相关论文: On $q$-analogues of some series for $\pi$ and $\pi…

We obtain $q$-analogues of several series for powers of $\pi$. For example, the identity $$\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^3}=\frac{\pi^3}{32}$$ has the following $q$-analogue: \begin{equation*}…

We give $q$-analogues of the following two Ramanujan-type formulas for $1/\pi$: \begin{align*} \sum_{k=0}^\infty (6k+1)\frac{(\frac{1}{2})_k^3}{k!^3 4^k} =\frac{4}{\pi} \quad\text{and}\quad \sum_{k=0}^\infty…

It is well known that $\zeta(2)=\pi^2/6$ as discovered by Euler. In this paper we present the following two $q$-analogues of this celebrated formula:…

Motivated by Ramanujan-type series and Zeilberger-type series, in this paper we investigate two new types of series for powers of $\pi$. For example, we prove that…

Let $H_n^{(2)}$ denote the second-order harmonic number $\sum_{0<k\le n}1/k^2$ for $n=0,1,2,\ldots$. In this paper we obtain the following identity: $$\sum_{k=1}^\infty\frac{2^kH_{k-1}^{(2)}}{k\binom{2k}k}=\frac{\pi^3}{48}.$$ We explain how…

In this paper, we evaluate some series of the form $$\sum_{k=1}^\infty\frac{ak^2+bk+c}{k(3k-1)(3k-2)m^k\binom{4k}k}.$$ For example, we prove that $$\sum_{k=1}^\infty\frac{(5k^2-4k+1)8^{k}}{k(3k-1)(3k-2)\binom{4k}k}=\frac{3}2\pi$$ and…

Let $E_n$ be the $n$-th Euler number and $(a)_n=a(a+1)\cdots (a+n-1)$ the rising factorial. Let $p>3$ be a prime. In 2012, Sun proved the that $$ \sum^{(p-1)/2}_{k=0}(-1)^k(4k+1)\frac{(\frac{1}{2})_k^3}{k!^3} \equiv…

Let $$ S_{m,n}(q):=\sum_{k=1}^{n}\frac{1-q^{2k}}{1-q^2} (\frac{1-q^k}{1-q})^{m-1}q^{\frac{m+1}{2}(n-k)}. $$ Generalizing the formulas of Warnaar and Schlosser, we prove that there exist polynomials $P_{m,k}(q)\in\mathbb{Z}[q]$ such that $$…

By making use of the multiplicate form of the extended Carlitz inverse series relations, we establish two general `dual' theorems of Jackson's summation formula for well--poised $_8\phi_7$-series. Their duplicate forms under the partition…

We give a $q$-congruence whose specializations $q=-1$ and $q=1$ correspond to supercongruences (B.2) and (H.2) on Van Hamme's 1997 list: $$ \sum_{k=0}^{(p-1)/2}(-1)^k(4k+1)A_k\equiv p(-1)^{(p-1)/2}\pmod{p^3} \quad\text{and}\quad…

For any odd prime p we obtain q-analogues of Van Hamme's supercongruence: $$ \sum_{k=0}^{\frac{p-1}{2}}{2k\choose k}^3\frac{1}{64^k} \equiv 0 \pmod{p^2} \quad\text{for}\quad p\equiv 3\pmod 4, $$ and Rodriguez-Villegas' Beukers-like…

It is known that $\sum_{k=0}^\infty\binom{2k}{k}/((2k+1)4^k)=\pi/2$ and $\sum_{k=0}^\infty\binom{2k}{k}/((2k+1)16^k)=\pi/3$. In this paper we obtain their p-adic analogues such as…

Via symbolic computation we deduce 97 new type series for powers of $\pi$ related to Ramanujan-type series. Here are three typical examples: $$\sum_{k=0}^\infty \frac{P(k) \binom{2k}k\binom{3k}k…

In this paper we apply a formula of the very-well poised $_{2k+4}\phi_{2k+3}$ to write a $k$-tuple sum of $q$-series as a linear combination of terms wherein each term is a product of expressions of the form $\frac{1}{(qy,…

Let $p>3$ be a prime. We prove that $$\sum_{k=0}^{p-1}\binom{2k}{k}/2^k=(-1)^{(p-1)/2}-p^2E_{p-3} (mod p^3),$$ $$\sum_{k=1}^{(p-1)/2}\binom{2k}{k}/k=(-1)^{(p+1)/2}8/3*pE_{p-3} (mod p^2),$$…

In this paper, by constructing some identities, we prove some $q$-analogues of some congruences. For example, for any odd integer $n>1$, we show that \begin{gather*} \sum_{k=0}^{n-1} \frac{(q^{-1};q^2)_k}{(q;q)_k} q^k \equiv (-1)^{(n+1)/2}…

Let $(a;q)_n=\prod_{0\le k<n}(1-aq^k)$ for n=0,1,2,.... Define q-Euler numbers $E_n(q)$, q-Sali\'e 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}…

We evaluate some series with summands involving a single binomial coefficient $\binom{6k}{3k}$. For example, we prove that $$\sum_{k=0}^\infty\frac{(63k^2+78k+22)8^k}{(2k+1)(6k+1)(6k+5)\binom{6k}{3k}}=\frac{3\pi}2.$$ Motivated by Galois…

In this paper we mainly employ the Zeilberger algorithm to study congruences for sums of terms involving products of three binomial coefficients. Let $p>3$ be a prime. We prove that…

Let $p>3$ be a prime and let $a$ be a positive integer. We show that if $p\equiv1\pmod 4$ or $a>1$ then $$\sum_{k=0}^{\lfloor\frac34p^a\rfloor}\frac{\binom{2k}k^2}{16^k}\equiv\l(\frac{-1}{p^a}\r)\pmod{p^3}$$ with $(-)$ the Jacobi symbol,…