cp's OEIS Frontend

From Peter Bala, Apr 18 2023: (Start)

The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and all positive integers n and k.

Conjecture: the stronger supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for all primes p >= 3 and all positive integers n and k. (End)

Given an integer sequence {f(n) : n >= 0} with f(0) = 1, there is a unique power series F(x) with rational coefficients, where F(0) = 1, 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.