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A380581 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} 1/(1 - x^k)^(k^4) is the g.f. of A023873.

%I A380581 #18 Jul 31 2025 02:29:46
%S A380581 1,1,35,397,5075,67126,897911,12144945,165880531,2280262825,
%T A380581 31522512910,437730330357,6101414176535,85317965576325,
%U A380581 1196299277106675,16813979471920522,236812229975204563,3341448338530887015,47225228515043980715,668417245247747877735,9473101371364286661950,134416752857691389968377,1909344928242571795580255
%N A380581 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} 1/(1 - x^k)^(k^4) is the g.f. of A023873.
%C A380581 Given an integer sequence {f(n) : n >= 0} with f(0) = 1, there is a unique power series F(x) with rational coefficients, where F(0) = 1, such that f(n) = [x^n] F(x)^n. F(x) is given by F(x) = series_reversion(x/E(x)), where E(x) = exp(Sum_{n >= 1} f(n)*x^n/n). Furthermore, if the series E(x) has integer coefficients then the series F(x) also has integer coefficients and the sequence {f(n)} satisfies the Gauss congruences: f(n*p^r) == f(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r (by Stanley, Ch. 5, Ex. 5.2(a), p. 72 and the Lagrange inversion formula).
%C A380581 Thus the present sequence satisfies the Gauss congruences. In fact, stronger congruences appear to hold for this sequence.
%C A380581 We conjecture that a(p) == 1 (mod p^3) for all primes p >= 5 (checked up to p = 61).
%C A380581 More generally, we conjecture that the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 5 and positive integers n and r. Some examples are given below.
%C A380581 Let A >= 1 and B be integers. Define a sequence {u(n) : n >= 0} by u(n) = [x^(A*n)] G(x)^(B*n), so the present sequence is the case A = B = 1. We conjecture that the above supercongruences also hold for the sequence {u(n)}.
%D A380581 R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.
%F A380581 a(n) = [x^n] exp(n*Sum_{k >= 1} sigma_5(k)*x^k/k).
%e A380581 Examples of supercongruences:
%e A380581 a(7) - a(1) = 12144945 - 1 = (2^4)*(7^3)*2213 = 0 (mod 7^3)
%e A380581 a(3*7) - a(3) = 134416752857691389968377 - 397 = (2^2)*5*(7^3)*17*223*5168630662682423 == 0 (mod 7^3)
%e A380581 a(2*11) - a(2) = 1909344928242571795580255 - 35 = (2^2)*(3^4)*5*7*(11^4)*17*23* 29411951377843 == 0 (mod 11^4)
%p A380581 with(numtheory):
%p A380581 G(x) := series(exp(add(sigma[5](k)*x^k/k, k = 1..22)), x, 23):
%p A380581 seq(coeftayl(G(x)^n, x = 0, n), n = 0..22);
%Y A380581 Cf. A001160, A023873, A380290, A380582, A380583.
%K A380581 nonn,easy
%O A380581 0,3
%A A380581 _Peter Bala_, Jan 27 2025