OFFSET
0,3
COMMENTS
Given an integer sequence {f(n) : n >= 0} with f(0) = 1, there is a unique power series F(x), with rational coefficients, such that f(n) = [x^n] F(x)^n. F(x) is given by F(x) = series_reversion(x/E(x)), where E(x) = exp(Sum_{n >= 1} f(n)*x^n/n). Furthermore, if the series E(x) has integer coefficients then the series F(x) also has integer coefficients and the sequence {f(n)} satisfies the Gauss congruences: f(n*p^r) == f(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r (by Stanley, Ch. 5, Ex. 5.2(a), p. 72 and the Lagrange inversion formula).
Thus the present sequence satisfies the Gauss congruences. In fact, stronger congruences appear to hold for the present sequence.
We conjecture that a(p) == 1 (mod p^3) for all primes p >= 7 (checked up to p = 61).
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.
REFERENCES
R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.
LINKS
FORMULA
a(n) = [x^n] exp(n*Sum_{k >= 1} sigma_3(k)*x^k/k).
EXAMPLE
Examples of supercongruences:
a(7) - a(1) = 229811 - 1 = 2*5*(7^3)*67 == 0 (mod 7^3)
a(3*7) - a(3) = 849845499077000534 - 73 = (7^3)*29243*84727410689 == 0 (mod 7^3)
a(19) - a(1) = 13258065574274885 - 1 = (2^2)*11*(19^3)*29*26723*56687 == 0 (mod 19^3)
MAPLE
with(numtheory):
G(x) := series(exp(add(sigma[3](k)*x^k/k, k = 1..23)), x, 24):
seq(coeftayl(G(x)^n, x = 0, n), n = 0..23);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Jan 19 2025
STATUS
approved