A380582 -id:A380582 - cp's OEIS frontend

A206622 G.f.: Product_{n>0} ( (1+x^n)/(1-x^n) )^(n^2).

1, 2, 10, 36, 118, 376, 1148, 3376, 9654, 26894, 73192, 195188, 510948, 1315048, 3332720, 8326448, 20529526, 49998884, 120379574, 286726340, 676057144, 1578880480, 3654180236, 8385122192, 19085029540, 43103203626, 96630606968, 215105226728, 475608824400

Offset: 0

Paul D. Hanna, Feb 10 2012

nonn

Compare g.f. to: Product_{n>0} (1+x^n)/(1-x^n) = exp( Sum_{n>=1} (sigma(2*n) - sigma(n))*x^n/n ) which equals 1/theta_4(x) = 1/(1 + 2*Sum_{n>=1} (-x)^(n^2)).
Convolution of A023871 and A027998. - Vaclav Kotesovec, Aug 19 2015
In general, if g.f. = Product_{k>=1} ((1 + x^k)/(1 - x^k))^(c2*k^2 + c1*k + c0) and c2>0, then a(n) ~ exp(Pi * 2^(5/4) * c2^(1/4) * n^(3/4) / 3 + 7*c1 * Zeta(3) * sqrt(n) / (Pi^2 * sqrt(2*c2)) + (c0*Pi / (2^(5/4) * c2^(1/4)) - 49*c1^2 * Zeta(3)^2 / (2^(5/4) * c2^(5/4) * Pi^5)) * n^(1/4) + 22411 * c1^3 * Zeta(3)^3 / (196 * c2^2 * Pi^8) - 7*c0*c1 * Zeta(3) / (4*c2 * Pi^2) - c2 * Zeta(3) / (4*Pi^2) + c1/12) * Pi^(c1/12) * c2^(1/8 + c0/8 + c1/48) / (A^c1 * 2^(15/8 + 11*c0/8 + 7*c1/48) * n^(5/8 + c0/8 + c1/48)), where A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Nov 08 2017
Let A(x) denote the g.f. and let m be an integer. Define a sequence by u(n) = [x^n] A(x)^(m*n). We conjecture that the supercongruence u(n*p^r) == u(n*p^(r-1)) (mod p^(3*r)) holds for all positive integers n and r and all primes p >= 5. Cf. A380582. - Peter Bala, Jan 21 2025

			G.f.: A(x) = 1 + 2*x + 10*x^2 + 36*x^3 + 118*x^4 + 376*x^5 + 1148*x^6 +...
where A(x) = (1+x)/(1-x) * (1+x^2)^4/(1-x^2)^4 * (1+x^3)^9/(1-x^3)^9 *...
Also, A(x) = Euler transform of [2,7,18,28,50,63,98,112,162,175,...]:
A(x) = 1/((1-x)^2*(1-x^2)^7*(1-x^3)^18*(1-x^4)^28*(1-x^5)^50*(1-x^6)^63*...).

Seiichi Manyama, Table of n, a(n) for n = 0..9042 (terms 0..1000 from Vaclav Kotesovec)
Vaclav Kotesovec, A method of finding the asymptotics of q-series based on the convolution of generating functions, arXiv:1509.08708 [math.CO], Sep 30 2015, p. 23.

Cf. A156616, A206623, A206624, A001158 (sigma_3), A380582.

nmax = 40; CoefficientList[Series[Product[((1+x^k)/(1-x^k))^(k^2), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 19 2015 *)

{a(n)=polcoeff(prod(m=1,n+1,((1+x^m)/(1-x^m+x*O(x^n)))^(m^2)),n)}

{a(n)=polcoeff(exp(sum(m=1, n, (sigma(2*m, 3)-sigma(m, 3))/4*x^m/m)+x*O(x^n)), n)}

{a(n)=local(InvEulerGF=x*(2+7*x+12*x^2+7*x^3+2*x^4)/(1-x^2+x*O(x^n))^3);polcoeff(1/prod(k=1,n,(1-x^k+x*O(x^n))^polcoeff(InvEulerGF,k)),n)}
for(n=0,35,print1(a(n),", "))

G.f.: exp( Sum_{n>=1} (sigma_3(2*n) - sigma_3(n))/4 * x^n/n ), where sigma_3(n) is the sum of cubes of divisors of n (A001158).
The inverse Euler transform has g.f.: x*(2 + 7*x + 12*x^2 + 7*x^3 + 2*x^4)/(1-x^2)^3.
a(n) ~ exp(2^(5/4)*Pi*n^(3/4)/3 - Zeta(3)/(4*Pi^2)) / (2^(15/8) * n^(5/8)), where Zeta(3) = A002117. - Vaclav Kotesovec, Aug 19 2015
a(0) = 1, a(n) = (2/n)*Sum_{k=1..n} A007331(k)*a(n-k) for n > 0. - Seiichi Manyama, Apr 30 2017

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.

Original entry on oeis.org

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).

A380583 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^(2*k))/(1 - x^k))^(k^2).

Original entry on oeis.org

1, 1, 13, 82, 665, 5026, 40180, 319677, 2583401, 20965150, 171276238, 1405008925, 11571476120, 95601033542, 792038546739, 6577523807332, 54737967873385, 456368114019558, 3811136362823056, 31873576059000827, 266919720010452190, 2237944814420991135, 18784073017650350445

Offset: 0

Peter Bala, Jan 27 2025

The sequence satisfies the Gauss congruences: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and for all positive integers n and r. In fact, stronger congruences appear to hold for this sequence.
We conjecture that a(p) == 1 (mod p^3) for all primes p >= 7.
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.
Let A and B be integers. Let C be a positive integer. Define u(n) = [x^(C*n)] Product_{k >= 1} ((1 + - x^(2*k))^A * (1 + - x^k)^B)^(k^2). The present sequence is the case A = 1, B = -1 and C = 1, with the appropriate choice of signs. 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) = 319677 - 1 = (2^2)*(7^3)*233 == 0 (mod 7^3)
a(11) - a(1) = 1405008925 - 1 = (2^2)*3*(11^5)*727 == 0 (mod 11^5)
a(22) - a(2) = 18784073017650350445 - 13 = (2^5)*(11^3)*222773*1979699077 == 0 (mod 11^3)

Cf. A380290, A380581, A380582.

G(x) := series(mul( ( (1 + x^(2*k))/(1 - x^k) )^(k^2), k = 1..22), x, 23):
a:= n-> coeftayl(G(x)^n, x = 0, n):
seq(a(n), n = 0..22);

cp's OEIS Frontend

A206622 G.f.: Product_{n>0} ( (1+x^n)/(1-x^n) )^(n^2).

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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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A380583 a(n) = [x^n] G(x)^n, where G(x) = Product_{k >= 1} ((1 + x^(2*k))/(1 - x^k))^(k^2).

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