cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A380581 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} 1/(1 - x^k)^(k^4) is the g.f. of A023873.

Original entry on oeis.org

1, 1, 35, 397, 5075, 67126, 897911, 12144945, 165880531, 2280262825, 31522512910, 437730330357, 6101414176535, 85317965576325, 1196299277106675, 16813979471920522, 236812229975204563, 3341448338530887015, 47225228515043980715, 668417245247747877735, 9473101371364286661950, 134416752857691389968377, 1909344928242571795580255
Offset: 0

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Author

Peter Bala, Jan 27 2025

Keywords

Comments

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 this sequence.
We conjecture that a(p) == 1 (mod p^3) for all primes p >= 5 (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 >= 5 and positive integers n and r. Some examples are given below.
Let A >= 1 and B be integers. Define a sequence {u(n) : n >= 0} by u(n) = [x^(A*n)] G(x)^(B*n), so the present sequence is the case A = B = 1. We conjecture that the above supercongruences also hold for the sequence {u(n)}.

Examples

			Examples of supercongruences:
a(7) - a(1) = 12144945 - 1 = (2^4)*(7^3)*2213 = 0 (mod 7^3)
a(3*7) - a(3) = 134416752857691389968377 - 397 = (2^2)*5*(7^3)*17*223*5168630662682423 == 0 (mod 7^3)
a(2*11) - a(2) = 1909344928242571795580255 - 35 = (2^2)*(3^4)*5*7*(11^4)*17*23* 29411951377843 == 0 (mod 11^4)
		

References

  • R. P. Stanley. Enumerative combinatorics. Vol. 2, volume 62 of Cambridge Studies in Advanced Mathematics. Cambridge University Press, Cambridge, 1999.

Crossrefs

Programs

  • Maple
    with(numtheory):
    G(x) := series(exp(add(sigma[5](k)*x^k/k, k = 1..22)), x, 23):
    seq(coeftayl(G(x)^n, x = 0, n), n = 0..22);

Formula

a(n) = [x^n] exp(n*Sum_{k >= 1} sigma_5(k)*x^k/k).