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Hao Pan

Publications and source records attributed to Hao Pan.

At least 109 records · Page 6Linked to original sources
arXiv2016

On the generalized restricted sumsets in abelian groups

Suppose that $A$, $B$ and $S$ are non-empty subsets of a finite abelian group $G$. Then the generalized restricted sumset $$ A\stackrel{S}+B:=\{a+b:\,a\in A,\ b\in B,\ a-b\not\in S\} $$ contains at least $$ \min\{|A|+|B|-3|S|,p(G)\} $$ elements, where $p(G)$ is the least prime factor of $|G|$. Further, we also have $$ |A\stackrel{S}+B|\geq \min\{|A|+|B|-|S|-2,p(G)\}, $$ provided that both $|A|$ and $|B|$ are large with respect to $|S|$.

math.NT↗
arXiv2016

Ergodic recurrence and bounded gaps between primes

Let $(X,B_X,μ,T)$ be a measure-preserving probability system with $T$ is invertible. Suppose that $A\in B_X$ with $μ(A)>0$ and $ε>0$. For any $m\geq 1$, there exist infinitely many primes $p_0,p_1,\ldots,p_m$ with $p_0<\cdots 0$ is a constant only depending on $m$, $A$ and $ε$.

math.NT↗
arXiv2016

Identities involving weighted Catalan, Schroder and Motzkin paths

In this paper, we investigate the weighted Catalan, Motzkin and Schröder numbers together with the corresponding weighted paths. The relation between these numbers is illustrated by three equations, which also lead to some known and new interesting identities. To show these three equations, we provide combinatorial proofs. One byproduct is to find a bijection between two sets of Catalan paths: one consisting of those with $k$ valleys, and the other consisting of $k$ $\mathbf{N}$ steps in even positions.

math.CO↗
arXiv2016

Small gaps between the Piatetski-Shapiro primes

Suppose that $1<c<9/8$. For any $m\geq 1$, there exist infinitely many $n$ such that $$ \{[n^c],\ [(n+1)^c],\ \ldots,\ [(n+k_0)^c]\} $$ contains at least $m+1$ primes, if $k_0$ is sufficiently large (only depending on $m$).

math.NT↗
arXiv2015

Proof of a conjecture involving Sun polynomials

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) \in\mathbb{Z}[x],\quad\text{and}\\ &\sum_{k=0}^{n-1}(8k^2+12k+5)g_k(-1)\equiv 0\pmod{n}. \end{align*} The first one confirms a recent conjecture of Z.-W. Sun, while the second one partially answers another conjecture of Z.-W. Sun. We give three different proofs of the former. One of them depends on the following congruence: $$ {m+n-2\choose m-1}{n\choose m}{2n\choose n}\equiv 0\pmod{m+n}\quad\text{for $m,n\geqslant 1$.} $$

math.NT↗
arXiv2015

Bounded gaps between primes of the special form

For each $m\geq 1$, there exist infinitely many primes $p_1<p_2<\ldots<p_{m+1}$ such that $p_{m+1}-p_1=O(m^4e^{8m})$ and $p_j+2$ has at most $\frac{16m}{\log 2}+\frac{5\log m}{\log 2}+37$ prime divisors for each $j$.

math.NT↗
arXiv2015

Note on polynomial recurrence

Let $(X,μ,T_1,...,T_l)$ be a measure-preserving system with those $T_i$ are commuting. Suppose that the polynomials $p_1(t),...,p_{l}(t)\in\Z[t]$ with $p_j(0)=0$ have distinct degrees. Then for any $ε>0$ and $A\subseteq X$ with $μ(A)>0$, the set $$ \{n:\,μ(A\cap T_1^{-p_1(n)}A\cap...\cap T_l^{-p_l(n)}A)\geqμ(A)^{l+1}-ε\} $$ has bounded gaps.

math.CO↗
arXiv2014

Proof of three conjectures on congruences

In this paper we prove three conjectures on congruences involving central binomial coefficients or Lucas sequences. Let $p$ be an odd prime and let $a$ be a positive integer. We show that if $p\equiv 1\pmod{4}$ or $a>1$ then $$ \sum_{k=0}^{\lfloor\frac34p^a\rfloor}\binom{-1/2}k\equiv\left(\frac{2}{p^a}\right)\pmod{p^2}, $$ where $(-)$ denotes the Jacobi symbol. This confirms a conjecture of the second author. We also confirm a conjecture of R. Tauraso by showing that $$\sum_{k=1}^{p-1}\frac{L_k}{k^2}\equiv0\pmod{p}\quad {\rm provided}\ \ p>5,$$ where the Lucas numbers $L_0,L_1,L_2,\ldots$ are defined by $L_0=2,\ L_1=1$ and $L_{n+1}=L_n+L_{n-1}\ (n=1,2,3,\ldots)$. Our third theorem states that if $p\not=5$ then we can determine $F_{p^a-(\frac{p^a}5)}$ mod $p^3$ in the following way: $$\sum_{k=0}^{p^a-1}(-1)^k\binom{2k}k\equiv\left(\frac{p^a}5\right)\left(1-2F_{p^a-(\frac{p^a}5)}\right)\ \pmod{p^3},$$ which appeared as a conjecture in a paper of Sun and Tauraso in 2010.

math.NT↗
arXiv2013

A note on the Polignac numbers

Suppose that $k\geq 3.5\times 10^6$ and $\hH=\{h_1,\ldots,h_{k_0}\}$ is admissible. Then for any $m\geq 1$, the set $$ \{m(h_j-h_i):\,h_i<h_j\} $$ contains at least one Polignac number.

math.NT↗