A380583 -id:A380583 - cp's OEIS frontend

A380290 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} 1/(1 - x^k)^(k^2) is the g.f. of A023871.

1, 1, 11, 73, 539, 3976, 30107, 229811, 1771803, 13749742, 107305836, 841211966, 6619647419, 52258136399, 413682035393, 3282569032273, 26101575743771, 207930807629248, 1659134361686186, 13258065574274885, 106084302933126364, 849845499077000534, 6815530442695480418, 54712839001004065090

Offset: 0

Peter Bala, Jan 19 2025

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.

			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)

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

Vaclav Kotesovec, Table of n, a(n) for n = 0..1000
Peter Bala, Notes on A380290 and A380291

Cf. A023871, A001158, A023873, A283263, A380291, A380581, A380582, A380583.

Cf. A008485, A255672, A386720.

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);

Table[SeriesCoefficient[Product[1/(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[DivisorSigma[3, k]*x^k/k, {k, 1, n}]], {x, 0, n}], {n, 0, 25}] (* Vaclav Kotesovec, Jul 30 2025 *)

a(n) = [x^n] exp(n*Sum_{k >= 1} sigma_3(k)*x^k/k).

a(n) ~ c * d^n / sqrt(n), where d = 8.20432131153340331179513077696629277558952852444670658917204305357709... and c = 0.2513708881073263860977360125648021910598660424705749139651716452651... - Vaclav Kotesovec, Jul 30 2025

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

Peter Bala, Jan 27 2025

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 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)

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

Cf. A001160, A023873, A380290, A380582, A380583.

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);

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

A380582 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^k)/(1 - x^k))^(k^2) is the g.f. of A206622.

Original entry on oeis.org

1, 2, 24, 236, 2432, 25752, 277152, 3019088, 33186816, 367378814, 4089875024, 45741207228, 513537853952, 5784253405192, 65332622356032, 739706089046736, 8392732289277952, 95401363286044260, 1086232605119042424, 12386037358495697292, 141422619808922418432, 1616691574828234720352

Offset: 0

Peter Bala, Jan 27 2025

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, B be integers and let C be a positive integer. Define u(n) = [x^(C*n)] Product_{k >= 1} ((1 + x^k)^A * (1 - x^k)^B)^(k^2). The present sequence is the case A = 1, B = -1 and C = 1. We conjecture that the above supercongruences also hold for the sequence {u(n)} for all primes p >= 7.

			Examples of supercongruences:
a(7) - a(1) = 3019088 - 2 = 2*(3^3)*(7^3)*163 == 0 (mod 7^3)
a(13) - a(1) = 5784253405192 - 2 = 2*5*(13^4)*20252279 == 0 (mod 13^4)
a(2*11) - a(2) = 18501616629347623668448 - 24 = (2^3)*(11^3)*17*1951*4243*9817*1257719 == 0 (mod 11^3)
a(5^2) - a(5) = 1884578634304981694792832319004 - 256504 = (2^2)*(5^6)*193381* 155926684363405438573 == 0 (mod 5^6)

Cf. A001158, A015128, A206622, A380290, A380581, A380583.

with(numtheory):
G(x) := series(exp(add( (1/4)*(sigma[3](2*k) - sigma[3](k))*x^k/k, k = 1..23 )),x,24):
seq(coeftayl(G(x)^n, x = 0, n), n = 0..23);

a(n) = [x^n] exp( n*Sum_{k >= 1} (sigma_3(2*k) - sigma_3(k))/4 * x^k/k ).

cp's OEIS Frontend

A380290 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} 1/(1 - x^k)^(k^2) is the g.f. of A023871.

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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.

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A380582 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^k)/(1 - x^k))^(k^2) is the g.f. of A206622.

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Examples

Crossrefs

Programs

Maple

Formula