Two q-congruences from Carlitz’s formula

Abstract

Recently, using a formula of Carlitz, the second author proved a q-congruence conjectured by Tauraso. In this note we utilize Carlitz’s formula to prove two more similar q-congruences. We also propose a conjectural q-congruence that refines one of our q-congruences.

Introduction

In 2010, Sun [14, Corollary 1.1] proved that, for any odd prime p and positive integer r,

$$\begin{aligned} \sum _{k=0}^{p^{r}-1}\frac{1}{2^k}{2k\atopwithdelims ()k}\equiv (-1)^{(p^r-1)/2}\pmod {p^2} . \end{aligned}$$

(1.1)

This congruence modulo p was first given by Sun and Tauraso [15, Corollary 1.1]. Recently, by using a transformation formula of Carlitz [1] and the symmetry of cyclotomic polynomials, the second author proved the following q-analogue of (1.1): for any positive odd integer n,

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q;q^2)_k}{(q;q)_{k}}q^k \equiv (-1)^{(n-1)/2}q^{(n^2-1)/4} \pmod {\Phi _n(q)^2}, \end{aligned}$$

(1.2)

where $(a;q)_n=(1-a)(1-aq)\ldots (1-aq^{n-1})$ is the q-shifted factorial and $\Phi _n(q)$ denotes the n-th cyclotomic polynomial in q. The q-congruence (1.2) was observed by Tauraso [16] for primes n, and its weaker form modulo $\Phi _n(q)$ was first established by the second author and Zeng [10, Corollary 4.2].

The aim of this note is to give the following two q-congruences by making use of Carlitz’s transformation formula again.

Theorem 1.1

Let $n>1$ be an odd integer. Then

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-1};q^2)_k}{(q;q)_k}q^k&\equiv (-1)^{(n+1)/2}q^{(n^2-1)/4} \pmod {\Phi _n(q)}, \end{aligned}$$

(1.3)

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{3};q^2)_k}{(q;q)_k}q^k&\equiv (-1)^{(n+1)/2}q^{-(n-3)^2/4} \pmod {\Phi _n(q)}. \end{aligned}$$

(1.4)

Letting $n=p^r$ be an odd prime power and then letting $q\rightarrow 1$ in this theorem, we obtain the following conclusion.

Corollary 1.2

Let p be an odd prime and $r\geqslant 1$. Then

$$\begin{aligned} \sum _{k=0}^{p^r-1}\frac{1}{2^k(2k-1)}{2k\atopwithdelims ()k} \equiv (-1)^{(p^r-1)/2}\pmod {p}, \end{aligned}$$

(1.5)

$$\begin{aligned} \sum _{k=0}^{p^r-1}\frac{2k+1}{2^k}{2k\atopwithdelims ()k} \equiv (-1)^{(p^r+1)/2}\pmod {p}. \end{aligned}$$

(1.6)

Numerical calculation indicates that the following generalization of Corollary 1.2 should be true.

Conjecture 1.3

Let p be an odd prime. Then

$$\begin{aligned} \sum _{k=0}^{p-1}\frac{1}{2^k(2k-1)}{2k\atopwithdelims ()k}&\equiv (-1)^{(p-1)/2}+2p\pmod {p^2}, \end{aligned}$$

(1.7)

$$\begin{aligned} \sum _{k=0}^{p-1}\frac{2k+1}{2^k}{2k\atopwithdelims ()k}&\equiv (-1)^{(p+1)/2}+2p\pmod {p^2}. \end{aligned}$$

(1.8)

Moreover, for $r\geqslant 2$, the congruences (1.5) and (1.6) also hold modulo $p^2$.

Meanwhile, we have the following conjectural refinement of (1.3), which is also a q-analogue of (1.7).

Conjecture 1.4

Let $n>1$ be an odd integer. Then

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-1};q^2)_k}{(q;q)_k}q^k \equiv (-1)^{(n+1)/2}q^{(n^2-1)/4}-(1+q)[n]\pmod {\Phi _n(q)^2}, \end{aligned}$$

(1.9)

where $[n]=1+q+\cdots +q^{n-1}$ is the q-integer.

However, we did not find the corresponding q-analogue of (1.8). Some other recent q-analogues of congruences can be found in [3,4,5,6,7,8,9, 11,12,13] using different techniques.

It is easy to see that $\Phi _n(q^2)=\Phi _n(q)\Phi _n(-q)$ for odd n. Thus, the q-congruence (1.3) is equivalent to

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-2};q^4)_k}{(q^2;q^2)_k}q^{2k} \equiv (-1)^{(n+1)/2}q^{(n^2-1)/2} \pmod {\Phi _n(q)}, \end{aligned}$$

(1.10)

since the left-hand side of (1.10) is an even function in q. For the same reason, the q-congruence (1.4) is equivalent to

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{6};q^4)_k}{(q^2;q^2)_k}q^{2k} \equiv (-1)^{(n+1)/2}q^{-(n-3)^2/2} \pmod {\Phi _n(q)}. \end{aligned}$$

(1.11)

We shall prove (1.10) and (1.11) using the aforementioned transformation formula due to Carlitz [1].

Proof of the theorem

Proof of (1.3)

Letting $q\mapsto q^{-1}$, we see that the congruence (1.10), which is an equivalent form of (1.3), can be written as

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-2};q^4)_k q^{3k-k^2}}{(q^2;q^2)_{k}} \equiv (-1)^{(n+1)/2}q^{(1-n^2)/2}\pmod {\Phi _n(q)}. \end{aligned}$$

(2.1)

On the other hand, performing substitutions $a\mapsto q^{-1}$, $b\mapsto -q^{-1}$, $q\mapsto q^2$, and $n\mapsto n-1$ in Carlitz’s formula (see [1] or [2, Exercise 1.17]):

$$\begin{aligned} \sum _{k=0}^{n}\frac{(a;q)_k (b;q)_k}{(q;q)_k} (-ab)^{n-k}q^{(n-k)(n+k-1)/2} =\sum _{k=0}^{n}\frac{(a;q)_{n+1}(-b)^k q^{k\atopwithdelims ()2}}{(q;q)_k (q;q)_{n-k}(1-aq^{n-k})}, \end{aligned}$$

(2.2)

we get

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-2};q^4)_k q^{n^2-5n-k^2+3k+4}}{(q^2;q^2)_{k}} =\sum _{k=0}^{n-1}\frac{(q^{-1};q^2)_{n}q^{k^2-2k}}{(q^2;q^2)_k (q^2;q^2)_{n-k-1}(1-q^{2n-2k-3})}. \end{aligned}$$

(2.3)

Since n is an odd integer greater than 1, the q-shifted factorial $(q^{-1};q^2)_{n}$ contains the factor $1-q^n$ and is therefore divisible by $\Phi _n(q)$. Moreover, the expression $(q^2;q^2)_k (q^2;q^2)_{n-k-1}$ is relatively prime to $\Phi _n(q)$ for $0\leqslant k\leqslant n-1$. Therefore, each summand on the right-hand side of (2.3) is congruent to 0 modulo $\Phi _n(q)$ except for $k=(n-3)/2$. Namely, modulo $\Phi _n(q)$, the identity (2.3) reduces to

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{-2};q^4)_k q^{3k-k^2+4}}{(q^2;q^2)_{k}}&\equiv \frac{(q^{-1};q^2)_n q^{(n-3)(n-7)/4}}{(q^2;q^2)_{(n-3)/2} (q^2;q^2)_{(n+1)/2} (1-q^n) }\nonumber \\&={n-1\brack \frac{n-1}{2}}_{q^2} {2n-1\brack n-1}\frac{(1-q^{-1})(1-q^{n-1}) q^{(n-3)(n-7)/4} }{(-q;q)_{n-1}^2 (1-q^{2n-1})(1-q^{n+1})}, \end{aligned}$$

(2.4)

where the q-binomial coefficients ${n\brack k}$ are defined by

$$\begin{aligned} {n\brack k}={n\brack k}_q ={\left\{ \begin{array}{ll}\displaystyle \frac{(q;q)_n}{(q;q)_k (q;q)_{n-k}} &{}\text {if }0\leqslant k\leqslant n, \\ 0 &{}\text {otherwise.} \end{array}\right. } \end{aligned}$$

In view of $q^n\equiv 1\pmod {\Phi _n(q)}$, we have

$$\begin{aligned} \frac{(1-q^{-1})(1-q^{n-1})}{(1-q^{2n-1})(1-q^{n+1})}\equiv -q^{-1}\pmod {\Phi _n(q)}. \end{aligned}$$

Furthermore, a special case of a q-analogue of Morley’s congruence [12, (1.5)] gives

$$\begin{aligned} {n-1\brack \frac{n-1}{2}}_{q^2}\equiv (-1)^{(n-1)/2}q^{(1-n^2)/4}(-q;q)_{n-1}^2 \pmod {\Phi _n(q)^2} \end{aligned}$$

(2.5)

(the modulus $\Phi _n(q)$ case is enough for us), and we have

$$\begin{aligned} {2n-1\brack n-1}=\prod _{k=1}^{n-1}\frac{1-q^{2n-k}}{1-q^k} \equiv \prod _{k=1}^{n-1}\frac{1-q^{-k}}{1-q^k}= (-1)^{n-1}q^{-n(n-1)/2} \pmod {\Phi _n(q)}. \end{aligned}$$

(2.6)

Substituting the above three q-congruences into (2.4), we arrive at (2.1). $\square$

Proof of (1.4)

The proof is very similar to that of (1.3). Replacing q by $q^{-1}$, we see that the congruence (1.10), the equivalent form of (1.4), can be written as

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{6};q^4)_k q^{-k^2-5k}}{(q^2;q^2)_{k}} \equiv (-1)^{(n+1)/2}q^{(n-3)^2/2}\pmod {\Phi _n(q)}. \end{aligned}$$

(2.7)

This time we substitute $a\mapsto q^{3}$, $b\mapsto -q^{3}$, $q\mapsto q^2$, and $n\mapsto n-1$ in (2.2) to obtain

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{6};q^4)_k q^{n^2+3n-k^2-5k-4}}{(q^2;q^2)_{k}} =\sum _{k=0}^{n-1}\frac{(q^{3};q^2)_{n}q^{k^2+2k}}{(q^2;q^2)_k (q^2;q^2)_{n-k-1}(1-q^{2n-2k+1})}. \end{aligned}$$

(2.8)

For any odd integer n greater than 1, the q-shifted factorial $(q^{3};q^2)_{n}$ has the factor $1-q^n$, and $1-q^{2n-2k+1}\not \equiv 0\pmod {\Phi _n(q)}$ for $0\leqslant k\leqslant n-1$ and $k\ne (n+1)/2$. Thus, modulo $\Phi _n(q)$, the identity (2.8) reduces to

$$\begin{aligned} \sum _{k=0}^{n-1}\frac{(q^{6};q^4)_k q^{-k^2-5k-4}}{(q^2;q^2)_{k}}&\equiv \frac{(q^{3};q^2)_n q^{(n+1)(n+5)/4}}{(q^2;q^2)_{(n+1)/2} (q^2;q^2)_{(n-3)/2} (1-q^n) }\\&={n-1\brack \frac{n-1}{2}}_{q^2} {2n-1\brack n-1}\frac{(1-q^{2n+1})(1-q^{n-1}) q^{(n+1)(n+5)/4} }{(-q;q)_{n-1}^2 (1-q)(1-q^{n+1})}. \end{aligned}$$

Using the q-congruences $q^n\equiv 1\pmod {\Phi _n(q)}$, (2.5) and (2.6), we deduce (2.7) from the above q-congruence immediately.$\square$