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Related papers: A remark about positive polynomials

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In this paper, we prove that if $f(x)=\sum_{k=0}^n{n\choose k}a_kx^k$ is a polynomial with real zeros only, then the sequence $\{a_k\}_{k=0}^n$ satisfies the following inequalities $a_{k+1}^2(1-\sqrt{1-c_k})^2/a_k^2…

Combinatorics · Mathematics 2020-12-08 J. J. F Guo

The multi-variable Schmidt polynomials are defined by $$ S_n^{(r)}(x_0,\ldots,x_n):=\sum_{k=0}^n {n+k \choose 2k}^{r}{2k\choose k} x_k. $$ We prove that, for any positive integers $m$, $n$, $r$, and $\varepsilon=\pm 1$, all the coefficients…

Number Theory · Mathematics 2014-12-19 Qi-Fei Chen , Victor J. W. Guo

We present several new inequalities for trigonometric sums. Among others, we show that the inequality $$ \sum_{k=1}^n (n-k+1)(n-k+2)k\sin(kx) > \frac{2}{9} \sin(x) \bigl( 1+2\cos(x) \bigr)^2 $$ holds for all $n\geq 1$ and $x\in (0,…

Classical Analysis and ODEs · Mathematics 2023-07-11 Horst Alzer , Man Kam Kwong

For any positive integers $m$ and $\alpha$, we prove that $$\sum_{k=0}^{n-1}\epsilon^k(2k+1)A_k^{(\alpha)}(x)^m\equiv0\pmod{n}, $$ where $\epsilon\in\{1,-1\}$ and $$…

Number Theory · Mathematics 2011-08-09 Hao Pan

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…

Number Theory · Mathematics 2011-08-03 Zhi-Wei Sun

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…

Number Theory · Mathematics 2014-04-29 Zhi-Wei Sun

We study divisibility properties of certain sums and alternating sums involving binomial coefficients and powers of integers. For example, we prove that for all positive integers $n_1,..., n_m$, $n_{m+1}=n_1$, and any nonnegative integer…

Number Theory · Mathematics 2012-04-10 Victor J. W. Guo , Jiang Zeng

The Sun polynomials $g_n(x)$ are defined by \begin{align*} g_n(x)=\sum_{k=0}^n{n\choose k}^2{2k\choose k}x^k. \end{align*} We prove that, for any positive integer $n$, there hold \begin{align*} &\frac{1}{n}\sum_{k=0}^{n-1}(4k+3)g_k(x)…

Number Theory · Mathematics 2015-12-29 Victor J. W. Guo , Guo-Shuai Mao , Hao Pan

Let $n$ be any nonnegative integer and \[ D_n^{(h)}(x)=\sum_{k=0}^{n}\binom{n+k}{2k}^{h}\binom{2k}{k}^{h}{x}^{k} \text{ and } S_{n}^{(h)}(x)=\sum_{k=0}^{n}\binom{n+k}{2k}^{h}C_{k}^{h}{x}^{k} \] be the generalized Delannoy polynomials and…

Number Theory · Mathematics 2025-03-18 Lin-Yue Li , Rong-Hua Wang

In this paper, we investigate some congruences involving sums of $\frac{d^{-k}{x\choose k}{x+k\choose k}}{{2k \choose k}}$, where $x$ be a $p$-adic integer, $k$ be a non-negative integer, and $d$ $(d\neq 0)$ be a rational number.

Number Theory · Mathematics 2025-12-02 Wei-Wei Qi

Let ${\mathcal P}_k$ denote the set of all algebraic polynomials of degree at most $k$ with real coefficients. Let ${\mathcal P}_{n,k}$ be the set of all algebraic polynomials of degree at most $n+k$ having exactly $n+1$ zeros at $0$. Let…

Classical Analysis and ODEs · Mathematics 2018-09-21 Tamás Erdélyi

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…

Number Theory · Mathematics 2015-03-12 Zhi-Hong Sun

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 $$…

Number Theory · Mathematics 2024-07-30 Chen Wang , Hui-Li Han

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,…

Number Theory · Mathematics 2018-09-25 Guo-Shuai Mao , Zhi-Wei Sun

Let p be a prime and let a be a positive integer. In this paper we determine $\sum_{k=0}^{p^a-1}\binom{2k}{k+d}/m^k$ and $\sum_{k=1}^{p-1}\binom{2k}{k+d}/(km^{k-1})$ modulo $p$ for all d=0,...,p^a, where m is any integer not divisible by p.…

Number Theory · Mathematics 2010-04-02 Zhi-Wei Sun , Roberto Tauraso

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*}…

Number Theory · Mathematics 2016-04-19 Song Guo , Victor J. W. Guo

For a real number $k$, define $\pi_k(x) = \sum_{p\le x} p^k$. When $k>0$, we prove that $$ \pi_k(x) - \pi(x^{k+1}) = \Omega_{\pm}\left(\frac{x^{\frac12+k}}{\log x} \log\log\log x\right) $$ as $x\to\infty$, and we prove a similar result when…

Number Theory · Mathematics 2022-09-27 Lawrence C. Washington

We prove effective finiteness results concerning polynomial values of the sums $$ b^k +\left(a+b\right)^k + \cdots + \left(a\left(x-1\right) + b\right)^k $$ and $$ b^k - \left(a+b\right)^k + \left(2a+b\right)^k - \ldots + (-1)^{x-1}…

Number Theory · Mathematics 2024-04-26 András Bazsó

For positive integers $n>k$, let $P_{n,k}(x)=\displaystyle\sum_{j=0}^k \binom{n}{j}x^j $ be the polynomial obtained by truncating the binomial expansion of $(1+x)^n$ at the $k^{th}$ stage. These polynomials arose in the investigation of…

Number Theory · Mathematics 2013-06-05 Sudesh K. Khanduja , Ramneek Khassa , Shanta Laishram

We prove that for $k\ge 1$, all coefficients in the expansion of the series $$\sum_{n\ge 0} \frac{(q^{2n+2}, q^{2n+2k}; q^2)_\infty}{(q^{2n+1};q^2)_\infty^2} q^{2n}$$ are positive, by $q$-hypergeometric means. This confirms a recent…

Combinatorics · Mathematics 2026-01-28 Shane Chern , Chun Wang
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