pith. sign in

arxiv: 1408.5381 · v10 · pith:WDZKAUXMnew · submitted 2014-08-21 · 🧮 math.NT · math.CO

Two new kinds of numbers and related divisibility results

classification 🧮 math.NT math.CO
keywords binommathbbpmodfrac1numbersquadequiv1frac
0
0 comments X
read the original abstract

We mainly introduce two new kinds of numbers given by $$R_n=\sum_{k=0}^n\binom nk\binom{n+k}k\frac1{2k-1}\quad\ (n=0,1,2,...)$$ and $$S_n=\sum_{k=0}^n\binom nk^2\binom{2k}k(2k+1)\quad\ (n=0,1,2,...).$$ We find that such numbers have many interesting arithmetic properties. For example, if $p\equiv1\pmod 4$ is a prime with $p=x^2+y^2$ (where $x\equiv1\pmod 4$ and $y\equiv0\pmod 2$), then $$R_{(p-1)/2}\equiv p-(-1)^{(p-1)/4}2x\pmod{p^2}.$$ Also, $$\frac1{n^2}\sum_{k=0}^{n-1}S_k\in\mathbb Z\ \ {and}\ \ \frac1n\sum_{k=0}^{n-1}S_k(x)\in\mathbb Z[x]\quad\text{for all}\ n=1,2,3,...,$$ where $S_k(x)=\sum_{j=0}^k\binom kj^2\binom{2j}j(2j+1)x^j$. For any positive integers $a$ and $n$, we show that, somewhat surprisingly, $$\frac1{n^2}\sum_{k=0}^{n-1}(2k+1)\binom{n-1}k^a\binom{-n-1}k^a\in\mathbb Z\ \ {and} \ \ \frac 1n\sum_{k=0}^{n-1}\frac{\binom{n-1}k^a\binom{-n-1}k^a}{4k^2-1}\in\mathbb Z.$$ We also solve a conjecture of V.J.W. Guo and J. Zeng, and pose several conjectures for further research.

This paper has not been read by Pith yet.

discussion (0)

Sign in with ORCID, Apple, or X to comment. Anyone can read and Pith papers without signing in.