A372212 a(n) = [x^n] f(x)^n, where f(x) = (1 - x^7)^7/((1 - x^2)^2 * (1 - x^5)^5).

1, 0, 4, 0, 36, 25, 364, 441, 3876, 6561, 43779, 91839, 513900, 1245699, 6201199, 16645750, 76379940, 220760742, 955328863, 2916666288, 12090544611, 38466060066, 154437142545, 506976137710, 1987270052460, 6681958793775, 25724578443321, 88104794553729

Offset

0,3

Comments

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.

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.

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.

References

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

Links

Table of n, a(n) for n=0..27.

Formula

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.

Example

Supercongruences:
a(11) = 91839 = 3*(11^3)*23 == 0 (mod 11^3).
a(2*11) - a(2) = 154437142545 - 4 = (11^3)*2671*43441 == 0 (mod 11^3).

Maple

f(x) := (1 - x^7)^7/((1 - x^2)^2*(1 - x^5)^5):
seq(coeftayl(f(x)^n, x = 0, n), n = 0..27);

Crossrefs

Cf. A351858, A372211, A372213.

Sequence in context: A271292 A271094 A110947 * A270167 A270159 A270213

Adjacent sequences: A372208 A372210 A372211 * A372213 A372214 A372215

Keyword

nonn,easy

Author

Peter Bala, Apr 22 2024

Status

approved