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%I #12 Oct 13 2022 12:58:13
%S 27,20577,60353937,287798988897,1782634331587527,13011500170881726987,
%T 106321024671550496694837,943479109706472533832704097,
%U 8916177779855571182824077866307,88547154924474394601268826256953077,915376390434997094066775480671975209017
%N a(n) = A005258(n)^3 * A005258(n-1).
%C The Apéry numbers B(n) = A005258(n) satisfy the supercongruences B(p) == 3 (mod p^3) and B(p-1) == 1 (mod p^3) for all primes p >= 5 (see, for example, Straub, Example 3.4). It follows that a(p) == 27 (mod p^3) for all primes p >= 5. We conjecture that, in fact, the stronger congruence a(p) == 27 (mod p^5) holds for all primes p >= 3 (checked up to p = 251). Compare with the congruence B(p) + B(p-1) == 4 (mod p^5) conjectured to hold for all primes p >= 5. See A352655.
%C Conjecture: for r >= 2, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for all primes p >= 5. - _Peter Bala_, Oct 13 2022
%H Armin Straub, <a href="https://arxiv.org/abs/1401.0854">Multivariate Apéry numbers and supercongruences of rational functions</a>, arXiv:1401.0854 [math.NT] (2014).
%e Example of a supercongruence:
%e a(7) - a(1) = 106321024671550496694837 - 27 = 2*(3^3)*5*(7^5)* 11*18143* 117398731273 == 0 (mod 7^5)
%p A005258 := n -> add(binomial(n,k)^2*binomial(n+k,k), k = 0..n):
%p seq(A005258(n)^3*A005258(n-1), n = 1..20);
%Y Cf. A005258, A212334, A339946, A352655, A357507, A357508, A357509.
%K nonn,easy
%O 1,1
%A _Peter Bala_, Oct 01 2022
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