Related papers: Identities involving Bernoulli and Euler polynomia…

We establish two general identities for Bernoulli and Euler polynomials, which are of a new type and have many consequences. The most striking result in this paper is as follows: If $n$ is a positive integer, $r+s+t=n$ and $x+y+z=1$, then…

Let $p>3$ be a prime, and let $a$ be a rational p-adic integer. Let $\{B_n(x)\}$ and $\{E_n(x)\}$ denote the Bernoulli polynomials and Euler polynomials, respectively. In this paper we show that $$\sum_{k=0}^{p-1}\binom…

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

Let $\{B_n\}$, $\{B_n(x)\}$ and $\{E_n(x)\}$ be the Bernoulli numbers, Bernoulli polynomials and Euler polynomials, respectively. In this paper we mainly establish formulas for $\sum_{6\mid k-3}\binom nkB_{n-k}(x)$, $\sum_{6\mid k}\binom…

Let $p$ be an odd prime and let $x$ be a $p$-adic integer. In this paper, we establish supercongruences for $$ \sum_{k=0}^{p-1}\frac{\binom{x}{k}\binom{x+k}{k}(-4)^k}{(dk+1)\binom{2k}{k}}\pmod{p^2} $$ and $$…

We produce formulas for $$\sum_{j=1}^{2^{n-2}}\frac{1}{\sin^s\left(\frac{(2j-1)\pi}{2^n}\right)}$$ in terms of Generalized Bernoulli and Euler polynomials and use one of the formulas to produce a nice integral representation of the Riemann…

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…

We present a new proof of Euler's formulas for $\zeta(2k)$, where $k = 1,2,3,...$, which uses only the defining properties of the Bernoulli polynomials, obtaining the value of $\zeta(2k)$ by summing a telescoping series. Only basic…

In this paper we obtain a new curious identity involving trigonometric functions. Namely, for any positive odd integer $n$ we prove that $$\sum_{k=1}^n(-1)^k(\cot kx)\sin k(n-k)x=\frac{1-n}2,$$ which is equivalent to the identity…

Let $p>3$ be a prime, and let $a$ be a rational p-adic integer with $a\not\equiv 0\pmod p$. In this paper we establish congruences for $$\sum_{k=1}^{(p-1)/2}\frac{\binom ak\binom{-1-a}k}k, \quad\sum_{k=0}^{(p-1)/2}k\binom ak\binom{-1-a}k…

The polynomials $d_n(x)$ are defined by \begin{align*} d_n(x) &= \sum_{k=0}^n{n\choose k}{x\choose k}2^k. \end{align*} We prove that, for any prime $p$, the following congruences hold modulo $p$: \begin{align*}…

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 mainly prove two conjectural supercongruences of Sun by using the following identity $$ \sum_{k=0}^n\binom{2k}{k}^2\binom{2n-2k}{n-k}^2=16^n\sum_{k=0}^n\frac{\binom{n+k}{k}\binom{n}{k}\binom{2k}{k}^2}{(-16)^k} $$ which…

We present a new formula for the Bernoulli numbers as the following integral $$B_{2m} =\frac{(-1)^{m-1}}{2^{2m+1}} \int_{-\infty}^{+\infty} (\frac{d^{m-1}}{dx^{m-1}} {sech}^2 x)^2dx. $$ This formula is motivated by the results of Fairlie…

In this paper we deduce some new supercongruences modulo powers of a prime $p>3$. Let $d\in\{0,1,\ldots,(p-1)/2\}$. We show that $$\sum_{k=0}^{(p-1)/2}\frac{\binom{2k}k\binom{2k}{k+d}}{8^k}\equiv 0\ (\mbox{mod}\ p)\ \ \ \mbox{if}\ d\equiv…

In this paper we study recurrences concerning the combinatorial sum $[n,r]_m=\sum_{k\equiv r (mod m)}\binom {n}{k}$ and the alternate sum $\sum_{k\equiv r (mod m)}(-1)^{(k-r)/m}\binom{n}{k}$, where m>0, $n\ge 0$ and r are integers. For…

Let $p>3$ be a prime. In this paper, we obtain the congruences for $$\sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^3}{(-8)^k},\ \sum_{k=0}^{p-1}\frac{w(k)\binom{2k}k^2\binom{3k}k}{(-192)^k},\…

In this paper, we prove two conjectures of Z.-W. Sun: $$2n\binom{2n}n\big|\sum_{k=0}^{n-1}(3k+1)\binom{2k}k^3{16}^{n-1-k}\ \mbox{for}\ \mbox{all}\ n=2,3,\cdots,$$ and $$\sum_{k=0}^{(p-1)/2}\frac{3k+1}{16^k}\binom{2k}{k}^3\equiv…

We present some simple proofs of the well-known expressions for \[ \zeta(2k) = \sum_{m=1}^\infty \frac{1}{m^{2k}}, \qquad \beta(2k+1) = \sum_{m=0}^\infty \frac{(-1)^m}{(2m+1)^{2k+1}}, \] where $k = 1,2,3,\dots$, in terms of the Bernoulli…

In this paper, we provide some novel binomial convolution related to symmetric functions, as well as convolution sums without the binomial symbol. Moreover we give some new convolution sums of Bernoulli, Euler, and Genocchi numbers and…