相关论文: On the $p$-adic transcendence of $\sum_{k=1}^\inft…

Summation of the $p$-adic functional series $\sum \varepsilon^n \, n! \, P_k^\varepsilon (n; x)\, x^n ,$ where $P_k^\varepsilon (n; x)$ is a polynomial in $x$ and $n$ with rational coefficients, and $\varepsilon = \pm 1$, is considered. The…

Let $p$ be an odd prime, and let $\sum_{n=0}^{\infty} a_{n}X^{n}\in\mathbb{F}_p[[X]]$ be the reduction modulo $p$ of the Artin-Hasse exponential. We obtain a polynomial expression for $a_{kp}$ in terms of those $a_{rp}$ with $r<k$, for even…

In this paper, we confirm several conjectures posed by Sun recently; for example, we prove that for any odd prime $p$ we have $$ \sum_{k=0}^{p-1}A_k\equiv\begin{cases}4x^2-2p\pmod{p^2}\quad&\text{if $p=x^2+2y^2\ (x,y\in\mathbb{Z})$},\\…

The Ap\'ery polynomials are given by $$A_n(x)=\sum_{k=0}^n\binom nk^2\binom{n+k}k^2x^k\ \ (n=0,1,2,\ldots).$$ (Those $A_n=A_n(1)$ are Ap\'ery numbers.) Let $p$ be an odd prime. We show that…

Let $p>5$ be a prime. Motivated by the known formulae $\sum_{k=1}^\infty(-1)^k/(k^3\binom{2k}{k})=-2\zeta(3)/5$ and $\sum_{k=0}^\infty \binom{2k}{k}^2/((2k+1)16^k)=4G/\pi$$ (where $G=\sum_{k=0}^\infty(-1)^k/(2k+1)^2$ is the Catalan…

We show that for any prime prime $p\not=2$ $$\sum_{k=1}^{p-1} {(-1)^k\over k}{-{1\over 2} \choose k} \equiv -\sum_{k=1}^{(p-1)/2}{1\over k} \pmod{p^3}$$ by expressing the l.h.s. as a combination of alternating multiple harmonic sums.

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…

In this paper we prove three results conjectured by Z.-W. Sun. Let $p$ be an odd prime and let $h\in \mathbb{Z}$ with $2h-1\equiv0\pmod{p^{}}$. For $a\in\mathbb{Z}^{+}$ and $p^a>3$, we show that \begin{align}\notag…

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…

We consider summation of some finite and infinite functional p-adic series with factorials. In particular, we are interested in the infinite series which are convergent for all primes p, and have the same integer value for an integer…

Let $p>3$ be a prime. For any $p$-adic integer $a$, we determine $$\sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}kH_k,\ \ \sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}kH_k^{(2)},\ \ \sum_{k=0}^{p-1}\binom{-a}k\binom{a-1}k\frac{H_k^{(2)}}{2k+1}$$ modulo…

Let $p$ be a prime with $p>3$, and let $a,b$ be two rational $p-$integers. In this paper we present general congruences for $\sum_{k=0}^{p-1}\binom ak\binom{-1-a}k\frac p{k+b}\pmod {p^2}$. For $n=0,1,2,\ldots$ let $D_n$ and $b_n$ be Domb…

We prove that the sum of the series $\sum_{n=0}^{\infty}\, p^{v_p(n!)}$ is a $p$-adic irrational for all primes $p$, where $v_p(n!)$ denotes the exponent of the highest power of $p$ dividing $n!$.

We investigate various properties of p-adic differential equations which have as a solution an analytic function of the form $F_k (x) = \sum_{n\geq 0} n! P_k (n) x^n$, where $P_k (n) = n^k + C_{k-1} n^{k-1} + ...+ C_0$ is a polynomial in n…

Let $m$ and $n>0$ be integers. Suppose that $p$ is a prime dividing $m-4$ but not dividing $m$. We show that $\nu_p(\sum_{k=0}^{n-1}\frac{\binom{2k}k}{m^k})$ and $\nu_p(\sum_{k=0}^{n-1}\binom{n-1}{k}(-1)^k\frac{\binom{2k}k}{m^k})$ are at…

Let f(n)= Sum binomial(n,k)^(-1). First, we show that f:N to Q_p is nowhere continuous in the p-adic topology. If x is a p-adic integer, we say that f(x) is p-definable if lim f(x_j) exists in Q_p, where x_j denotes the jth partial sum for…

We study $p$-adic Euler's series $E_p(t) = \sum_{k=0}^{\infty}k!t^k$ at a point $p^a$, $a \in \mathbb{Z}_{\ge 1}$, and use Pad\'e approximations to prove a lower bound for the $p$-adic absolute value of the expression $cE_p\left(\pm…

Let $p$ be an odd prime, and let $m$ be an integer with $p\nmid m$. In this paper show that $$\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak\binom{-1-a}k}{m^k} \equiv 0\pmod p \quad\hbox{implies}\quad\sum_{k=0}^{p-1}\frac{\binom{2k}k\binom ak…

Let $p$ be any prime and let $a$ and $n$ be positive integers with $p\nmid n$. We show that $$\sum_{k=1}^{p^a-1}\frac{B_k}{(-n)^k}\equiv a(-1)^{n-1}D_{n-1}\pmod {p},$$ where $B_0,B_1,\ldots$ are the Bell numbers and $D_0,D_1,\ldots$ are the…

For all integers $n \geq k \geq 1$, define $H(n,k) := \sum 1 / (i_1 \cdots i_k)$, where the sum is extended over all positive integers $i_1 < \cdots < i_k \leq n$. These quantities are closely related to the Stirling numbers of the first…