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A376464 a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2*A108625(n, n-k).

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%I A376464 #7 Sep 29 2024 09:20:15
%S A376464 1,7,307,20167,1667251,157682257,16284029959,1790285284039,
%T A376464 206214335367859,24625438825569757,3026463150129458557,
%U A376464 380777119628689277809,48850700118326352063559,6371139828304049908181989,842708419059535649574061807,112830751689834347753141396167,15268616117602266095803172887219
%N A376464 a(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2*A108625(n, n-k).
%C A376464 The sequence of Apéry numbers A005259 defined by A005259(n) = Sum_{k = 0..n} binomial(n, k)^2*binomial(n+k, k)^2 satisfies the pair of supercongruences
%C A376464 1) A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r
%C A376464 and
%C A376464 2) A005259(n*p^r - 1) == A005259(n*p^(r-1) - 1) (mod p^(3*r)) for all primes p >= 5 and all positive integers n and r.
%C A376464 We conjecture that the present sequence satisfies the same pair of supercongruences. Some examples are given below.
%F A376464 Examples of supercongruences:
%F A376464 a(11) - a(1) = 380777119628689277809 - 7 = 2*(3^3)*7*(11^3)*(19^2)*83*103*587*417773 == 0 (mod 11^3).
%F A376464 a(10) - a(0) = 3026463150129458557 - 1 = (2^2)*3*(11^3)*17*19*191*251*12236761 == 0 (mod 11^3).
%p A376464 A108625(n, k) := add(binomial(n, i)^2 * binomial(n+k-i, k-i), i = 0..k):
%p A376464 a(n) := add(binomial(n, k)^2*binomial(n+k, k)^2*A108625(n, n-k), k = 0..n):
%p A376464 seq(a(n), n = 0..25);
%Y A376464 Cf. A005259, A376458 - A376466.
%K A376464 nonn,easy
%O A376464 0,2
%A A376464 _Peter Bala_, Sep 24 2024