On the mod $p^7$ determination of ${2p-1\choose p-1}$

In this paper we prove that for any prime $p \geq 11$ holds $(\binom{2 p - 1}{p - 1}) \equiv 1 - 2 p \sum_{k = 1}^{p - 1} \frac{1}{k} + 4 p^{2} \sum_{1 \leq i < j \leq p - 1} \frac{1}{i j} (m o d p^{7}) .$ This is a generalization of the famous Wolstenholme’s theorem which asserts that $(\binom{2 p - 1}{p - 1}) \equiv 1 (p^{3})$ for all primes $p \geq 5$ . Our proof is elementary and it does not use a standard technique involving the classic formula for the power sums in terms of the Bernoulli numbers. Notice that the above congruence reduced modulo $p^{6}$ , $p^{5}$ and $p^{4}$ yields related congruences obtained by R. Tauraso, J. Zhao and J.W.L. Glaisher, respectively.

On the mod $p^{7}$ determination of $(\binom{2 p - 1}{p - 1})$

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On the mod $p^{7}$ determination of $(\binom{2 p - 1}{p - 1})$

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On the mod p7p7 determination of (2p−1p−1)(p−12p−1​)

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On the mod p7p7 determination of (2p−1p−1)(p−12p−1​)

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On the mod $p^{7}$ determination of $(\binom{2 p - 1}{p - 1})$

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On the mod $p^{7}$ determination of $(\binom{2 p - 1}{p - 1})$

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