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.

1, 2, 24, 236, 2432, 25752, 277152, 3019088, 33186816, 367378814, 4089875024, 45741207228, 513537853952, 5784253405192, 65332622356032, 739706089046736, 8392732289277952, 95401363286044260, 1086232605119042424, 12386037358495697292, 141422619808922418432, 1616691574828234720352

Offset: 0

Author

Peter Bala, Jan 27 2025

Keywords

• nonn
• easy

Comments

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).

Thus the present sequence satisfies the Gauss congruences. In fact, stronger congruences appear to hold for this sequence.

We conjecture that a(p) == 1 (mod p^3) for all primes p >= 5 (checked up to p = 61).

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.

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.

Examples

			Examples of supercongruences:
a(7) - a(1) = 3019088 - 2 = 2*(3^3)*(7^3)*163 == 0 (mod 7^3)
a(13) - a(1) = 5784253405192 - 2 = 2*5*(13^4)*20252279 == 0 (mod 13^4)
a(2*11) - a(2) = 18501616629347623668448 - 24 = (2^3)*(11^3)*17*1951*4243*9817*1257719 == 0 (mod 11^3)
a(5^2) - a(5) = 1884578634304981694792832319004 - 256504 = (2^2)*(5^6)*193381* 155926684363405438573 == 0 (mod 5^6)
		

Crossrefs

Cf. A001158, A015128, A206622, A380290, A380581, A380583.

Programs

Maple

with(numtheory):
G(x) := series(exp(add( (1/4)*(sigma[3](2*k) - sigma[3](k))*x^k/k, k = 1..23 )),x,24):
seq(coeftayl(G(x)^n, x = 0, n), n = 0..23);

Formula

a(n) = [x^n] exp( n*Sum_{k >= 1} (sigma_3(2*k) - sigma_3(k))/4 * x^k/k ).