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a(n) = [x^n] f(x)^n, where f(x) = (1 - x^7)^7/((1 - x^2)^2 * (1 - x^5)^5).
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%I #6 Apr 25 2024 13:43:52

%S 1,0,4,0,36,25,364,441,3876,6561,43779,91839,513900,1245699,6201199,

%T 16645750,76379940,220760742,955328863,2916666288,12090544611,

%U 38466060066,154437142545,506976137710,1987270052460,6681958793775,25724578443321,88104794553729

%N a(n) = [x^n] f(x)^n, where f(x) = (1 - x^7)^7/((1 - x^2)^2 * (1 - x^5)^5).

%C Let G(x) be a formal power series with integer coefficients. The sequence defined by g(n) = [x^n] G(x)^n satisfies the Gauss congruences: g(n*p^r) == g(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.

%C We conjecture that in this case the stronger supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for primes p >= 11 and positive integers n and r. Some examples are given below. Cf. A351858.

%C More generally, if r is a positive integer and s an integer then the sequence defined by u(r,s; n) = [x^(r*n)] f(x)^(s*n) may satisfy the same supercongruences.

%D R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

%F The o.g.f. A(x) = 1 + 4*x^2 + 36*x^4 + 25*x^5 + ... is the diagonal of the bivariate rational function 1/(1 - t*f(x)) and hence is an algebraic function over the field of rational functions Q(x) by Stanley, Theorem 6.33, p. 197.

%e Supercongruences:

%e a(11) = 91839 = 3*(11^3)*23 == 0 (mod 11^3).

%e a(2*11) - a(2) = 154437142545 - 4 = (11^3)*2671*43441 == 0 (mod 11^3).

%p f(x) := (1 - x^7)^7/((1 - x^2)^2*(1 - x^5)^5):

%p seq(coeftayl(f(x)^n, x = 0, n), n = 0..27);

%Y Cf. A351858, A372211, A372213.

%K nonn,easy

%O 0,3

%A _Peter Bala_, Apr 22 2024