On q-Euler numbers, q-Salie numbers and q-Carlitz numbers

Abstract

Let $(a;q{)}_n={\prod }_{0\le k<n}(1-a{q}^k)$ for n=0,1,2,.... Define q-Euler numbers $E_n(q)$, q-Salié numbers $S_n(q)$ and q-Carlitz numbers $C_n(q)$ as follows: $$\sum _{n=0}^{\infty }{E}_n(q)\frac{x^n}{(q,q{)}_n}=1/\sum _{n=0}^{\infty }\frac{q^{n(2n-1)}{x}^{2n}}{(q;q{)}_{2n}},$$ $$\sum _{n=0}^{\infty }{S}_n(q)\frac{x^n}{(q;q{)}_n}=\sum _{n=0}^{\infty }\frac{q^{n(n-1)}{x}^{2n}}{(q;q{)}_{2n}}/\sum _{n=0}^{\infty }\frac{(-1{)}^n{q}^{n(2n-1)}{x}^{2n}}{(q;q{)}_{2n}},$$ $$\sum _{n=0}^{\infty }{C}_n(q)\frac{x^n}{(q;q{)}_n}=\sum _{n=0}^{\infty }\frac{q^{n(n-1)}{x}^{2n+1}}{(q;q{)}_{2n+1}}/\sum _{n=0}^{\infty }\frac{(-1{)}^n{q}^{n(2n+1)}{x}^{2n+1}}{(q;q{)}_{2n+1}}.$$ We show that $$E_{2n}(q)-E_{2n+2^{s}t}(q)=[2^s{]}_{q^t}(mod(1+q)[2^s{]}_{q^t})$$ for any nonnegative integers n,s,t with t odd, where $[k{]}_q=(1-q^k)/(1-q)$; this is a q-analogue of Stern's congruence $E_{2n+2^s}=E_{2n}+2^s(mod{2}^{s+1})$. We also prove that $(-q;q{)}_n={\prod }_{0<k\le n}(1+q^k)$ divides $S_{2n}(q)$ and the numerator of $C_{2n}(q)$; this extends Carlitz's result that $2^n$ divides the Salié number $S_{2n}$ and the numerator of the Carlitz number $C_{2n}$. Our result on q-Salié numbers implies a conjecture of Guo and Zeng.

19 pages

References in corpus (1)

• Some Arithmetic Properties of the q-Euler Numbers and q-Salié Numbers

Cited by in corpus (2)

• q-Analogue of Gauss' Divisibility Theorem
• q-Analogue of Wilson's theorem