On sums of binomial coefficients modulo $p^2$

Volume 127 / 2012

Zhi-Wei Sun Colloquium Mathematicum 127 (2012), 39-54 MSC: Primary 11B65; Secondary 05A10, 11A07, 11B39, 11E25, 11S99. DOI: 10.4064/cm127-1-3

Abstract

Let $p$ be an odd prime and let $a$ be a positive integer. In this paper we investigate the sum $\def\bi#1#2{\bigg({#1\atop#2}\bigg)}\sum_{k=0}^{p^a-1}\big({hp^a-1\atop k} \big) \big({{2k}\atop k}\big)/m^k$ mod $p^2$, where $h$ and $m$ are $p$-adic integers with $m\not\equiv0 \pmod{p}$. For example, we show that if $h\not\equiv0 \pmod{p}$ and $p^a>3$, then $$\def\bi#1#2{\bigg({#1\atop#2}\bigg)} \sum_{k=0}^{p^a-1}\bi{hp^a-1}k\bi{2k}k\biggl(-\frac h2\biggr)^k \equiv \bigg(\frac{1-2h}{p^a}\bigg)\bigg(1+h \bigg( \bigg(4-\frac 2h\bigg)^{p-1}-1\bigg)\bigg)\pmod{p^2}, $$ where $(\frac{\cdot}{\cdot})$ denotes the Jacobi symbol. Here is another remarkable congruence: If $p^a>3$ then $$\def\bi#1#2{\bigg({#1\atop#2}\bigg)} \sum_{k=0}^{p^a-1}\bi{p^a-1}k\bi{2k}k(-1)^k\equiv 3^{p-1} \bigg(\frac{p^a}3\bigg) \pmod{p^2}. $$

Authors

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

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