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

Author

Paul D. Hanna, Feb 10 2012

Keywords

• nonn

Comments

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

Examples

			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*...).
		

Links

• 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.

Crossrefs

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

Programs

Mathematica

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 *)

PARI

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

PARI

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

PARI

{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),", "))

Formula

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