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

%I A380583 #15 Jan 31 2025 04:26:49
%S A380583 1,1,13,82,665,5026,40180,319677,2583401,20965150,171276238,
%T A380583 1405008925,11571476120,95601033542,792038546739,6577523807332,
%U A380583 54737967873385,456368114019558,3811136362823056,31873576059000827,266919720010452190,2237944814420991135,18784073017650350445
%N A380583 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^(2*k))/(1 - x^k))^(k^2).
%C A380583 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.
%C A380583 We conjecture that a(p) == 1 (mod p^3) for all primes p >= 7.
%C A380583 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.
%C A380583 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.
%e A380583 Examples of supercongruences:
%e A380583 a(7) - a(1) = 319677 - 1 = (2^2)*(7^3)*233 == 0 (mod 7^3)
%e A380583 a(11) - a(1) = 1405008925 - 1 = (2^2)*3*(11^5)*727 == 0 (mod 11^5)
%e A380583 a(22) - a(2) = 18784073017650350445 - 13 = (2^5)*(11^3)*222773*1979699077 == 0 (mod 11^3)
%p A380583 G(x) := series(mul( ( (1 + x^(2*k))/(1 - x^k) )^(k^2), k = 1..22), x, 23):
%p A380583 a:= n-> coeftayl(G(x)^n, x = 0, n):
%p A380583 seq(a(n), n = 0..22);
%Y A380583 Cf. A380290, A380581, A380582.
%K A380583 nonn,easy
%O A380583 0,3
%A A380583 _Peter Bala_, Jan 27 2025