A128128 Expansion of chi(-q^3) / chi^3(-q) in powers of q where chi() is a Ramanujan theta function.
1, 3, 6, 12, 21, 36, 60, 96, 150, 228, 342, 504, 732, 1050, 1488, 2088, 2901, 3996, 5460, 7404, 9972, 13344, 17748, 23472, 30876, 40413, 52644, 68268, 88152, 113364, 145224, 185352, 235734, 298800, 377514, 475488, 597108, 747690, 933672, 1162824
Offset: 0
Keywords
Examples
G.f. = 1 + 3*q + 6*q^2 + 12*q^3 + 21*q^4 + 36*q^5 + 60*q^6 + 96*q^7 + ...
Links
- G. C. Greubel, Table of n, a(n) for n = 0..1000
- Shane Chern, Dennis Eichhorn, Shishuo Fu, and James A. Sellers, Convolutive sequences, I: Through the lens of integer partition functions, arXiv:2507.10965 [math.CO], 2025. See pp. 4, 11, 13.
- 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.
- Michael Somos, Introduction to Ramanujan theta functions
- Eric Weisstein's World of Mathematics, Ramanujan Theta Functions
Programs
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Mathematica
a[ n_] := SeriesCoefficient[ QPochhammer[ q^2]^3 QPochhammer[ q^3] / (QPochhammer[ q]^3 QPochhammer[ q^6]), {q, 0, n}]; (* Michael Somos, Feb 19 2015 *) nmax=60; CoefficientList[Series[Product[(1-x^(2*k))^3 * (1-x^(3*k)) / ((1-x^k)^3 * (1-x^(6*k))),{k,1,nmax}],{x,0,nmax}],x] (* Vaclav Kotesovec, Oct 13 2015 *)
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PARI
{a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A)^3 * eta(x^3 + A) / (eta(x + A)^3 * eta(x^6 + A)), n))};
Formula
Expansion of eta(q^2)^3 * eta(q^3) / (eta(q)^3 * eta(q^6)) in powers of q.
Euler transform of period 6 sequence [ 3, 0, 2, 0, 3, 0, ...].
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = u^2 + v - 2*u*v^2.
G.f. A(x) satisfies 0 = f(A(x), A(x^3)) where f(u, v) = (u + u^2 + u^3) - v^3*(1 - 2*u + 4*u^2).
G.f. A(x) satisfies 0 = f(A(x), A(x^5)) where f(u, v) = u^6 + v^6 - 16*u^5*v^5 + 20*u^4*v^4 + 10*u^2*v^2*(u^3 + v^3) - 20*u^3*v^3 - 5*u*v*(u^3 + v^3) + 5*u^2*v^2 - u*v.
Expansion of b(q^2) / b(q) in powers of q where b() is a cubic AGM theta function.
G.f. is a period 1 Fourier series which satisfies f(-1 / (18 t)) = (1/2) g(t) where q = exp(2 Pi i t) and g() is the g.f. for A062242.
a(n) = 3*A128129(n) unless n=0.
Convolution inverse of A141094. - Michael Somos, Feb 19 2015
a(n) ~ exp(2*sqrt(2*n)*Pi/3) / (2^(7/4) * sqrt(3) * n^(3/4)). - Vaclav Kotesovec, Oct 13 2015
Comments