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A380582 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^k)/(1 - x^k))^(k^2) is the g.f. of A206622.

%I A380582 #15 Jul 31 2025 02:29:58
%S A380582 1,2,24,236,2432,25752,277152,3019088,33186816,367378814,4089875024,
%T A380582 45741207228,513537853952,5784253405192,65332622356032,
%U A380582 739706089046736,8392732289277952,95401363286044260,1086232605119042424,12386037358495697292,141422619808922418432,1616691574828234720352
%N A380582 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^k)/(1 - x^k))^(k^2) is the g.f. of A206622.
%C A380582 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 A380582 Thus the present sequence satisfies the Gauss congruences. In fact, stronger congruences appear to hold for this sequence.
%C A380582 We conjecture that a(p) == 1 (mod p^3) for all primes p >= 5 (checked up to p = 61).
%C A380582 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 A380582 Let A, B be integers and let C be a positive integer. Define u(n) = [x^(C*n)] Product_{k >= 1} ((1 + x^k)^A * (1 - x^k)^B)^(k^2). The present sequence is the case A = 1, B = -1 and C = 1. We conjecture that the above supercongruences also hold for the sequence {u(n)} for all primes p >= 7.
%F A380582 a(n) = [x^n] exp( n*Sum_{k >= 1} (sigma_3(2*k) - sigma_3(k))/4 * x^k/k ).
%e A380582 Examples of supercongruences:
%e A380582 a(7) - a(1) = 3019088 - 2 = 2*(3^3)*(7^3)*163 == 0 (mod 7^3)
%e A380582 a(13) - a(1) = 5784253405192 - 2 = 2*5*(13^4)*20252279 == 0 (mod 13^4)
%e A380582 a(2*11) - a(2) = 18501616629347623668448 - 24 = (2^3)*(11^3)*17*1951*4243*9817*1257719 == 0 (mod 11^3)
%e A380582 a(5^2) - a(5) = 1884578634304981694792832319004 - 256504 = (2^2)*(5^6)*193381* 155926684363405438573 == 0 (mod 5^6)
%p A380582 with(numtheory):
%p A380582 G(x) := series(exp(add( (1/4)*(sigma[3](2*k) - sigma[3](k))*x^k/k, k = 1..23 )),x,24):
%p A380582 seq(coeftayl(G(x)^n, x = 0, n), n = 0..23);
%Y A380582 Cf. A001158, A015128, A206622, A380290, A380581, A380583.
%K A380582 nonn,easy
%O A380582 0,2
%A A380582 _Peter Bala_, Jan 27 2025