A380583 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^(2*k))/(1 - x^k))^(k^2).

1, 1, 13, 82, 665, 5026, 40180, 319677, 2583401, 20965150, 171276238, 1405008925, 11571476120, 95601033542, 792038546739, 6577523807332, 54737967873385, 456368114019558, 3811136362823056, 31873576059000827, 266919720010452190, 2237944814420991135, 18784073017650350445

Offset: 0

Author

Peter Bala, Jan 27 2025

Keywords

• nonn
• easy

Comments

The sequence satisfies the Gauss congruences: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and for all positive integers n and r. In fact, stronger congruences appear to hold for this sequence.

We conjecture that a(p) == 1 (mod p^3) for all primes p >= 7.

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.

Let A and B be integers. Let C be a positive integer. Define u(n) = [x^(C*n)] Product_{k >= 1} ((1 + - x^(2*k))^A * (1 + - x^k)^B)^(k^2). The present sequence is the case A = 1, B = -1 and C = 1, with the appropriate choice of signs. 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) = 319677 - 1 = (2^2)*(7^3)*233 == 0 (mod 7^3)
a(11) - a(1) = 1405008925 - 1 = (2^2)*3*(11^5)*727 == 0 (mod 11^5)
a(22) - a(2) = 18784073017650350445 - 13 = (2^5)*(11^3)*222773*1979699077 == 0 (mod 11^3)
		

Crossrefs

Cf. A380290, A380581, A380582.

Programs

Maple

G(x) := series(mul( ( (1 + x^(2*k))/(1 - x^k) )^(k^2), k = 1..22), x, 23):
a:= n-> coeftayl(G(x)^n, x = 0, n):
seq(a(n), n = 0..22);