English

Two new kinds of numbers and related divisibility results

Number Theory 2018-11-13 v9 Combinatorics

Abstract

We mainly introduce two new kinds of numbers given by Rn=k=0n(nk)(n+kk)12k1 (n=0,1,2,...)R_n=\sum_{k=0}^n\binom nk\binom{n+k}k\frac1{2k-1}\quad\ (n=0,1,2,...) and Sn=k=0n(nk)2(2kk)(2k+1) (n=0,1,2,...).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 p1(mod4)p\equiv1\pmod 4 is a prime with p=x2+y2p=x^2+y^2 (where x1(mod4)x\equiv1\pmod 4 and y0(mod2)y\equiv0\pmod 2), then R(p1)/2p(1)(p1)/42x(modp2).R_{(p-1)/2}\equiv p-(-1)^{(p-1)/4}2x\pmod{p^2}. Also, 1n2k=0n1SkZ  and  1nk=0n1Sk(x)Z[x]for all n=1,2,3,...,\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 Sk(x)=j=0k(kj)2(2jj)(2j+1)xjS_k(x)=\sum_{j=0}^k\binom kj^2\binom{2j}j(2j+1)x^j. For any positive integers aa and nn, we show that, somewhat surprisingly, 1n2k=0n1(2k+1)(n1k)a(n1k)aZ  and  1nk=0n1(n1k)a(n1k)a4k21Z.\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.

Keywords

Cite

@article{arxiv.1408.5381,
  title  = {Two new kinds of numbers and related divisibility results},
  author = {Zhi-Wei Sun},
  journal= {arXiv preprint arXiv:1408.5381},
  year   = {2018}
}

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32 pages