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## Colloquium Mathematicum

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## Two new kinds of numbers and related divisibility results

### Volume 154 / 2018

Colloquium Mathematicum 154 (2018), 241-273 MSC: Primary 11A07, 11B65; Secondary 05A10, 05A30, 11B75, 11E25. DOI: 10.4064/cm7405-1-2018 Published online: 14 September 2018

#### Abstract

We mainly introduce two new kinds of numbers given by \begin{alignat*}2 R_n&=\sum_{k=0}^n\left(n\atop k\right)\left({n+k}\atop k\right)\frac1{2k-1}&\ \quad&(n=0,1,2,\ldots),\\ S_n&=\sum_{k=0}^n\left(n\atop k\right)^2\left({2k}\atop k\right)(2k+1)&\quad\ &(n=0,1,2,\ldots). \end{alignat*} 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, $$\frac1{n^2}\sum_{k=0}^{n-1}S_k\in\mathbb Z\quad \text{and}\quad \frac1n\sum_{k=0}^{n-1}S_k(x)\in\mathbb Z[x]\ \quad\text{for all } n=1,2,\ldots,$$ 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, $$\frac1{n^2}\sum_{k=0}^{n-1}(2k+1)\left( {n-1}\atop k\right)^a\left( {-n-1}\atop k\right)^a\in\mathbb Z\quad \text{and}\quad \frac1n\sum_{k=0}^{n-1}\frac{\binom{n-1}k^a\binom{-n-1}k^a}{4k^2-1}\in\mathbb Z.$$ We also solve a conjecture of V. J. W. Guo and J. Zeng, and pose several conjectures for further research.

#### Authors

• Zhi-Wei SunDepartment of Mathematics
Nanjing University
Nanjing 210093, People’s Republic of China
e-mail

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