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A380290 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} 1/(1 - x^k)^(k^2) is the g.f. of A023871.

%I A380290 #31 Jul 31 2025 07:03:36
%S A380290 1,1,11,73,539,3976,30107,229811,1771803,13749742,107305836,841211966,
%T A380290 6619647419,52258136399,413682035393,3282569032273,26101575743771,
%U A380290 207930807629248,1659134361686186,13258065574274885,106084302933126364,849845499077000534,6815530442695480418,54712839001004065090
%N A380290 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} 1/(1 - x^k)^(k^2) is the g.f. of A023871.
%C A380290 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 A380290 Thus the present sequence satisfies the Gauss congruences. In fact, stronger congruences appear to hold for the present sequence.
%C A380290 We conjecture that a(p) == 1 (mod p^3) for all primes p >= 7 (checked up to p = 61).
%C A380290 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 >= 7 and positive integers n and r. Some examples are given below.
%D A380290 R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.
%H A380290 Vaclav Kotesovec, <a href="/A380290/b380290.txt">Table of n, a(n) for n = 0..1000</a>
%H A380290 Peter Bala, <a href="/A380290/a380290.pdf">Notes on A380290 and A380291</a>
%F A380290 a(n) = [x^n] exp(n*Sum_{k >= 1} sigma_3(k)*x^k/k).
%F A380290 a(n) ~ c * d^n / sqrt(n), where d = 8.20432131153340331179513077696629277558952852444670658917204305357709... and c = 0.2513708881073263860977360125648021910598660424705749139651716452651... - _Vaclav Kotesovec_, Jul 30 2025
%e A380290 Examples of supercongruences:
%e A380290 a(7) - a(1) = 229811 - 1 = 2*5*(7^3)*67 == 0 (mod 7^3)
%e A380290 a(3*7) - a(3) = 849845499077000534 - 73 = (7^3)*29243*84727410689 == 0 (mod 7^3)
%e A380290 a(19) - a(1) = 13258065574274885 - 1 = (2^2)*11*(19^3)*29*26723*56687 == 0 (mod 19^3)
%p A380290 with(numtheory):
%p A380290 G(x) := series(exp(add(sigma[3](k)*x^k/k, k = 1..23)),x,24):
%p A380290 seq(coeftayl(G(x)^n, x = 0, n), n = 0..23);
%t A380290 Table[SeriesCoefficient[Product[1/(1 - x^k)^(n*k^2), {k, 1, n}], {x, 0, n}], {n, 0, 25}] (* _Vaclav Kotesovec_, Jul 30 2025 *)
%t A380290 (* or *)
%t A380290 Table[SeriesCoefficient[Exp[n*Sum[DivisorSigma[3, k]*x^k/k, {k, 1, n}]], {x, 0, n}], {n, 0, 25}] (* _Vaclav Kotesovec_, Jul 30 2025 *)
%Y A380290 Cf. A023871, A001158, A023873, A283263, A380291, A380581, A380582, A380583.
%Y A380290 Cf. A008485, A255672, A386720.
%K A380290 nonn,easy
%O A380290 0,3
%A A380290 _Peter Bala_, Jan 19 2025