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 all primes p >= 5 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 for primes p >= 7.
Even more generally, if a, b and m are a positive integers, with m = a + b, then the sequences whose n-th term equals [x^n] (1 - x^m)^m/((1 - x^a)^a * (1 - x^b)^b) or [x^n] (1 - x^m)^m/((1 + x^a)^a * (1 + x^b)^b) may both satisfy the above supercongruences for sufficiently large primes p depending on m.
The sequence of central binomial coefficients A000984 corresponds to the case m = 2 and a = b = 1.
REFERENCES
R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.
FORMULA
The o.g.f. A(x) = 1 + 4*x^2 + 9*x^3 + 36*x^4 + ... 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) = 419265 = (3^2)*5*7*11^3 == 0 (mod 11^3).
a(23) = 6541552959219 = (3^2)*(23^3)*59738573 == 0 (mod 23^3).
a(2*7) - a(2) = 26081381 - 4 = (7^3)*76039 == 0 (mod 7^3).
MAPLE
f(x) := (1 - x^5)^5/((1 - x^2)^2*(1 - x^3)^3):
seq(coeftayl(f(x)^n, x = 0, n), n = 0..27);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Apr 22 2024
STATUS
approved