A380291 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} (1 + x^k)^(k^2) is the g.f. of A027998.

1, 1, 9, 64, 425, 3026, 21672, 157095, 1149289, 8464240, 62683134, 466307865, 3482008904, 26083955002, 195932407939, 1475267031164, 11131100990825, 84140066313620, 637054366975740, 4830417047590165, 36674477204674750, 278779034863684377, 2121418004609211361, 16159262748227985561

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, 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 supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold 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

Vaclav Kotesovec, Table of n, a(n) for n = 0..1000

Formula

a(n) = [x^n] exp(n*Sum_{k >= 1} s_3(k)*x^k/k), where s_3(n) = Sum_{d divides n} (-1)^(n/d+1)*d^3 = A078307(n).

a(n) ~ c * d^n / sqrt(n), where d = 7.7846790125019502578773343468308844201627754275100035492213697757399421948... and c = 0.2484592487737716543953469621097743519172686743284742545545347906986158... - Vaclav Kotesovec, Jul 30 2025

Example

Examples of supercongruences:
a(7) - a(1) = 157095 - 1 = 2*(7^3)*229 == 0 (mod 7^3)
a(11) - a(1) = 466307865 - 1 = (2^3)*(11^3)*43793 == 0 (mod 11^3)
a(3*7) - a(3) = 278779034863684377 - 64 = (7^4)*43*26891*100413601 == 0 (mod 7^4)

Maple

with(numtheory):
s_3 := n-> add((-1)^(n/d+1)*d^3, d in divisors(n)):
G(x) := series(exp(add(s_3(k)*x^k/k, k = 1..23)), x, 24):
seq(coeftayl(G(x)^n, x = 0, n), n = 0..23);

Mathematica

Table[SeriesCoefficient[Product[(1 + x^k)^(n*k^2), {k, 1, n}], {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Jul 30 2025 *)
(* or *)
Table[SeriesCoefficient[Exp[n*Sum[Sum[(-1)^(k/d + 1)*d^3, {d, Divisors[k]}]*x^k/k, {k, 1, n}]], {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Jul 30 2025 *)

Crossrefs

Cf. A027998, A078307, A380290.

Cf. A270913, A270922.

Sequence in context: A143631 A083328 A000846 * A357649 A231822 A049684

Adjacent sequences: A380288 A380289 A380290 * A380292 A380293 A380294

Keyword

nonn,easy

Author

Peter Bala, Jan 19 2025

Status

approved