x−1}{2}})^{p−1}}{2\sqrt{x^{2} − 1}}

= \frac{(\sqrt{\frac{x+1}{2}} − \sqrt{\frac{x−1}{2}}) (\sqrt{\frac{x+1}{2}} + \sqrt{\frac{x−1}{2}})^{p} − (\sqrt{\frac{x+1}{2}} + \sqrt{\frac{x−1}{2}}) (\sqrt{\frac{x+1}{2}} − \sqrt{\frac{x−1}{2}})^{p}}{2\sqrt{x^{2} − 1}} .

Simplifying as in the previous case, we get modulo p,

w_{⌊ \frac{p}{4} ⌋}(2x^{2} − 1) ≡ \frac{1}{2} ((\frac{x − 1}{2})^{\frac{p−1}{2}} − (\frac{x + 1}{2})^{\frac{p−1}{2}} ) ≡ \frac{1}{2} (\frac{2}{p}) ((\frac{x − 1}{p}) − (\frac{x + 1}{p})) ,

and the first congruence of the lemma follows. Similarly, to prove the second congruence, we consider two cases. If p ≡ 1 (mod 4), then we get

w_{⌊ \frac{3p}{4} ⌋}(2x^{2} − 1) = w_{\frac{3(p−1)}{4}} (2x^{2} − 1) = \frac{α^{\frac{3p−1}{2}} − α^{− \frac{3p−1}{2}}}{α − α^{−1}}

= \frac{(\sqrt{\frac{x+1}{2}} + \sqrt{\frac{x−1}{2}})^{3p−1} − (\sqrt{\frac{x+1}{2}} − \sqrt{\frac{x−1}{2}})^{3p−1}}{2\sqrt{x^{2} − 1}}

= \frac{(\sqrt{\frac{x+1}{2}} − \sqrt{\frac{x−1}{2}}) (\sqrt{\frac{x+1}{2}} + \sqrt{\frac{x−1}{2}})^{3p} − (\sqrt{\frac{x+1}{2}} + \sqrt{\frac{x−1}{2}}) (\sqrt{\frac{x+1}{2}} − \sqrt{\frac{x−1}{2}})^{3p}}{2\sqrt{x^{2} − 1}} .

Simplifying the right-hand side modulo p, we obtain

\frac{1}{2} (\frac{x − 1}{2})^{\frac{3p−1}{2}} − \frac{1}{2} (\frac{x + 1}{2})^{\frac{3p−1}{2}} + \frac{3}{2} (\frac{x − 1}{2})^{\frac{p−1}{2}} (\frac{x + 1}{2})^{p} − \frac{3}{2} (\frac{x + 1}{2})^{\frac{p−1}{2}} (\frac{x − 1}{2})^{p}

and therefore,

w_{⌊ \frac{3p}{4} ⌋}(2x^{2} − 1) ≡ \frac{1}{2} (\frac{2}{p}) ((\frac{x − 1}{p}) (1 + 2x) + (\frac{x + 1}{p}) (1 − 2x)) (mod p).

If p ≡ 3 (mod 4), then

\begin{aligned}w_{⌊ \frac{3p}{4} ⌋}(2x^{2} − 1) & = w_{\frac{3p−1}{4}} (2x^{2} − 1) = \frac{α^{\frac{3p+1}{2}} − α^{− \frac{3p+1}{2}}}{α − α^{−1}} \\ & = \frac{(\sqrt{\frac{x+1}{2}} + \sqrt{\frac{x−1}{2}})^{3p+1} − (\sqrt{\frac{x+1}{2}} − \sqrt{\frac{x−1}{2}})^{3p+1}}{2\sqrt{x^{2} − 1}} . \end{aligned}

Simplifying the right-hand side modulo p, we get

\frac{1}{2} (\frac{x − 1}{2})^{\frac{3p−1}{2}} + \frac{1}{2} (\frac{x + 1}{2})^{\frac{3p−1}{2}} + \frac{3}{2} (\frac{x − 1}{2})^{\frac{p−1}{2}} (\frac{x + 1}{2})^{p} + \frac{3}{2} (\frac{x + 1}{2})^{\frac{p−1}{2}} (\frac{x − 1}{2})^{p}

and therefore,

w_{⌊ \frac{3p}{4} ⌋}(2x^{2} − 1) ≡ \frac{1}{2} (\frac{2}{p}) ((\frac{x − 1}{p}) (2x + 1) + (\frac{x + 1}{p}) (2x − 1)) (mod p),

as required. If x = ±1, then, by (10), we have w_{⌊ \frac{p}{4} ⌋}(1) = 2⌊\frac{p}{4}⌋ + 1 ≡ (−1)^{(p−1)/2}/2 (mod p) and w_{⌊ \frac{3p}{4} ⌋}(1) = 2⌊\frac{3p}{4}⌋ + 1 ≡ (−1)^{(p+1)/2}/2 (mod p), which completes the proof of the lemma.

From Lemma 8 and Corollary 4 we immediately deduce the following result.

Theorem 9. Let p be a prime, p > 3, and let t ∈ D_{p}. Then

∑_{k=0}^{p−1} C_{2k} (\frac{1 − t^{2}}{16} )^{k} ≡ \frac{1}{2} (\frac{2}{p}) ((\frac{1 − t}{p}) (1 − t) + (\frac{1 + t}{p}) (1 + t)) (mod p),
∑_{k=0}^{p−1} \binom{4k}{2k}t^{2k} ≡ \frac{1}{2} ((\frac{1 − 4t}{p}) (1 + 2t) + (\frac{1 + 4t}{p}) (1 − 2t)) (mod p).

Proof. From Corollary 4 we have

∑_{k=0}^{p−1} C_{2k} (\frac{1 − t^{2}}{16} )^{k} ≡ \frac{1}{2} (\frac{−1}{p}) (3w_{⌊ \frac{p}{4} ⌋}(2t^{2} − 1) − w_{⌊ \frac{3p}{4} ⌋}(2t^{2} − 1)) (mod p)

and

∑_{k=0}^{p−1} \binom{4k}{2k}t^{2k} ≡ \frac{1}{4} (\frac{−2}{p}) (3w_{⌊ \frac{p}{4} ⌋}(32t^{2} − 1) + w_{⌊ \frac{3p}{4} ⌋}(32t^{2} − 1)) (mod p).

Now by Lemma 8 with x replaced by 4t for the last congruence, we conclude the proof. Theorem 10. Let p be a prime, p > 3, and let a, b ∈ ℤ, ab ≢ 0 (mod p), and a ≢ b (mod p). Then we have the following congruences modulo p:

∑_{k=0}^{p−1} C_{2k} \frac{(a − b)^{2k}}{(−64ab)^{k}} ≡ \begin{cases}\frac{(ab)^{\frac{p−1}{4}}}{2(a − b)} ((3a + b) (\frac{b}{p}) − (3b + a) (\frac{a}{p})) , & if p ≡ 1 (mod 4); \\ \frac{(ab)^{\frac{p+1}{4}}}{2(b − a)} (\frac{3a + b}{a} (\frac{b}{p}) − \frac{3b + a}{b} (\frac{a}{p})) , & if p ≡ 3 (mod 4), \end{cases}

∑_{k=0}^{p−1} \binom{4k}{2k}\frac{(a + b)^{2k}}{(64ab)^{k}} ≡ \begin{cases}\frac{(\frac{2}{p})(ab)^{\frac{p−1}{4}}}{4(a − b)} ((3a − b) (\frac{b}{p}) − (3b − a) (\frac{a}{p})) , & if p ≡ 1 (mod 4); \\ \frac{(\frac{2}{p})(ab)^{\frac{p+1}{4}}}{4(b − a)} (\frac{3a − b}{a} (\frac{b}{p}) − \frac{3b − a}{b} (\frac{a}{p})) , & if p ≡ 3 (mod 4). \end{cases}

Proof. By Corollary 4, we have

∑_{k=0}^{p−1} C_{2k} \frac{(a − b)^{2k}}{(−64ab)^{k}} ≡ \frac{1}{2} (\frac{−1}{p}) (3w_{⌊ \frac{p}{4} ⌋} (\frac{a^{2} + b^{2}}{2ab} ) − w_{⌊ \frac{3p}{4} ⌋} (\frac{a^{2} + b^{2}}{2ab} )) (mod p), (19)
∑_{k=0}^{p−1} \binom{4k}{2k}\frac{(a + b)^{2k}}{(64ab)^{k}} ≡ \frac{1}{4} (\frac{−2}{p}) (3w_{⌊ \frac{p}{4} ⌋} (\frac{a^{2} + b^{2}}{2ab} ) + w_{⌊ \frac{3p}{4} ⌋} (\frac{a^{2} + b^{2}}{2ab} )) (mod p). (20)

From (10) we obtain

w_{n} (\frac{a^{2} + b^{2}}{2ab} ) = \frac{a(a/b)^{n} − b(b/a)^{n}}{a − b} .

If p ≡ 1 (mod 4), then we have modulo p,

w_{⌊ \frac{p}{4} ⌋} (\frac{a^{2} + b^{2}}{2ab} ) = w_{\frac{p−1}{4}} (\frac{a^{2} + b^{2}}{2ab} ) = \frac{a(a/b)^{\frac{p−1}{4}} − b(b/a)^{\frac{p−1}{4}}}{a − b} ≡ \frac{a(\frac{b}{p}) − b(\frac{a}{p})}{a − b} (ab)^{\frac{p−1}{4}} ,

w_{⌊ \frac{3p}{4} ⌋} (\frac{a^{2} + b^{2}}{2ab} ) = \frac{a(a/b)^{\frac{3(p−1)}{4}} − b(b/a)^{\frac{3(p−1)}{4}}}{a − b} ≡ \frac{a(\frac{a}{p}) − b(\frac{b}{p})}{a − b} (ab)^{\frac{p−1}{4}} .

If p ≡ 3 (mod 4), then

w_{⌊ \frac{p}{4} ⌋} (\frac{a^{2} + b^{2}}{2ab} ) = w_{\frac{p−3}{4}} (\frac{a^{2} + b^{2}}{2ab} ) = \frac{a(a/b)^{\frac{p−3}{4}} − b(b/a)^{\frac{p−3}{4}}}{a − b} ≡ \frac{(\frac{b}{p}) − (\frac{a}{p})}{a − b} (ab)^{\frac{p+1}{4}} ,
w_{⌊ \frac{3p}{4} ⌋} (\frac{a^{2} + b^{2}}{2ab} ) = \frac{a(a/b)^{\frac{3p−1}{4}} − b(b/a)^{\frac{3p−1}{4}}}{a − b} ≡ \frac{\frac{a}{b}(\frac{a}{p}) − \frac{b}{a}(\frac{b}{p})}{a − b} (ab)^{\frac{p+1}{4}} .

Now substituting the above congruences in (19) and (20), we conclude the proof.

4 Congruences involving second-order Catalan numbers

In this section, we will deal with a particular case of Theorem 1 when m = 3. This case leads to congruences containing second-order Catalan numbers C_{n}^{(2)} (sequence \underline{A001764} in the OEIS [7]) and binomial coefficients \binom{3n}{n} (sequence \underline{A005809}).

Theorem 11. Let p be a prime greater than 3, and let t ∈ D_{p}. Then

∑_{k=0}^{⌊p/3⌋} C_{k}^{(2)}t^{k} ≡ 3 (\frac{p}{3}) w_{⌊ \frac{p}{3} ⌋}(1 − 27t/2) (mod p),  (21)

∑_{k=(p−1)/2}^{⌊2p/3⌋} C_{k}^{(2)}t^{k} ≡ − (\frac{p}{3}) (w_{⌊ \frac{2p}{3} ⌋}(1 − 27t/2) + w_{⌊ \frac{p}{3} ⌋}(1 − 27t/2)) (mod p),

∑_{k=0}^{⌊p/3⌋} \binom{3k}{k}t^{k} ≡ (\frac{p}{3}) w_{⌊ \frac{p}{3} ⌋}(27t/2 − 1) (mod p),  (22)

∑_{k=(p+1)/2}^{⌊2p/3⌋} \binom{3k}{k}t^{k} ≡ \frac{1}{3} (\frac{p}{3}) (w_{⌊ \frac{2p}{3} ⌋}(27t/2 − 1) − w_{⌊ \frac{p}{3} ⌋}(27t/2 − 1)) (mod p). (23)

Corollary 12. Let p be a prime greater than 3, and let t ∈ D_{p}. Then

∑_{k=0}^{p−1} C_{k}^{(2)}t^{k} ≡ (\frac{p}{3}) (2w_{⌊ \frac{p}{3} ⌋}(1 − 27t/2) − w_{⌊ \frac{2p}{3} ⌋}(1 − 27t/2)) (mod p),

∑_{k=0}^{p−1} \binom{3k}{k}t^{k} ≡ \frac{1}{3} (\frac{p}{3}) (2w_{⌊ \frac{p}{3} ⌋}(27t/2 − 1) + w_{⌊ \frac{2p}{3} ⌋}(27t/2 − 1)) (mod p).

Using the exact values of w_{n} from (17) and (18), we immediately get numerical congruences at the points t = 4/27, 2/27, 1/27, 1/9.

Corollary 13. Let p be a prime greater than 3. Then

∑_{k=0}^{p−1} C_{k}^{(2)} (\frac{4}{27})^{k} ≡ 1,  ∑_{k=0}^{⌊p/3⌋} C_{k}^{(2)} (\frac{4}{27})^{k} ≡ 3 (mod p),

∑_{k=0}^{p−1} \binom{3k}{k} (\frac{4}{27})^{k} ≡ \frac{1}{9},  ∑_{k=0}^{⌊p/3⌋} \binom{3k}{k} (\frac{4}{27})^{k} ≡ \frac{1}{3} (mod p),  (24)

∑_{k=0}^{p−1} C_{k}^{(2)} (\frac{2}{27})^{k} ≡ 2 (\frac{3}{p}) − 1,  ∑_{k=0}^{⌊p/3⌋} C_{k}^{(2)} (\frac{2}{27})^{k} ≡ 3 (\frac{3}{p}) (mod p),

∑_{k=0}^{p−1} \binom{3k}{k} (\frac{2}{27})^{k} ≡ \frac{2}{3} (\frac{3}{p}) + \frac{1}{3},  ∑_{k=0}^{⌊p/3⌋} \binom{3k}{k} (\frac{2}{27})^{k} ≡ (\frac{3}{p}) (mod p).  (25)

Remark 14. Z. W. Sun [12, Thm. 3.1] gave another proof of the first congruence in (24) based on third-order recurrences. Z. H. Sun [11, Rem. 3.1] proved the second congruence in (24) as well as the second congruence in (25) with the help of Lucas sequences.

Corollary 15. Let p be a prime, p > 3. Then

∑_{k=0}^{⌊p/3⌋} \frac{C_{k}^{(2)}}{27^{k}} ≡ \begin{cases}−6 (mod p), & if p ≡ ±4 (mod 9); \\ 3 (mod p), & otherwise, \end{cases}

∑_{k=0}^{p−1} \frac{C_{k}^{(2)}}{27^{k}} ≡ \begin{cases}1 (mod p), & if p ≡ ±1 (mod 9); \\ 4 (mod p), & if p ≡ ±2 (mod 9); \\ −5 (mod p), & if p ≡ ±4 (mod 9), \end{cases}

∑_{k=0}^{p−1} \binom{3k}{k}\frac{1}{27^{k}} ≡ \begin{cases}1 (mod p), & if p ≡ ±1 (mod 9); \\ −2/3 (mod p), & if p ≡ ±2 (mod 9); \\ −1/3 (mod p), & if p ≡ ±4 (mod 9). \end{cases}

Corollary 16. Let p be a prime, p > 3. Then

∑_{k=0}^{p−1} \frac{C_{k}^{(2)}}{9^{k}} ≡ \begin{cases}−2 (mod p), & if p ≡ ±2 (mod 9); \\ 1 (mod p), & otherwise, \end{cases}  (26)

∑_{k=0}^{⌊p/3⌋} \frac{C_{k}^{(2)}}{9^{k}} ≡ \begin{cases}3 (mod p), & if p ≡ ±1 (mod 9); \\ −3 (mod p), & if p ≡ ±2 (mod 9); \\ 0 (mod p), & if p ≡ ±4 (mod 9), \end{cases}

∑_{k=0}^{p−1} \binom{3k}{k}\frac{1}{9^{k}} ≡ \begin{cases}1 (mod p), & if p ≡ ±1 (mod 9); \\ 0 (mod p), & if p ≡ ±2 (mod 9); \\ −1 (mod p), & if p ≡ ±4 (mod 9). \end{cases}  (27)

Remark 17. Z. W. Sun [12, Thm. 1.5] provided another proof of congruences (26) and (27) by using cubic residues and third-order recurrences.

Lemma 18. For any x ≠ 1, −1/2, we have

w_{n}(4x^{3} − 3x) = \frac{α^{3n+2} − α^{−3n−1}}{α^{2} − α^{−1}} , where α = x + \sqrt{x^{2} − 1}.

Proof. Starting with w_{3n}(x), by (10), we get

w_{3n}(x) = \frac{(α + 1)α^{3n} − (α^{−1} + 1)α^{−3n}}{α − α^{−1}}

= \frac{(α^{3} + 1)α^{3n} − (α^{−3} + 1)α^{−3n}}{α^{3} − α^{−3}} · \frac{α^{3} − α^{−3}}{α − α^{−1}}

+ \frac{α^{3n+1} − α^{3n+3} − α^{−3n−1} + α^{−3n−3}}{α − α^{−1}}

= w_{n}(4x^{3} − 3x)(α^{2} + 1 + α^{−2})

+ \frac{α^{3n+1} − α^{3n+3} − α^{−3n−1} + α^{−3n−3}}{α − α^{−1}} .

Comparing the right and left-hand sides, we obtain

w_{n}(4x^{3} − 3x)(α^{2} + 1 + α^{−2}) = \frac{α^{3n} + α^{3n+3} − α^{−3n} − α^{−3n−3}}{α − α^{−1}} = \frac{(α^{3} + 1)(α^{3n} − α^{−3n−3})}{α − α^{−1}}

and therefore,

w_{n}(4x^{3} − 3x) = \frac{(α^{3} + 1)(α^{3n} − α^{−3n−3})}{α^{3} − α^{−3}} = \frac{α^{3}(α^{3n} − α^{−3n−3})}{α^{3} − 1} = \frac{α^{3n+2} − α^{−3n−1}}{α^{2} − α^{−1}} .

Lemma 19. Let p be a prime, p > 3, and let x ∈ D_{p}. Then

(2x + 1) · w_{⌊ \frac{p}{3} ⌋}(4x^{3} − 3x) ≡ (\frac{p}{3}) x + (\frac{x^{2} − 1}{p}) (x + 1) (mod p),
(2x + 1) · w_{⌊ \frac{2p}{3} ⌋}(4x^{3} − 3x) ≡ (\frac{p}{3}) (1 − 2x^{2}) + 2 (\frac{x^{2} − 1}{p}) x(x + 1) (mod p).

Proof. First we suppose that x ≢ 1, −1/2 (mod p). Then by Lemma 18, if p ≡ 1 (mod 3), we have

w_{⌊ \frac{p}{3} ⌋}(4x^{3} − 3x) = w_{\frac{p−1}{3}} (4x^{3} − 3x) = \frac{α^{p+1} − α^{−p}}{α^{2} − α^{−1}} .  (28)

For p-powers of α and α^{−1}, we easily obtain

α^{±p} = (x ± \sqrt{x^{2} − 1})^{p} ≡ x^{p} ± (\sqrt{x^{2} − 1})^{p} ≡ x ± \sqrt{x^{2} − 1}(x^{2} − 1)^{\frac{p−1}{2}}
≡ x ± (\frac{x^{2} − 1}{p}) \sqrt{x^{2} − 1} (mod p).  (29)

Substituting (29) into (28) and simplifying, we get

\begin{aligned}w_{⌊ \frac{p}{3} ⌋}(4x^{3} − 3x) & ≡ \frac{(x + \sqrt{x^{2} − 1})(x + (\frac{x^{2}−1}{p})\sqrt{x^{2} − 1}) − x + (\frac{x^{2}−1}{p})\sqrt{x^{2} − 1}}{(2x + 1)(x − 1 + \sqrt{x^{2} − 1})} \\ & ≡ \frac{x + (\frac{x^{2}−1}{p})(x + 1)}{2x + 1} (mod p). \end{aligned}

If p ≡ 2 (mod 3), then, by Lemma 18 and (29), we have

w_{⌊ \frac{p}{3} ⌋}(4x^{3} − 3x) = w_{\frac{p−2}{3}} (4x^{3} − 3x) = \frac{α^{p} − α^{1−p}}{α^{2} − α^{−1}}

≡ \frac{x + (\frac{x^{2}−1}{p})\sqrt{x^{2} − 1} − (x + \sqrt{x^{2} − 1})(x − (\frac{x^{2}−1}{p})\sqrt{x^{2} − 1})}{(2x + 1)(x − 1 + \sqrt{x^{2} − 1})}

≡ \frac{−x + (\frac{x^{2}−1}{p})(x + 1)}{2x + 1} (mod p)

and the first congruence of the lemma follows. Similarly, if p ≡ 1 (mod 3), then we have

\begin{aligned}w_{⌊ \frac{2p}{3} ⌋}(4x^{3} − 3x) & = w_{\frac{2(p−1)}{3}} (4x^{3} − 3x) = \frac{α^{2p} − α^{1−2p}}{α^{2} − α^{−1}} \\ & ≡ \frac{(x + (\frac{x^{2}−1}{p})\sqrt{x^{2} − 1})^{2} − (x + \sqrt{x^{2} − 1})(x − (\frac{x^{2}−1}{p})\sqrt{x^{2} − 1})^{2}}{(2x + 1)(x − 1 + \sqrt{x^{2} − 1})} (mod p). \end{aligned}

Simplifying, we easily find

w_{⌊ \frac{2p}{3} ⌋}(4x^{3} − 3x) ≡ \frac{1 − 2x^{2} + 2(\frac{x^{2}−1}{p})x(x + 1)}{2x + 1} (mod p).

If p ≡ 2 (mod 3), then

\begin{aligned}w_{⌊ \frac{2p}{3} ⌋}(4x^{3} − 3x) & = w_{\frac{2p−1}{3}} (4x^{3} − 3x) = \frac{α^{2p+1} − α^{−2p}}{α^{2} − α^{−1}} \\ & ≡ \frac{(x + \sqrt{x^{2} − 1})(x + (\frac{x^{2}−1}{p})\sqrt{x^{2} − 1})^{2} − (x − (\frac{x^{2}−1}{p})\sqrt{x^{2} − 1})^{2}}{(2x + 1)(x − 1 + \sqrt{x^{2} − 1})} (mod p), \end{aligned}

and after simplification we get

w_{⌊ \frac{2p}{3} ⌋}(4x^{3} − 3x) ≡ \frac{2x^{2} − 1 + 2(\frac{x^{2}−1}{p})x(x + 1)}{2x + 1} (mod p),

as desired.

Finally, if x ≡ 1 (mod p), then, by (10), we have 3w_{⌊ \frac{p}{3} ⌋}(1) = 3(2⌊p/3⌋ + 1) ≡ (\frac{p}{3}) (mod p) and 3w_{⌊ \frac{2p}{3} ⌋}(1) = 3(2⌊2p/3⌋ + 1) ≡ −(\frac{p}{3}) (mod p), which coincide with the right-hand sides of the required congruences when x ≡ 1 (mod p).

If x ≡ −1/2 (mod p), then the congruences become trivial and the proof is complete.

Lemma 20. Let p be a prime, p > 3, and let x ∈ D_{p}. Then we have modulo p,

(2x + 1) · w_{⌊ \frac{p}{6} ⌋}(4x^{3} − 3x) ≡ (\frac{2x − 2}{p}) x + (\frac{−6x − 6}{p}) (x + 1),
(2x + 1) · w_{⌊ \frac{5p}{6} ⌋}(4x^{3} − 3x) ≡(\frac{2x − 2}{p}) x(4x^{2} + 2x − 1)−(\frac{−6x − 6}{p}) (x + 1)(4x^{2} − 2x − 1).

Proof. First we suppose that x ≢ 1, −1/2 (mod p). If p ≡ 1 (mod 6), then, by Lemma 18, we have

w_{⌊ \frac{p}{6} ⌋}(4x^{3} − 3x) = w_{\frac{p−1}{6}} (4x^{3} − 3x) = \frac{α^{\frac{p+1}{2} +1} − α^{− \frac{p+1}{2}}}{α^{2} − α^{−1}} .

Substituting α = x + \sqrt{x^{2} − 1} = (\sqrt{(x + 1)/2} + \sqrt{(x − 1)/2})^{2}, we have

w_{⌊ \frac{p}{6} ⌋}(4x^{3} − 3x) = \frac{(x + \sqrt{x^{2} − 1})(\sqrt{x + 1} + \sqrt{x − 1})^{p+1} − (\sqrt{x + 1} − \sqrt{x − 1})^{p+1}}{2^{(p+1)/2}(α^{2} − α^{−1})}

= \frac{(x + \sqrt{x^{2} − 1})(\sqrt{x + 1} + \sqrt{x − 1})^{p} − 1/2(\sqrt{x + 1} − \sqrt{x − 1})^{p+2}}{2^{(p+1)/2}(2x + 1)\sqrt{x − 1}}

≡ \frac{(x + \sqrt{x^{2} − 1})((x + 1)^{\frac{p}{2}} + (x − 1)^{\frac{p}{2}} ) −(x − \sqrt{x^{2} − 1})((x + 1)^{\frac{p}{2}} −(x − 1)^{\frac{p}{2}} )}{2^{(p+1)/2}(2x + 1)\sqrt{x − 1}}

≡ \frac{x(x − 1)^{\frac{p−1}{2}} + (x + 1)^{\frac{p+1}{2}}}{2^{(p−1)/2}(2x + 1)} ≡ \frac{(\frac{2x−2}{p})x + (\frac{2x+2}{p})(x + 1)}{2x + 1} (mod p).

Since (\frac{−3}{p}) = (\frac{p}{3}) = 1, we get the desired congruence in this case.

If p ≡ 5 (mod 6), then we have

w_{⌊ \frac{p}{6} ⌋}(4x^{3} − 3x) = w_{\frac{p−5}{6}} (4x^{3} − 3x) = \frac{α^{\frac{p−1}{2}} − α^{− \frac{p−3}{2}}}{α^{2} − α^{−1}}

and therefore,

\begin{aligned}w_{⌊ \frac{p}{6} ⌋}(4x^{3} − 3x) & ≡ \frac{(x − \sqrt{x^{2} − 1})((x + 1)^{\frac{p}{2}} + (x − 1)^{\frac{p}{2}} ) −(x + \sqrt{x^{2} − 1})((x + 1)^{\frac{p}{2}} −(x − 1)^{\frac{p}{2}} )}{2^{(p+1)/2}(2x + 1)\sqrt{x − 1}} \\ & ≡ \frac{x(x − 1)^{\frac{p−1}{2}} − (x + 1)^{\frac{p+1}{2}}}{2^{(p−1)/2}(2x + 1)} ≡ \frac{(\frac{2x−2}{p})x − (\frac{2x+2}{p})(x + 1)}{2x + 1} (mod p), \end{aligned}

as desired in view of the fact that (\frac{−3}{p}) = (\frac{p}{3}) = −1.

The similar analysis can be applied for evaluating w_{⌊ \frac{5p}{6} ⌋}(4x^{3} − 3x) modulo p. If p ≡ 1 (mod 6), then

w_{⌊ \frac{5p}{6} ⌋}(4x^{3} − 3x) = w_{\frac{5(p−1)}{6}} (4x^{3} − 3x) = \frac{α^{\frac{5p−1}{2}} − α^{− \frac{5p−3}{2}}}{α^{2} − α^{−1}} .

Simplifying, we obtain

w_{⌊ \frac{5p}{6} ⌋}(4x^{3} − 3x) = \frac{(x − \sqrt{x^{2} − 1})(\sqrt{x + 1} + \sqrt{x − 1})^{5p}−(x + \sqrt{x^{2} − 1})(\sqrt{x + 1} − \sqrt{x − 1})^{5p}}{2^{(5p+1)/2}(2x + 1)\sqrt{x − 1}}

≡ \frac{(x − \sqrt{x^{2} − 1})((x + 1)^{\frac{p}{2}} + (x − 1)^{\frac{p}{2}} )^{5} − (x + \sqrt{x^{2} − 1})((x + 1)^{\frac{p}{2}} − (x − 1)^{\frac{p}{2}} )^{5}}{2^{(5p+1)/2}(2x + 1)\sqrt{x − 1}}

≡ \frac{(\frac{2x−2}{p})x(4x^{2} + 2x − 1) − (\frac{2x+2}{p})(x + 1)(4x^{2} − 2x − 1)}{2x + 1} (mod p),

as desired. If p ≡ 5 (mod 6), then

w_{⌊ \frac{5p}{6} ⌋}(4x^{3} − 3x) = w_{\frac{5p−1}{6}} (4x^{3} − 3x) = \frac{α^{\frac{5p+3}{2}} − α^{− \frac{5p+1}{2}}}{α^{2} − α^{−1}}

= \frac{(\sqrt{x + 1} + \sqrt{x − 1})^{5p+2} − (\sqrt{x + 1} − \sqrt{x − 1})^{5p+2}}{2^{(5p+3)/2}(2x + 1)\sqrt{x − 1}}

≡ \frac{(x + \sqrt{x^{2} − 1})((x + 1)^{\frac{p}{2}} + (x − 1)^{\frac{p}{2}} )^{5} − (x − \sqrt{x^{2} − 1})((x + 1)^{\frac{p}{2}} − (x − 1)^{\frac{p}{2}} )^{5}}{8 · 2^{(p−1)/2}(2x + 1)\sqrt{x − 1}}

≡ \frac{(\frac{2x−2}{p})x(4x^{2} + 2x − 1) + (\frac{2x+2}{p})(x + 1)(4x^{2} − 2x − 1)}{2x + 1} (mod p),

and the congruence is true. If x ≡ 1 (mod 6), then, by (10), we have 3w_{⌊ \frac{p}{6} ⌋}(1) = 3(2⌊p/6⌋ + 1) = 2(\frac{p}{3}) (mod p) and 3w_{⌊ \frac{5p}{6} ⌋}(1) = 3(2⌊5p/6⌋+1) = −2(\frac{p}{3}) (mod p), which prove the lemma in this case too. Finally, if x ≡ −1/2 (mod p), we get the trivial congruences 0 ≡ 0, and the proof is complete.

Theorem 21. Let p be a prime, p > 3, and let t ∈ D_{p}.

If t ≢ 0 (mod p), then

∑_{k=1}^{⌊p/3⌋} C_{k}^{(2)}(t^{2}(t + 1))^{k} ≡ \frac{1 + t}{2t} − \frac{1 − 3t}{2t} (\frac{(1 + t)(1 − 3t)}{p}) (mod p),

∑_{k=1}^{p−1} C_{k}^{(2)}(t^{2}(t + 1))^{k} ≡ \frac{(1 + t)(1 − 3t)}{2t} (1 − (\frac{(1 + t)(1 − 3t)}{p})) (mod p).  (30)

If 3t + 2 ≢ 0 (mod p), then

∑_{k=1}^{⌊p/3⌋} \binom{3k}{k}(t^{2}(t + 1))^{k} ≡ \frac{3(t + 1)}{2(3t + 2)} ((\frac{(1 + t)(1 − 3t)}{p}) − 1) (mod p),  (31)

∑_{k=1}^{p−1} \binom{3k}{k}(t^{2}(t + 1))^{k} ≡ \frac{3(t + 1)^{2}}{2(3t + 2)} ((\frac{(1 + t)(1 − 3t)}{p}) − 1) (mod p).  (32)

Proof. From (21), Corollary 12 and Lemma 19 we have modulo p,

∑_{k=0}^{⌊p/3⌋} C_{k}^{(2)} (\frac{2(1 − x)(2x + 1)^{2}}{27} )^{k} ≡ 3 (\frac{p}{3}) w_{⌊ \frac{p}{3} ⌋}(4x^{3} − 3x) ≡ \frac{3x}{2x + 1} + \frac{3x + 3}{2x + 1} (\frac{3 − 3x^{2}}{p} )

and

\begin{aligned}∑_{k=0}^{p−1} C_{k}^{(2)} (\frac{2(1 − x)(2x + 1)^{2}}{27} )^{k} & ≡ (\frac{p}{3}) (2w_{⌊ \frac{p}{3} ⌋}(4x^{3} − 3x) − w_{⌊ \frac{2p}{3} ⌋}(4x^{3} − 3x)) \\ & ≡ \frac{2x^{2} + 2x − 1}{2x + 1} + \frac{2(1 − x^{2})}{2x + 1} (\frac{3 − 3x^{2}}{p} ) (mod p) \end{aligned}

for any x ∈ D_{p} such that 2x + 1 ≢ 0 (mod p). Replacing x by (−1 − 3t)/2 with t ≢ 0 (mod p), we get the first two congruences of the theorem.

Similarly, from (22), Corollary 12 and Lemma 19 for any x ∈ D_{p} with 2x+1 ≢ 0 (mod p), we have

∑_{k=0}^{⌊p/3⌋} \binom{3k}{k}(\frac{2(x + 1)(2x − 1)^{2}}{27} )^{k}≡ (\frac{p}{3}) w_{⌊ \frac{p}{3} ⌋}(4x^{3} − 3x) ≡ \frac{x}{2x + 1} + \frac{x + 1}{2x + 1} (\frac{3 − 3x^{2}}{p} ) ,

∑_{k=0}^{p−1} \binom{3k}{k}(\frac{2(x + 1)(2x − 1)^{2}}{27} )^{k}≡ \frac{1}{3} (\frac{p}{3}) (2w_{⌊ \frac{p}{3} ⌋}(4x^{3} − 3x) + w_{⌊ \frac{2p}{3} ⌋}(4x^{3} − 3x))

≡ \frac{1 + 2x − 2x^{2}}{3(2x + 1)} + \frac{2(x + 1)^{2}}{3(2x + 1)} (\frac{3 − 3x^{2}}{p} ) (mod p).

This implies that

∑_{k=1}^{⌊p/3⌋} \binom{3k}{k} (\frac{2(x + 1)(2x − 1)^{2}}{27} )^{k}≡ \frac{x + 1}{2x + 1} ((\frac{3 − 3x^{2}}{p} ) − 1) (mod p),

∑_{k=1}^{p−1} \binom{3k}{k} (\frac{2(x + 1)(2x − 1)^{2}}{27} )^{k}≡ \frac{2(x + 1)^{2}}{3(2x + 1)} ((\frac{3 − 3x^{2}}{p} ) − 1) (mod p).

Replacing x by (3t + 1)/2, we derive the other two congruences of the theorem.

Remark 22. Note that Z. H. Sun [9, Thm. 2.3] proved congruence (31) by another method using cubic congruences. If we put t = −c/(c + 1) in (30) and (32), we recover corresponding congruences of Z. W. Sun [12, Thm. 1.1] proved by applying properties of third-order recurrences.

5 Cubic residues and non-residues and their application to congruences

We begin with a brief review of basic facts from the theory of cubic residues that will be needed later in this section. Let ω = e^{2πi/3} = (−1 + i\sqrt{3})/2. We consider the ring of the Eisenstein integers ℤ[ω] = {a + bω : a, b ∈ ℤ}. To define the cubic residue symbol, we recall arithmetic properties of the ring ℤ[ω] including description of its units and primes [4, Chapter 9].

If α = a + bω ∈ ℤ[ω], the norm of α is defined by the formula N(α) = α\overline{α} = a^{2} − ab + b^{2}, where \overline{α} = a + b\overline{ω} = a + bω^{2} = (a − b) − bω is the complex conjugate of α. Note that the norm is a nonnegative integer always congruent to 0 or 1 modulo 3. It is well known that ℤ[ω] is a unique factorization domain. The units of ℤ[ω] are ±1, ±ω, ±ω^{2}.

Let p be a prime in ℤ, then p in ℤ[ω] falls into three categories [1, Prop. 4.7]: (i) if p = 3, then 3 = −ω^{2}(1 − ω)^{2}, where 1 − ω is prime in ℤ[ω] and N(1 − ω) = (1 − ω)(1 − ω^{2}) = 3; (ii) if p ≡ 2 (mod 3), then p remains prime in ℤ[ω] and N(p) = p^{2}; (iii) if p ≡ 1 (mod 3), then p splits into the product of two conjugate non-associate primes in ℤ[ω], p = π\overline{π} and N(π) = π\overline{π} = p. Moreover, every prime in ℤ[ω] is associated with one of the primes listed in (i) − (iii).

An analog of Fermat’s little theorem is true in ℤ[ω]: if π is a prime and π ∤ α, then

α^{N(π)−1} ≡ 1 (mod π).

Note that if π is a prime such that N(π) ≠ 3, then N(π) ≡ 1 (mod 3) and the expression α^{\frac{N(π)−1}{3}} is well defined in ℤ[ω], i.e., α^{\frac{N(π)−1}{3}} ≡ ω^{j} (mod π) for a unique unit ω^{j}. This leads to the definition of the cubic residue character of α modulo π [4, p. 112]:

(\frac{α}{π})_{3} = \begin{cases}0, & if π|α; \\ ω^{j}, & if α^{\frac{N(π)−1}{3}} ≡ ω^{j} (mod π). \end{cases}  (33)

The cubic residue character has formal properties similar to those of the Legendre symbol [4, Prop. 9.3.3]:

(i) The congruence x^{3} ≡ α (mod π) is solvable in ℤ[ω] if and only if (\frac{α}{π})_{3} = 1, i.e., iff α is a cubic residue modulo π;

(ii) (\frac{αβ}{π} )_{3} = (\frac{α}{π})_{3} ( \frac{β}{π})_{3};

(iii) \overline{(\frac{α}{π})}_{3} = (\frac{\overline{α}}{\overline{π}})_{3};

(iv) If π and θ are associates, then (\frac{α}{π})_{3} = (\frac{α}{θ})_{3};

(v) If α ≡ β (mod π), then (\frac{α}{π})_{3} = (\frac{β}{π})_{3}.

Let π = a + bω ∈ ℤ[ω]. We say that π is primary if π ≡ 2 (mod 3), that is equivalent to a ≡ 2 (mod 3) and b ≡ 0 (mod 3). If π ∈ ℤ[ω], N(π) > 1 and π ≡ ±2 (mod 3), we may decompose π = ±π_{1} … π_{r}, where π, …, π_{r} are primary primes [4, p. 135]. For α ∈ ℤ[ω], the cubic Jacobi symbol (\frac{α}{π})_{3} is defined by

(\frac{α}{π})_{3} = (\frac{α}{π_{1}} )_{3} … (\frac{α}{π_{r}} )_{3} .

Now let p be a prime. We define a cubic residue modulo p in ℤ. We say that m ∈ ℤ is a cubic residue modulo p if the congruence x^{3} ≡ m (mod p) has an integer solution, otherwise m is called a cubic non-residue modulo p. If p = 3, then by Fermat’s little theorem, m^{3} ≡ m (mod 3) for all integers m, so x^{3} ≡ m (mod 3) always has a solution. If p ≡ 2 (mod 3), then every integer m is a cubic residue modulo p. Indeed, we have 2p − 1 ≡ 0 (mod 3) and by Fermat’s little theorem, m ≡ m^{2p−1} = (m^{\frac{2p−1}{3}} )^{3} (mod p). So the only interesting case which remains is when a prime p ≡ 1 (mod 3).

If a prime p ≡ 1 (mod 3), then it is well known that there are unique integers L and |M| such that 4p = L^{2} + 27M^{2} with L ≡ 1 (mod 3). In this case, p splits into the product of primes of ℤ[ω], p = π\overline{π}, where we can write π in the form

π = \frac{1}{2}(L + 3M\sqrt{−3}) = \frac{L + 3M}{2} + 3Mω.

It is easy to see that (\frac{L}{3M} )^{2} ≡ −3 (mod p) and therefore for any integer m coprime to p by Euler’s criterion [5, 15], we have one of the three possibilities

m^{(p−1)/3} ≡ 1, (−1 − L/(3M))/2 or (−1 + L/(3M))/2 (mod p).

Moreover, m^{(p−1)/3} ≡ 1 (mod p) if and only if m is a cubic residue modulo p. When m is a prime and a cubic non-residue modulo p, Williams [15] found a method how to choose the sign of M so that m^{(p−1)/3} ≡ (−1 − L/(3M))/2 (mod p). To classify cubic residues and non-residues in ℤ, Sun [8] introduced three subsets

C_{j}(m) = \{c ∈ D_{m} |(\frac{c + 1 + 2ω}{m} )_{3} = ω^{j} \} , j = 0, 1, 2, m ∈ ℕ, m ≢ 0 (mod 3),

of D_{m}, which posses the following properties:

(i) C_{0}(m) ∪ C_{1}(m) ∪ C_{2}(m) = {c ∈ D_{m} | gcd(c^{2} + 3, m) = 1}; (ii) c ∈ C_{0}(m) if and only if − c ∈ C_{0}(m); (iii) c ∈ C_{1}(m) if and only if − c ∈ C_{2}(m); (iv) If c, c′ ∈ D_{m} and cc′ ≡ −3 (mod m), then c ∈ C_{j}(m) if and only if c′ ∈ C_{j}(m). Using these sets, Z. H. Sun proved the following criterion of cubic residuacity in ℤ: Let

p be a prime of the form p ≡ 1 (mod 3) and hence 4p = L^{2} + 27M^{2} for some L, M ∈ ℤ and L ≡ 1 (mod 3). If q is a prime with q|M, then q^{(p−1)/3} ≡ 1 (mod p). If q ∤ M and j ∈ {0, 1, 2}, then

q^{(p−1)/3} ≡ ((−1 − L/(3M))/2)^{j} (mod p) if and only if L/(3M) ∈ C_{j}(q).  (34)

Sun [9] gave a simple criterion in terms of values of the sum ∑_{k=1}^{⌊p/3⌋} \binom{3k}{k} (\frac{4}{9(c^{2}+3)})^{k} modulo a

prime p for c ∈ C_{j}(p) and conjectured a similar criterion in terms of the sum ∑_{k=(p+1)/2}^{⌊2p/3⌋} \binom{3k}{k}t^{k}.

In this section, using our formulas from Theorem 11, we address this question of Sun (see Theorem 25 below). First, we will need the following statement.

Lemma 23. ([8, Lemma 2.2]) Let p be a prime, p ≠ 3, and let c ∈ D_{p}.

(i) If p ≡ 1 (mod 3) and so p splits into the product of primes, p = π\overline{π} with π ∈ ℤ[ω] and π ≡ 2 (mod 3), then

\begin{aligned}(\frac{c + 1 + 2ω}{p} )_{3} & = (\frac{(c^{2} + 3)(c − 1 − 2ω)}{π})_{3} , \\ (\frac{c − 1 − 2ω}{p} )_{3} & = (\frac{c + 1 + 2ω}{p} )_{3}^{−1} = (\frac{(c^{2} + 3)(c + 1 + 2ω)}{π})_{3} . \end{aligned}

(ii) If p ≡ 2 (mod 3), then

\begin{aligned}(\frac{c + 1 + 2ω}{p} )_{3} & ≡ (c^{2} + 3)^{(p−2)/3}(c + 1 + 2ω)^{(p+1)/3} (mod p), \\ (\frac{c − 1 − 2ω}{p} )_{3} & = (\frac{c + 1 + 2ω}{p} )_{3}^{−1} ≡ (c^{2} + 3)^{(p−2)/3}(c − 1 − 2ω)^{(p+1)/3} (mod p). \end{aligned}

Now we prove the following criterion.

Theorem 24. Let p be a prime, p > 3, and let c ∈ D_{p} with c^{2} ≢ −3 (mod p). Then

c ∑_{k=(p+1)/2}^{⌊2p/3⌋} \binom{3k}{k} (\frac{4}{9(c^{2} + 3)})^{k} ≡ \begin{cases}0 (mod p), & if c ∈ C_{0}(p); \\ 1 (mod p), & if c ∈ C_{1}(p); \\ −1 (mod p), & if c ∈ C_{2}(p). \end{cases}

Proof. By (23), we have

∑_{k=(p+1)/2}^{⌊2p/3⌋} \binom{3k}{k} (\frac{4}{9(c^{2} + 3)})^{k} ≡ \frac{1}{3} (\frac{p}{3}) (w_{⌊ \frac{2p}{3} ⌋}(\frac{3 − c^{2}}{3 + c^{2}} ) − w_{⌊ \frac{p}{3} ⌋}(\frac{3 − c^{2}}{3 + c^{2}} )) (mod p).

(35)

From (10) it easily follows that

w_{n}(\frac{3 − c^{2}}{3 + c^{2}} ) = \frac{(−1)^{n}}{2c(c^{2} + 3)^{n}} ((c − 1 − 2ω)^{2n+1} + (c + 1 + 2ω)^{2n+1}) .  (36)

If p ≡ 1 (mod 3), then p splits into the product of primes in ℤ[ω], p = π\overline{π} with π ≡ 2 (mod 3) and, by (36), we have

w_{⌊ \frac{p}{3} ⌋}(\frac{3 − c^{2}}{3 + c^{2}} ) = \frac{1}{2c(c^{2} + 3)^{(p−1)/3}} ((c − 1 − 2ω)^{2(p−1)/3+1} + (c + 1 + 2ω)^{2(p−1)/3+1}) . (37)

By (33) and Lemma 23, we have

(c^{2} + 3)^{2(p−1)/3}(c − 1 − 2ω)^{2(p−1)/3} ≡ (\frac{(c^{2} + 3)(c − 1 − 2ω)}{π})^{2} = (\frac{c + 1 + 2ω}{p} )_{3}^{2} (mod π)

(38)

and

(c^{2} + 3)^{2(p−1)/3}(c + 1 + 2ω)^{2(p−1)/3} ≡ (\frac{(c^{2} + 3)(c + 1 + 2ω)}{π})^{2} = (\frac{c + 1 + 2ω}{p} )_{3}^{−2} (mod π).

(39)

Substituting (38) and (39) into (37), we get

w_{⌊ \frac{p}{3} ⌋}(\frac{3 − c^{2}}{3 + c^{2}} )≡ \frac{1}{2c} ((c − 1 − 2ω)(\frac{c + 1 + 2ω}{p} )_{3}^{2}+ (c + 1 + 2ω) (\frac{c + 1 + 2ω}{p} )_{3}^{−2}) (mod π)

and therefore,

w_{⌊ \frac{p}{3} ⌋}(\frac{3 − c^{2}}{3 + c^{2}} ) ≡ \begin{cases}1 (mod π), & if c ∈ C_{0}(p); \\ −\frac{3+c}{2c} (mod π), & if c ∈ C_{1}(p); \\ \frac{3−c}{2c} (mod π), & if c ∈ C_{2}(p). \end{cases}  (40)

Since both sides of the above congruence are rational, the congruence is also true modulo p = π\overline{π}. Similarly, if p ≡ 2 (mod 3), then

w_{⌊ \frac{p}{3} ⌋}(\frac{3 − c^{2}}{3 + c^{2}} ) = \frac{−1}{2c(c^{2} + 3)^{(p−2)/3}} ((c − 1 − 2ω)^{2(p+1)/3−1} + (c + 1 + 2ω)^{2(p+1)/3−1}) . (41)

Now, by Lemma 23, we have

(c + 1 + 2ω)^{2(p+1)/3} ≡ (c^{2} + 3)^{−2(p−2)/3} (\frac{c + 1 + 2ω}{p} )_{3}^{2} (mod p)

and

(c − 1 − 2ω)^{2(p+1)/3} ≡ (c^{2} + 3)^{−2(p−2)/3} (\frac{c + 1 + 2ω}{p} )_{3}^{−2} (mod p).

Substituting the above congruences into (41) and noticing that c^{2}+3 = (c+1+2ω)(c−1−2ω), we get

w_{⌊ \frac{p}{3} ⌋}(\frac{3 − c^{2}}{3 + c^{2}} ) ≡ \frac{−1}{2c} ((c + 1 + 2ω) (\frac{c + 1 + 2ω}{p} )_{3}^{−2} + (c − 1 − 2ω) (\frac{c + 1 + 2ω}{p} )_{3}^{2})

and therefore,

w_{⌊ \frac{p}{3} ⌋}(\frac{3 − c^{2}}{3 + c^{2}} ) ≡ \begin{cases}−1 (mod p), & if c ∈ C_{0}(p); \\ \frac{3+c}{2c} (mod p), & if c ∈ C_{1}(p); \\ −\frac{3−c}{2c} (mod p), & if c ∈ C_{2}(p). \end{cases}  (42)

Combining congruences (40) and (42), we obtain that for all primes p > 3,

(\frac{p}{3}) w_{⌊ \frac{p}{3} ⌋}(\frac{3 − c^{2}}{3 + c^{2}} ) ≡ \begin{cases}1 (mod p), & if c ∈ C_{0}(p); \\ −\frac{3+c}{2c} (mod p), & if c ∈ C_{1}(p); \\ \frac{3−c}{2c} (mod p), & if c ∈ C_{2}(p). \end{cases}  (43)

Applying the similar argument for evaluation of w_{⌊ \frac{2p}{3} ⌋}(\frac{3−c^{2}}{3+c^{2}} ), we see that if p ≡ 1 (mod 3), then 2p ≡ 2 (mod 3) and therefore,

\begin{aligned}w_{⌊ \frac{2p}{3} ⌋} & (\frac{3 − c^{2}}{3 + c^{2}} ) = \frac{1}{2c(c^{2} + 3)^{2(p−1)/3}} ((c − 1 − 2ω)^{4(p−1)/3+1} + (c + 1 + 2ω)^{4(p−1)/3+1}) \\ & ≡ \frac{1}{2c} ((c − 1 − 2ω) (\frac{c + 1 + 2ω}{p} )_{3} + (c + 1 + 2ω) (\frac{c + 1 + 2ω}{p} )_{3}^{−1}) (mod π), \end{aligned}

which implies

w_{⌊ \frac{2p}{3} ⌋}(\frac{3 − c^{2}}{3 + c^{2}} ) ≡ \begin{cases}1 (mod p), & if c ∈ C_{0}(p); \\ \frac{3−c}{2c} (mod p), & if c ∈ C_{1}(p); \\ −\frac{3+c}{2c} (mod p), & if c ∈ C_{2}(p). \end{cases}  (44)

If p ≡ 2 (mod 3), then 2p ≡ 1 (mod 3) and we have

\begin{aligned}w_{⌊ \frac{2p}{3} ⌋} & (\frac{3 − c^{2}}{3 + c^{2}} ) = \frac{−1}{2c(c^{2} + 3)^{(2p−1)/3}} ((c + 1 + 2ω)^{4(p+1)/3−1} + (c − 1 − 2ω)^{4(p+1)/3−1}) \\ & ≡ \frac{−1}{2c} ((c − 1 − 2ω) (\frac{c + 1 + 2ω}{p} )_{3} + (c + 1 + 2ω) (\frac{c + 1 + 2ω}{p} )_{3}^{−1}) (mod p), \end{aligned}

and therefore,

w_{⌊ \frac{2p}{3} ⌋}(\frac{3 − c^{2}}{3 + c^{2}} ) ≡ \begin{cases}−1 (mod p), & if c ∈ C_{0}(p); \\ −\frac{3−c}{2c} (mod p), & if c ∈ C_{1}(p); \\ \frac{3+c}{2c} (mod p), & if c ∈ C_{2}(p). \end{cases}  (45)

Combining (44) and (45), we see that for all primes p > 3,

(\frac{p}{3}) w_{⌊ \frac{2p}{3} ⌋}(\frac{3 − c^{2}}{3 + c^{2}} ) ≡ \begin{cases}1 (mod p), & if c ∈ C_{0}(p); \\ \frac{3−c}{2c} (mod p), & if c ∈ C_{1}(p); \\ −\frac{3+c}{2c} (mod p), & if c ∈ C_{2}(p). \end{cases}  (46)

Now, by (43), (46) and (35), the congruence of the theorem easily follows.

From Theorem 24 and criterion (34) we deduce the following result confirming a question of Z. H. Sun [9, Conj. 2.1].

Theorem 25. Let q be a prime, q ≡ 1 (mod 3) and so 4q = L^{2} + 27M^{2} with L, M ∈ ℤ and L ≡ 1 (mod 3). Let p be a prime with p ≠ 2, 3, q, and let p ∤ LM. Then

∑_{k=(p+1)/2}^{⌊2p/3⌋} \binom{3k}{k}\frac{M^{2k}}{q^{k}} ≡ \begin{cases}0 (mod p), & if p^{\frac{q−1}{3}} ≡ 1 (mod q); \\ ±\frac{3M}{L} (mod p), & if p^{\frac{q−1}{3}} ≡ \frac{−1±9M/L}{2} (mod q) \end{cases}

and

∑_{k=(p+1)/2}^{⌊2p/3⌋} \binom{3k}{k}\frac{L^{2k}}{(27q)^{k}} ≡ \begin{cases}0 (mod p), & if p^{\frac{q−1}{3}} ≡ 1 (mod q); \\ ±\frac{L}{9M} (mod p), & if p^{\frac{q−1}{3}} ≡ \frac{−1±L/(3M)}{2} (mod q). \end{cases}

Proof. To prove the first congruence, we put c = \frac{L}{3M} in Theorem 24. Then c(c^{2} + 3) ≢ 0 (mod p), \frac{4}{9(c^{2}+3)} = \frac{M^{2}}{q} and we have

∑_{k=(p+1)/2}^{⌊2p/3⌋} \binom{3k}{k}\frac{M^{2k}}{q^{k}} ≡ \begin{cases}0 (mod p), & if L/(3M) ∈ C_{0}(p); \\ \frac{3M}{L} (mod p), & if L/(3M) ∈ C_{1}(p); \\ −\frac{3M}{L} (mod p), & if L/(3M) ∈ C_{2}(p). \end{cases}

Now applying (34) and taking into account that L/(3M) ≡ −9M/L (mod q), we get the result.

To prove the second congruence, we put c = −9M/L in Theorem 24. Then c(c^{2} + 3) ≢ 0 (mod p), \frac{4}{9(c^{2}+3)} = \frac{L^{2}}{27q} and we have

∑_{k=(p+1)/2}^{⌊2p/3⌋} \binom{3k}{k}\frac{L^{2k}}{(27q)^{k}} ≡ \begin{cases}0 (mod p), & if −9M/L ∈ C_{0}(p); \\ −\frac{L}{9M} (mod p), & if −9M/L ∈ C_{1}(p); \\ \frac{L}{9M} (mod p), & if −9M/L ∈ C_{2}(p). \end{cases}  (47)

By (iv), we know that −9M/L ∈ C_{j}(p) if and only if L/(3M) ∈ C_{j}(p). This together with (47) and (34) implies the required congruence.

From Corollary 12 and formulas (43) and (46) we get the following statement.

Theorem 26. Let p be a prime, p > 3, and let c ∈ D_{p} with c^{2} ≢ −3 (mod p). Then

∑_{k=0}^{p−1} \binom{3k}{k} (\frac{4}{9(c^{2} + 3)})^{k} ≡ \begin{cases}1 (mod p), & if c ∈ C_{0}(p); \\ −\frac{1+c}{2c} (mod p), & if c ∈ C_{1}(p); \\ \frac{1−c}{2c} (mod p), & if c ∈ C_{2}(p) \end{cases}

and

∑_{k=0}^{p−1} C_{k}^{(2)} (\frac{4c^{2}}{27(c^{2} + 3)})^{k} ≡ \begin{cases}1 (mod p), & if c ∈ C_{0}(p); \\ −\frac{9+c}{2c} (mod p), & if c ∈ C_{1}(p); \\ \frac{9−c}{2c} (mod p), & if c ∈ C_{2}(p). \end{cases}

From Theorem 26 and criterion (34) we get the following congruences.

Theorem 27. Let q be a prime, q ≡ 1 (mod 3) and so 4q = L^{2} + 27M^{2} with L, M ∈ ℤ and L ≡ 1 (mod 3). Let p be a prime with p ≠ 2, 3, q, and let p ∤ LM. Then

∑_{k=0}^{p−1} \binom{3k}{k}\frac{M^{2k}}{q^{k}} ≡ \begin{cases}1 (mod p), & if p^{\frac{q−1}{3}} ≡ 1 (mod q); \\ \frac{±3M−L}{2L} (mod p), & if p^{\frac{q−1}{3}} ≡ \frac{−1±L/(3M)}{2} (mod q), \end{cases}

∑_{k=0}^{p−1} \binom{3k}{k}\frac{L^{2k}}{(27q)^{k}} ≡ \begin{cases}1 (mod p), & if p^{\frac{q−1}{3}} ≡ 1 (mod q); \\ \frac{±L−9M}{18M} (mod p), & if p^{\frac{q−1}{3}} ≡ \frac{−1±9M/L}{2} (mod q) \end{cases}

and

∑_{k=0}^{p−1} C_{k}^{(2)} \frac{M^{2k}}{q^{k}} ≡ \begin{cases}1 (mod p), & if p^{\frac{q−1}{3}} ≡ 1 (mod q); \\ \frac{±L−M}{2M} (mod p), & if p^{\frac{q−1}{3}} ≡ \frac{−1±9M/L}{2} (mod q), \end{cases}

∑_{k=0}^{p−1} C_{k}^{(2)} \frac{L^{2k}}{(27q)^{k}} ≡ \begin{cases}1 (mod p), & if p^{\frac{q−1}{3}} ≡ 1 (mod q); \\ \frac{±27M−L}{2L} (mod p), & if p^{\frac{q−1}{3}} ≡ \frac{−1±L/(3M)}{2} (mod q). \end{cases}

Proof. Substituting consequently c = L/(3M) and then c = −9M/L in Theorem 26 and following the same line of reasoning as in the proof of Theorem 25, we get the above congruences.

In particular, setting q = 7, 19, 31, 37 in Theorem 27, we get the following numerical congruences.

Corollary 28. Let p be a prime, p ≠ 2, 3, 7. Then

∑_{k=0}^{p−1} \binom{3k}{k}\frac{1}{189^{k}} ≡ \begin{cases}−2 (mod p), & if p ≡ ±2 (mod 7); \\ 1 (mod p), & otherwise, \end{cases}
∑_{k=0}^{p−1} \frac{C_{k}^{(2)}}{189^{k}} ≡ \begin{cases}1 (mod p), & if p ≡ ±1 (mod 7); \\ −14 (mod p), & if p ≡ ±2 (mod 7); \\ 13 (mod p), & if p ≡ ±3 (mod 7). \end{cases}

Corollary 29. Let p be a prime, p ≠ 2, 3, 7, 19. Then

∑_{k=0}^{p−1} \binom{3k}{k}\frac{1}{19^{k}} ≡ \begin{cases}1 (mod p), & if p ≡ ±1, ±7, ±8 (mod 19); \\ −2/7 (mod p), & if p ≡ ±2, ±3, ±5 (mod 19); \\ −5/7 (mod p), & if p ≡ ±4, ±6, ±9 (mod 19). \end{cases}

Corollary 30. Let p be a prime, p ≠ 2, 3, 31. Then

∑_{k=0}^{p−1} \binom{3k}{k} (\frac{4}{31})^{k} ≡ \begin{cases}1 (mod p), & if p ≡ ±1, ±2, ±4, ±8, ±15 (mod 31); \\ −5/4 (mod p), & if p ≡ ±3, ±6, ±7, ±12, ±14 (mod 31); \\ 1/4 (mod p), & if p ≡ ±5, ±9, ±10, ±11, ±13 (mod 31). \end{cases}

Corollary 31. Let p be a prime, p ≠ 2, 3, 11, 37. Then

∑_{k=0}^{p−1} \binom{3k}{k}\frac{1}{37^{k}} ≡ \begin{cases}1 (mod p), & if p ≡ ±1, ±6, ±8, ±10, ±11, ±14, (mod 37); \\ −4/11 (mod p), & if p ≡ ±2, ±9, ±12, ±15, ±16, ±17 (mod 37); \\ −7/11 (mod p), & if p ≡ ±3, ±4, ±5, ±7, ±13, ±18 (mod 37), \end{cases}
∑_{k=0}^{p−1} \frac{C_{k}^{(2)}}{37^{k}} ≡ \begin{cases}1 (mod p), & if p ≡ ±1, ±6, ±8, ±10, ±11, ±14, (mod 37); \\ −6 (mod p), & if p ≡ ±2, ±9, ±12, ±15, ±16, ±17 (mod 37); \\ 5 (mod p), & if p ≡ ±3, ±4, ±5, ±7, ±13, ±18 (mod 37). \end{cases}

6 Polynomial congruences involving S_{n}

In this section, we will deal with a particular case of Theorem 1 when m = 6. In this case, we get polynomial congruences containing the sequence S_{k} (OEIS \underline{A176898}) and also (2k +1)S_{k}.

Theorem 32. Let p be a prime greater than 3, and let t ∈ D_{p}. Then

∑_{k=0}^{⌊p/6⌋} S_{k}t^{k} ≡ \frac{3}{4} (\frac{p}{3}) w_{⌊ \frac{p}{6} ⌋}(1 − 216t) (mod p),  (48)

∑_{k=(p−1)/2}^{⌊5p/6⌋} S_{k}t^{k} ≡ −\frac{1}{8} (\frac{p}{3}) (w_{⌊ \frac{5p}{6} ⌋}(1 − 216t) + w_{⌊ \frac{p}{6} ⌋}(1 − 216t)) (mod p),

∑_{k=0}^{⌊p/6⌋}(2k + 1)S_{k}t^{k} ≡ \frac{1}{2}(−1)^{\frac{p−1}{2}} w_{⌊ \frac{p}{6} ⌋}(216t − 1) (mod p),  (49)

∑_{k=(p+1)/2}^{⌊5p/6⌋}(2k + 1)S_{k}t^{k} ≡ \frac{1}{12}(−1)^{\frac{p−1}{2}} (w_{⌊ \frac{5p}{6} ⌋}(216t − 1) − w_{⌊ \frac{p}{6} ⌋}(216t − 1)) (mod p).

Corollary 33. Let p be a prime greater than 3, and let t ∈ D_{p}. Then

∑_{k=0}^{p−1} S_{k}t^{k} ≡ \frac{1}{8} (\frac{p}{3}) (5w_{⌊ \frac{p}{6} ⌋}(1 − 216t) − w_{⌊ \frac{5p}{6} ⌋}(1 − 216t)) (mod p),

∑_{k=0}^{p−1}(2k + 1)S_{k}t^{k} ≡ \frac{1}{12}(−1)^{\frac{p−1}{2}} (w_{⌊ \frac{5p}{6} ⌋}(216t − 1) + 5w_{⌊ \frac{p}{6} ⌋}(216t − 1)) (mod p).

Taking into account (17), we get the following explicit congruences. Note that the first congruence below confirms a conjecture of Z. W. Sun [13, Conj. 2].

Corollary 34. Let p be a prime greater than 3. Then

∑_{k=0}^{p−1} \frac{S_{k}}{108^{k}} ≡ \frac{1}{2} (\frac{3}{p}) ,  ∑_{k=0}^{⌊p/6⌋} \frac{S_{k}}{108^{k}} ≡ \frac{3}{4} (\frac{3}{p}) (mod p),

∑_{k=0}^{p−1} \frac{S_{k}}{216^{k}} ≡ \frac{1}{2} (\frac{2}{p}) ,  ∑_{k=0}^{⌊p/6⌋} \frac{S_{k}}{216^{k}} ≡ \frac{3}{4} (\frac{2}{p}) (mod p),

∑_{k=0}^{p−1} \frac{(2k + 1)S_{k}}{108^{k}} ≡ \frac{2}{9} (\frac{3}{p}) ,  ∑_{k=0}^{⌊p/6⌋} \frac{(2k + 1)S_{k}}{108^{k}} ≡ \frac{1}{3} (\frac{3}{p}) (mod p),

∑_{k=0}^{p−1} \frac{(2k + 1)S_{k}}{216^{k}} ≡ \frac{1}{2} (\frac{6}{p}) ,  ∑_{k=0}^{⌊p/6⌋} \frac{(2k + 1)S_{k}}{216^{k}} ≡ \frac{1}{2} (\frac{6}{p}) (mod p).

From Corollary 33 and (18) we get the following congruences.

(\frac{3}{p}) ∑_{k=0}^{p−1} \frac{S_{k}}{432^{k}} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1 (mod 9); \\ −11/8 (mod p), & if p ≡ ±2 (mod 9); \\ 7/8 (mod p), & if p ≡ ±4 (mod 9), \end{cases}
∑_{k=0}^{p−1} S_{k} (\frac{3}{432})^{k} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1 (mod 9); \\ 1/8 (mod p), & if p ≡ ±2 (mod 9); \\ −5/8 (mod p), & if p ≡ ±4 (mod 9), \end{cases}
(\frac{p}{3}) ∑_{k=0}^{p−1} \frac{(2k + 1)S_{k}}{432^{k}} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1 (mod 9); \\ −1/12 (mod p), & if p ≡ ±2 (mod 9); \\ −5/12 (mod p), & if p ≡ ±4 (mod 9), \end{cases}
∑_{k=0}^{p−1}(2k + 1)S_{k} (\frac{3}{432})^{k} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1 (mod 9); \\ −3/4 (mod p), & if p ≡ ±2 (mod 9); \\ 1/4 (mod p), & if p ≡ ±4 (mod 9). \end{cases}

The following theorem provides two families of polynomial congruences. Theorem 36. Let p be a prime, p > 3, and let t ∈ D_{p}. If t ≢ 0 (mod p), then the following congruences hold modulo p:

∑_{k=0}^{⌊p/6⌋} S_{k}(t^{2}(4t + 1))^{k}≡\frac{1 + 12t}{32t}(\frac{1 + 4t}{p})− \frac{1 − 12t}{32t}(\frac{1 − 12t}{p}) ,
∑_{k=0}^{p−1} S_{k}(t^{2}(4t + 1))^{k}≡ \frac{(1 + 12t)(1 + 4t)(1 − 6t)}{32t}(\frac{1 + 4t}{p})− \frac{(1−12t)(24t^{2} + 6t + 1)}{32t}(\frac{1 − 12t}{p}).

If 6t + 1 ≢ 0 (mod p), then we have modulo p,

∑_{k=0}^{⌊p/6⌋}(2k + 1)S_{k}(t^{2}(4t + 1))^{k} ≡ \frac{1 + 12t}{8(1 + 6t)} (\frac{1 − 12t}{p}) + \frac{3(1 + 4t)}{8(1 + 6t)} (\frac{1 + 4t}{p}) ,
∑_{k=0}^{p−1}(2k + 1)S_{k}(t^{2}(4t + 1))^{k} ≡ \frac{(1 + 12t)(24t^{2} + 6t + 1)}{8(1 + 6t)} (\frac{1 − 12t}{p})

+ \frac{3(1 − 6t)(1 + 4t)^{2}}{8(1 + 6t)} (\frac{1 + 4t}{p}) .

Proof. From (48), Corollary 33 and Lemma 20 for any x ∈ D_{p} with 2x + 1 ≢ 0 (mod p), we have

∑_{k=0}^{⌊p/6⌋} S_{k} (\frac{(1 − x)(2x + 1)^{2}}{216} )^{k} ≡ \frac{3}{4} (\frac{p}{3}) w_{⌊ \frac{p}{6} ⌋}(4x^{3} − 3x)

≡ \frac{3(x + 1)}{4(2x + 1)} (\frac{2x + 2}{p}) + \frac{3x}{4(2x + 1)} (\frac{6 − 6x}{p}) (mod p),

∑_{k=0}^{p−1} S_{k} (\frac{(1 − x)(2x + 1)^{2}}{216} )^{k} ≡ \frac{1}{8} (\frac{p}{3}) (5w_{⌊ \frac{p}{6} ⌋}(4x^{3} − 3x) − w_{⌊ \frac{5p}{6} ⌋}(4x^{3} − 3x))

≡ \frac{(x + 1)(2x^{2} − x + 2)}{4(2x + 1)} (\frac{2x + 2}{p}) − \frac{x(x − 1)(2x + 3)}{4(2x + 1)} (\frac{6 − 6x}{p}) (mod p).

Now replacing x by (−12t − 1)/2, we get the first two congruences of the theorem.

Similarly, from (49), Corollary 33 and Lemma 20 for any x ∈ D_{p} such that 2x + 1 ≢ 0 (mod p), we have

∑_{k=0}^{⌊p/6⌋}(2k + 1)S_{k} (\frac{(x + 1)(2x − 1)^{2}}{216} )^{k} ≡ \frac{1}{2}(−1)^{\frac{p−1}{2}} w_{⌊ \frac{p}{6} ⌋}(4x^{3} − 3x)

≡ \frac{x}{2(2x + 1)} (\frac{2 − 2x}{p}) + \frac{x + 1}{2(2x + 1)} (\frac{6x + 6}{p}) (mod p),

∑_{k=0}^{p−1}(2k + 1)S_{k} (\frac{(x + 1)(2x − 1)^{2}}{216} )^{k} ≡ \frac{1}{12}(−1)^{\frac{p−1}{2}} (5w_{⌊ \frac{p}{6} ⌋}(4x^{3} − 3x) + w_{⌊ \frac{5p}{6} ⌋}(4x^{3} − 3x))

≡ \frac{x(2x^{2} + x + 2)}{6(2x + 1)} (\frac{2 − 2x}{p}) − \frac{(x + 1)^{2}(2x − 3)}{6(2x + 1)} (\frac{6x + 6}{p}) (mod p).

Replacing x by (12t + 1)/2, we conclude the proof.

The next theorem gives a criterion for c ∈ C_{j}(p) in terms of values of the sums ∑_{k=0}^{p−1} S_{k}t^{k} and ∑_{k=0}^{p−1}(2k + 1)S_{k}t^{k} modulo p.

Theorem 37. Let p be a prime, p > 3, and let c ∈ D_{p} with c^{2} ≢ −3 (mod p). Then

(\frac{3(c^{2} + 3)}{p}) · ∑_{k=0}^{p−1} S_{k} (\frac{c^{2}}{108(c^{2} + 3)})^{k} ≡ \begin{cases}1/2 (mod p), & if c ∈ C_{0}(p); \\ \frac{9−2c}{8c} (mod p), & if c ∈ C_{1}(p); \\ −\frac{9+2c}{8c} (mod p), & if c ∈ C_{2}(p) \end{cases}

and

(\frac{c^{2} + 3}{p}) · ∑_{k=0}^{p−1} \frac{(2k + 1)S_{k}}{36^{k}(3 + c^{2})^{k}} ≡ \begin{cases}1/2 (mod p), & if c ∈ C_{0}(p); \\ \frac{2−c}{4c} (mod p), & if c ∈ C_{1}(p); \\ −\frac{2+c}{4c} (mod p), & if c ∈ C_{2}(p). \end{cases}

Proof. From Corollary 33 we have

∑_{k=0}^{p−1} S_{k} (\frac{c^{2}}{108(3 + c^{2})})^{k} ≡ \frac{1}{8} (\frac{p}{3}) (5w_{⌊ \frac{p}{6} ⌋} (\frac{3 − c^{2}}{3 + c^{2}} ) − w_{⌊ \frac{5p}{6} ⌋} (\frac{3 − c^{2}}{3 + c^{2}} )) (mod p), (50)
∑_{k=0}^{p−1} \frac{(2k + 1)S_{k}}{36^{k}(3 + c^{2})^{k}} ≡ \frac{(−1)^{\frac{p−1}{2}}}{12} (5w_{⌊ \frac{p}{6} ⌋} (\frac{3 − c^{2}}{3 + c^{2}} ) + w_{⌊ \frac{5p}{6} ⌋} (\frac{3 − c^{2}}{3 + c^{2}} )) (mod p). (51)

If p ≡ 1 (mod 6), then p splits into the product of primes in ℤ[ω], p = π\overline{π} with π ≡ 2 (mod 3) and, by (36), we easily find

w_{⌊ \frac{p}{6} ⌋} (\frac{3 − c^{2}}{3 + c^{2}} ) = \frac{(−1)^{\frac{p−1}{6}}}{2c(c^{2} + 3)^{\frac{p−1}{6}}} ((c − 1 − 2ω)^{\frac{p−1}{3} +1} + (c + 1 + 2ω)^{\frac{p−1}{3} +1}) .

Applying Lemma 23, we have

w_{⌊ \frac{p}{6} ⌋} (\frac{3 − c^{2}}{3 + c^{2}} )≡ \frac{(−1)^{\frac{p−1}{6}}}{2c(c^{2} + 3)^{\frac{p−1}{2}}} ((c − 1 − 2ω) (\frac{c + 1 + 2ω}{p})_{3}+ (c + 1 + 2ω) (\frac{c + 1 + 2ω}{p})_{3}^{−1})

modulo π and therefore,

(−1)^{\frac{p−1}{2}} (\frac{c^{2} + 3}{p}) · w_{⌊ \frac{p}{6} ⌋} (\frac{3 − c^{2}}{3 + c^{2}} ) ≡ \begin{cases}1 (mod p), & if c ∈ C_{0}(p); \\ \frac{3−c}{2c} (mod p), & if c ∈ C_{1}(p); \\ −\frac{3+c}{2c} (mod p), & if c ∈ C_{2}(p). \end{cases}  (52)

If p ≡ 5 (mod 6), then

w_{⌊ \frac{p}{6} ⌋} (\frac{3 − c^{2}}{3 + c^{2}} ) = \frac{(−1)^{\frac{p−5}{6}}}{2c(c^{2} + 3)^{\frac{p−5}{6}}} ((c − 1 − 2ω)^{\frac{p+1}{3} −1} + (c + 1 + 2ω)^{\frac{p+1}{3} −1}) .

Now, by Lemma 23, we get

w_{⌊ \frac{p}{6} ⌋} (\frac{3 − c^{2}}{3 + c^{2}} )≡ \frac{(−1)^{\frac{p−5}{6}}}{2c(c^{2} + 3)^{\frac{p−1}{2}}} ((c + 1 + 2ω) (\frac{c + 1 + 2ω}{p})_{3}^{−1}+ (c − 1 − 2ω) (\frac{c + 1 + 2ω}{p})_{3})

modulo p and therefore,

(−1)^{\frac{p−5}{6}} (\frac{c^{2} + 3}{p}) · w_{⌊ \frac{p}{6} ⌋} (\frac{3 − c^{2}}{3 + c^{2}} ) ≡ \begin{cases}1 (mod p), & if c ∈ C_{0}(p); \\ \frac{3−c}{2c} (mod p), & if c ∈ C_{1}(p); \\ −\frac{3+c}{2c} (mod p), & if c ∈ C_{2}(p). \end{cases}  (53)

Comparing (52) and (53), we get that (52) holds for all primes p > 3.

Applying the similar argument for evaluation of w_{⌊ \frac{5p}{6} ⌋}(\frac{3−c^{2}}{3+c^{2}} ), we see that if p ≡ 1 (mod 6), then 5p ≡ 5 (mod 6) and we have

w_{⌊ \frac{5p}{6} ⌋} (\frac{3 − c^{2}}{3 + c^{2}} ) = \frac{(−1)^{\frac{p−1}{6}}}{2c(c^{2} + 3)^{\frac{5(p−1)}{6}}} ((c − 1 − 2ω)^{\frac{5(p−1)}{3} +1} + (c + 1 + 2ω)^{\frac{5(p−1)}{3} +1}) .

Now, by Lemma 23, we easily find

w_{⌊ \frac{5p}{6} ⌋} (\frac{3−c^{2}}{3+c^{2}} )≡ \frac{(−1)^{\frac{p−1}{6}}}{2c(c^{2} + 3)^{\frac{5(p−1)}{2}}} ((c − 1 − 2ω)(\frac{c + 1 + 2ω}{p})_{3}^{5}+(c + 1 + 2ω)(\frac{c + 1 + 2ω}{p})_{3}^{−5})

modulo π, which implies

(−1)^{\frac{p−1}{2}} (\frac{c^{2} + 3}{p}) · w_{⌊ \frac{5p}{6} ⌋} (\frac{3 − c^{2}}{3 + c^{2}} ) ≡ \begin{cases}1 (mod p), & if c ∈ C_{0}(p); \\ −\frac{3+c}{2c} (mod p), & if c ∈ C_{1}(p); \\ \frac{3−c}{2c} (mod p), & if c ∈ C_{2}(p). \end{cases}  (54)

Similarly, if p ≡ 5 (mod 6), then 5p ≡ 1 (mod 6) and we have

w_{⌊ \frac{5p}{6} ⌋} (\frac{3 − c^{2}}{3 + c^{2}} ) = \frac{(−1)^{\frac{5p−1}{6}}}{2c(c^{2} + 3)^{\frac{5p−1}{6}}} ((c − 1 − 2ω)^{\frac{5p+2}{3}} + (c + 1 + 2ω)^{\frac{5p+2}{3}} ) .

By Lemma 23, we readily get

w_{⌊ \frac{5p}{6} ⌋} (\frac{3−c^{2}}{3+c^{2}} ) ≡ \frac{(−1)^{\frac{5p−1}{6}}}{2c(c^{2} + 3)^{\frac{5p−1}{2}}} ((c + 1 + 2ω) (\frac{c + 1 + 2ω}{p})_{3}^{−5} +(c − 1 − 2ω) (\frac{c + 1 + 2ω}{p})_{3}^{5})

modulo p and therefore after simplification we obtain that (54) holds for all primes p > 3. Finally, substituting (52) and (54) into (50) and (51), we get the congruences of the theorem.

From Theorem 37 with c = −9M/L and c = L/3M and criterion (34) we get the following congruences.

Theorem 38. Let q be a prime, q ≡ 1 (mod 3) and so 4q = L^{2} + 27M^{2} with L, M ∈ ℤ and L ≡ 1 (mod 3). Let p be a prime with p ≠ 2, 3, q, and let p ∤ LM. Then

(\frac{q}{p}) ∑_{k=0}^{p−1} S_{k} \frac{M^{2k}}{(16q)^{k}} ≡ \begin{cases}1/2 (mod p), & if p^{\frac{q−1}{3}} ≡ 1 (mod q); \\ \frac{±L−2M}{8M} (mod p), & if p^{\frac{q−1}{3}} ≡ \frac{−1±L/(3M)}{2} (mod q), \end{cases}

(\frac{3q}{p} ) ∑_{k=0}^{p−1} S_{k} \frac{L^{2k}}{(432q)^{k}} ≡ \begin{cases}1/2 (mod p), & if p^{\frac{q−1}{3}} ≡ 1 (mod q); \\ \frac{±27M−2L}{8L} (mod p), & if p^{\frac{q−1}{3}} ≡ \frac{−1±9M/L}{2} (mod q) \end{cases}

and

(\frac{q}{p}) ∑_{k=0}^{p−1}(2k + 1)S_{k} \frac{M^{2k}}{(16q)^{k}} ≡ \begin{cases}1/2 (mod p), & if p^{\frac{q−1}{3}} ≡ 1 (mod q); \\ \frac{±6M−L}{4L} (mod p), & if p^{\frac{q−1}{3}} ≡ \frac{−1±9M/L}{2} (mod q), \end{cases}

(\frac{3q}{p} ) ∑_{k=0}^{p−1}(2k + 1)S_{k} \frac{L^{2k}}{(432q)^{k}} ≡ \begin{cases}1/2 (mod p), & if p^{\frac{q−1}{3}} ≡ 1 (mod q); \\ \frac{±2L−9M}{36M} (mod p), & if p^{\frac{q−1}{3}} ≡ \frac{−1±L/(3M)}{2} (mod q). \end{cases}

For example, if q = 7, then 4q = L^{2} + 27M^{2} with L = M = 1 and by Theorem 38, we get the following numerical congruences.

Corollary 39. Let p be a prime, p ≠ 2, 3, 7. Then

(\frac{7}{p}) ∑_{k=0}^{p−1} \frac{S_{k}}{112^{k}} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1 (mod 7); \\ −3/8 (mod p), & if p ≡ ±2 (mod 7); \\ −1/8 (mod p), & if p ≡ ±3 (mod 7), \end{cases}

(\frac{21}{p}) ∑_{k=0}^{p−1} \frac{S_{k}}{3024^{k}} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1 (mod 7); \\ 25/8 (mod p), & if p ≡ ±2 (mod 7); \\ −29/8 (mod p), & if p ≡ ±3 (mod 7), \end{cases}

(\frac{7}{p}) ∑_{k=0}^{p−1} \frac{(2k + 1)S_{k}}{112^{k}} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1 (mod 7); \\ 5/4 (mod p), & if p ≡ ±2 (mod 7); \\ −7/4 (mod p), & if p ≡ ±3 (mod 7), \end{cases}

(\frac{21}{p}) ∑_{k=0}^{p−1} \frac{(2k + 1)S_{k}}{3024^{k}} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1 (mod 7); \\ −11/36 (mod p), & if p ≡ ±2 (mod 7); \\ −7/36 (mod p), & if p ≡ ±3 (mod 7). \end{cases}

Similarly, setting q = 13, 19, 31 in Theorem 38, we obtain the following congruences.

Corollary 40. Let p be a prime, p ≠ 2, 3, 5, 13. Then

(\frac{13}{p}) ∑_{k=0}^{p−1} \frac{S_{k}}{208^{k}} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1, ±5 (mod 13); \\ −7/8 (mod p), & if p ≡ ±2, ±3 (mod 13); \\ 3/8 (mod p), & if p ≡ ±4, ±6 (mod 13), \end{cases}

(\frac{39}{p}) ∑_{k=0}^{p−1} S_{k} (\frac{25}{5616})^{k} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1, ±5 (mod 13); \\ 17/40 (mod p), & if p ≡ ±2, ±3 (mod 13); \\ −37/40 (mod p), & if p ≡ ±4, ±6 (mod 13), \end{cases}

(\frac{13}{p}) ∑_{k=0}^{p−1} \frac{(2k + 1)S_{k}}{208^{k}} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1, ±5 (mod 13); \\ 1/20 (mod p), & if p ≡ ±2, ±3 (mod 13); \\ −11/20 (mod p), & if p ≡ ±4, ±6 (mod 13), \end{cases}

(\frac{39}{p}) ∑_{k=0}^{p−1}(2k + 1)S_{k} (\frac{25}{5616})^{k} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1, ±5 (mod 13); \\ −19/36 (mod p), & if p ≡ ±2, ±3 (mod 13); \\ 1/36 (mod p), & if p ≡ ±4, ±6 (mod 13). \end{cases}

Corollary 41. Let p be a prime, p ≠ 2, 3, 7, 19. Then

(\frac{19}{p}) ∑_{k=0}^{p−1} \frac{S_{k}}{304^{k}} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1, ±7, ±8 (mod 19); \\ 5/8 (mod p), & if p ≡ ±2, ±3, ±5 (mod 19); \\ −9/8 (mod p), & if p ≡ ±4, ±6, ±9 (mod 19), \end{cases}

(\frac{19}{p}) ∑_{k=0}^{p−1} \frac{(2k + 1)S_{k}}{304^{k}} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1, ±7, ±8 (mod 19); \\ −13/28 (mod p), & if p ≡ ±2, ±3, ±5 (mod 19); \\ −1/28 (mod p), & if p ≡ ±4, ±6, ±9 (mod 19). \end{cases}

Corollary 42. Let p be a prime, p ≠ 2, 3, 31. Then

(\frac{31}{p}) ∑_{k=0}^{p−1} \frac{S_{k}}{124^{k}} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1, ±2, ±4, ±8, ±15 (mod 31); \\ −1/2 (mod p), & if p ≡ ±3, ±6, ±7, ±12, ±14 (mod 31); \\ 0 (mod p), & if p ≡ ±5, ±9, ±10, ±11, ±13 (mod 31), \end{cases}

(\frac{93}{p}) ∑_{k=0}^{p−1} \frac{S_{k}}{837^{k}} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1, ±2, ±4, ±8, ±15 (mod 31); \\ 23/16 (mod p), & if p ≡ ±3, ±6, ±7, ±12, ±14 (mod 31); \\ −31/16 (mod p), & if p ≡ ±5, ±9, ±10, ±11, ±13 (mod 31), \end{cases}

(\frac{31}{p}) ∑_{k=0}^{p−1} \frac{(2k + 1)S_{k}}{124^{k}} ≡ \begin{cases}−1 (mod p), & if p ≡ ±5, ±9, ±10, ±11, ±13 (mod 31); \\ 1/2 (mod p), & otherwise, \end{cases}

(\frac{93}{p}) ∑_{k=0}^{p−1} \frac{(2k + 1)S_{k}}{837^{k}} ≡ \begin{cases}1/2 (mod p), & if p ≡ ±1, ±2, ±4, ±8, ±15 (mod 31); \\ −13/36 (mod p), & if p ≡ ±3, ±6, ±7, ±12, ±14 (mod 31); \\ −5/36 (mod p), & if p ≡ ±5, ±9, ±10, ±11, ±13 (mod 31). \end{cases}

7 Closed form for a companion sequence of S_{n}

As we noticed in the Introduction, the sequence S_{n} can be defined explicitly by formula (1) or by the generating function (8). Sun [13] considered a companion sequence T_{n}, whose definition comes from a conjectural series expansion of trigonometric functions [13, Conj. 4]: there are positive integers T_{1}, T_{2}, T_{3}, … such that

∑_{k=0}^{∞} S_{k}x^{2k+1} + \frac{1}{24} − ∑_{k=1}^{∞} T_{k}x^{2k} = \frac{1}{12} cos (\frac{2}{3} arccos(6\sqrt{3}x))  (55)

for all real x with |x| ≤ 1/(6\sqrt{3}). The first few values of T_{n} are as follows:

1, 32, 1792, 122880, 9371648, 763363328, ….

In this section, we give an exact formula for T_{n}. It easily follows from the companion series expansion to (5) [6, p. 210,(12)]:

cos(a arcsin(z)) = F (−\frac{a}{2}, \frac{a}{2}; \frac{1}{2}; z^{2}) ,  |z| ≤ 1.  (56)

Proposition 43. The coefficients T_{k}, k ≥ 1, in expansion (55) are given by

T_{k} = \frac{16^{k−1}}{k} \binom{3k − 2}{2k − 1} = 16^{k−1} (2\binom{3k − 2}{k − 1} − \binom{3k − 2}{k}) .

Proof. Combining formulas (5) and (56) with the obvious trigonometric identity

arcsin(z) + arccos(z) = \frac{π}{2} ,

we get a transformation formula connecting both hypergeometric functions from (5) and (56):

cos (\frac{πa}{2})F (−\frac{a}{2}, \frac{a}{2}; \frac{1}{2}; z^{2})+sin (\frac{πa}{2}) azF (\frac{1 + a}{2}, \frac{1 − a}{2}; \frac{3}{2}; z^{2}) = cos(a arccos(z)), |z| ≤ 1.

Plugging in a = 2/3, we get

\frac{1}{2}F (−\frac{1}{3}, \frac{1}{3}; \frac{1}{2}; z^{2}) + \frac{z}{\sqrt{3}}F (\frac{1}{6}, \frac{5}{6}; \frac{3}{2}; z^{2}) = cos (\frac{2}{3} arccos(z)) , |z| ≤ 1.

Replacing z by 6\sqrt{3}x with |x| ≤ 1/(6\sqrt{3}) and taking into account that

F (\frac{1}{6}, \frac{5}{6}; \frac{3}{2}; 108x^{2}) = 2 ∑_{k=0}^{∞} S_{k}x^{2k},

we obtain

\frac{1}{24}F (−\frac{1}{3}, \frac{1}{3}; \frac{1}{2}; 108x^{2}) + ∑_{k=0}^{∞} S_{k}x^{2k+1} = \frac{1}{12} cos (\frac{2}{3} arccos(6\sqrt{3}x)) ,

which gives the following generating function for the companion sequence T_{n} :

\frac{1}{24} − ∑_{k=1}^{∞} T_{k}x^{2k} = \frac{1}{24}F (−\frac{1}{3}, \frac{1}{3}; \frac{1}{2}; 108x^{2}) .

Comparing coefficients of powers of x^{2}, we get a formula for T_{k},

T_{k} = −\frac{1}{24} \frac{(−1/3)_{k}(1/3)_{k}}{(1/2)_{k}k!} 108^{k} = \frac{16^{k−1}}{k} \binom{3k − 2}{2k − 1} = 16^{k−1} (2\binom{3k − 2}{k − 1} − \binom{3k − 2}{k}) ,

which shows that T_{k} ∈ ℕ for all positive integers k.

8 Acknowledgement

The authors would like to thank the editor-in-chief and the referees for their time and useful comments. The authors gratefully acknowledge support from the Research Immersion Fellowships of the Fields Institute.

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2010 Mathematics Subject Classification: Primary 11A07; Secondary 11A15, 11B37, 11B65, 33C05, 33C45. Keywords: Jacobi polynomial, higher-order Catalan number, polynomial congruence, cubic residue, generating function.

(Concerned with sequences \underline{A000108}, \underline{A001448}, \underline{A001764}, \underline{A005809}, \underline{A048990}, and \underline{A176898}.)

Received April 30 2015; revised version received December 16 2015. Published in Journal of Integer Sequences, December 16 2015.

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