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Super congruences and Euler numbers

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Abstract

Let p > 3 be a prime. A p-adic congruence is called a super congruence if it happens to hold modulo some higher power of p. The topic of super congruences is related to many fields including Gauss and Jacobi sums and hypergeometric series. We prove that

$$\begin{gathered} \sum\limits_{k = 0}^{p - 1} {\frac{{\left( {_k^{2k} } \right)}} {{2^k }}} \equiv \left( { - 1} \right)^{{{\left( {p - 1} \right)} \mathord{\left/ {\vphantom {{\left( {p - 1} \right)} 2}} \right. \kern-\nulldelimiterspace} 2}} - p^2 E_{p - 3} \left( {\bmod p^3 } \right), \hfill \\ \sum\limits_{k = 1}^{{{\left( {p - 1} \right)} \mathord{\left/ {\vphantom {{\left( {p - 1} \right)} 2}} \right. \kern-\nulldelimiterspace} 2}} {\frac{{\left( {_k^{2k} } \right)}} {k}} \equiv \left( { - 1} \right)^{{{\left( {p + 1} \right)} \mathord{\left/ {\vphantom {{\left( {p + 1} \right)} 2}} \right. \kern-\nulldelimiterspace} 2}} \frac{8} {3}pE_{p - 3} \left( {\bmod p^2 } \right), \hfill \\ \sum\limits_{k = 0}^{{{\left( {p - 1} \right)} \mathord{\left/ {\vphantom {{\left( {p - 1} \right)} 2}} \right. \kern-\nulldelimiterspace} 2}} {\frac{{\left( {_k^{2k} } \right)^2 }} {{16^k }}} \equiv \left( { - 1} \right)^{{{\left( {p - 1} \right)} \mathord{\left/ {\vphantom {{\left( {p - 1} \right)} 2}} \right. \kern-\nulldelimiterspace} 2}} + p^2 E_{p - 3} \left( {\bmod p^3 } \right), \hfill \\ \end{gathered}$$

where E 0,E 1,E 2, ... are Euler numbers. Our new approach is of combinatorial nature. We also formulate many conjectures concerning super congruences and relate most of them to Euler numbers or Bernoulli numbers. Motivated by our investigation of super congruences, we also raise a conjecture on 7 new series for π 2, π −2 and the constant \(K: = \sum\nolimits_{k = 1}^\infty {{{\left( {\tfrac{k} {3}} \right)} \mathord{\left/ {\vphantom {{\left( {\tfrac{k} {3}} \right)} {k^2 }}} \right. \kern-\nulldelimiterspace} {k^2 }}}\) (with (−) the Jacobi symbol), two of which are

$$\sum\limits_{k = 1}^\infty {\frac{{\left( {10k - 3} \right)8^k }} {{k^3 \left( {_k^{2k} } \right)^2 \left( {_k^{3k} } \right)}} = \frac{{\pi ^2 }} {2}} and \sum\limits_{k = 1}^\infty {\frac{{\left( {15k - 4} \right)\left( { - 27} \right)^{k - 1} }} {{k^3 \left( {_k^{2k} } \right)^2 \left( {_k^{3k} } \right)}} = K.}$$

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  1. Department of Mathematics, Nanjing University, Nanjing, 210093, China

    Zhi-Wei Sun

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Sun, ZW. Super congruences and Euler numbers. Sci. China Math. 54, 2509–2535 (2011). https://doi.org/10.1007/s11425-011-4302-x

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