A372213 a(n) = [x^n] f(x)^n, where f(x) = (1 - x^7)^7/((1 - x^3)^3 * (1 - x^4)^4).

1, 0, 0, 9, 16, 0, 171, 539, 528, 3654, 16500, 29282, 101851, 483340, 1215445, 3416634, 14564880, 44585475, 124007202, 462804166, 1555048516, 4547401595, 15500748802, 53459717443, 164998563675, 538593687500, 1845162146828, 5920282930815, 5920282930815

Offset

0,4

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..28.

Formula

The o.g.f. A(x) = 1 + 9*x^3 + 16*x^4 + 171*x^6 + ... 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) = 29282 = 2*(11^4) == 0 (mod 11^4).
a(13) = 483340 = (2^2)*5*11*(13^3) == 0 (mod 13^3).
a(2*11) = 15500748802 = 2*7*(11^4)*47*1609 == 0 (mod 11^4).

Maple

f(x) := (1 - x^7)^7/((1 - x^3)^3*(1 - x^4)^4):
seq(coeftayl(f(x)^n, x = 0, n), n = 0..30);

Crossrefs

Cf. A351858, A372211, A372212.

Sequence in context: A373332 A093595 A330429 * A215418 A133816 A048752

Adjacent sequences: A372210 A372211 A372212 * A372214 A372215 A372216

Keyword

nonn,easy

Author

Peter Bala, Apr 22 2024

Status

approved