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%I #10 May 11 2023 10:25:41
%S 1,5,123,3650,118059,4015380,141175410,5082313276,186243853995,
%T 6920379988871,260030830600748,9860709859708350,376821110248674594,
%U 14494688046084958080,560708803489098556632,21797478402692370515400,851057798310071946207915,33356751162583463626417872
%N a(n) = [x^n] E(x)^n, where E(x) = exp( Sum_{k >= 1} A005259(k)*x^k/k ).
%C Compare with A362723.
%C It is known that the sequence of Apéry numbers A005259 satisfies the Gauss congruences A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
%C One consequence is that the power series expansion of E(x) = exp( Sum_{k
%C >= 1} A005259(k)*x^k/k ) = 1 + 5*x + 49*x^2 + 685*x^3 + 11807*x^4 + ... has integer coefficients (see, for example, Beukers, Proposition, p. 143), and therefore a(n) = [x^n] E(x)^n is an integer.
%C In fact, the Apéry numbers satisfy stronger congruences than the Gauss congruences known as supercongruences: A005259(n*p^r) == A005259(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r (see Straub, Section 1).
%C We conjecture below that the present sequence satisfies supercongruences similar to (but weaker than) the above supercongruences satisfied by the Apéry numbers.
%C More generally, we inductively define a family of sequences {a(i,n) : n >= 0}, i >= 0, by setting a(0,n) = A005259(n) and, for i >= 1, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. In this notation the present sequence is {a(1,n)}.
%C We conjecture that the sequences {a(i,n) : n >= 0}, i >= 1, satisfy the supercongruences u(n*p^r) == u(n*p^(r-1)) (mod p^(2*r)) for all primes p >= 3, and positive integers n and r.
%H F. Beukers, <a href="https://doi.org/10.1016/0022-314X(85)90047-2">Some congruences for the Apery numbers</a>, Journal of Number Theory, Vol. 21, Issue 2, Oct. 1985, pp. 141-155. <a href="/A339710/a339710.pdf">local copy</a>
%H Armin Straub, <a href="http://dx.doi.org/10.2140/ant.2014.8.1985">Multivariate Apéry numbers and supercongruences of rational functions</a>, Algebra & Number Theory, Vol. 8, No. 8 (2014), pp. 1985-2008; <a href="https://arxiv.org/abs/1401.0854">arXiv preprint</a>, arXiv:1401.0854 [math.NT], 2014.
%F Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for all primes p >= 3 and positive integers n and r.
%p A005259 := proc(n) add(binomial(n,k)^2*binomial(n+k,k)^2, k = 0..n) end proc:
%p E(n,x) := series(exp(n*add((A005259(k)*x^k)/k, k = 1..20)), x, 21):
%p seq(coeftayl(E(n,x), x = 0, n), n = 0..20);
%Y Cf. A005259, A362722 - A362733.
%K nonn,easy
%O 0,2
%A _Peter Bala_, May 02 2023
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