A143752 Expansion of eta(q^3) * eta(q^4) * eta(q^5) * eta(q^60) / (eta(q) * eta(q^12) * eta(q^15) * eta(q^20)) in powers of q.
1, 1, 2, 2, 3, 3, 4, 5, 6, 7, 8, 10, 11, 14, 17, 20, 23, 27, 31, 36, 41, 48, 55, 63, 72, 82, 94, 106, 122, 137, 156, 175, 197, 222, 249, 280, 314, 352, 393, 439, 490, 546, 608, 676, 751, 834, 923, 1024, 1133, 1253, 1384, 1528, 1686, 1857, 2045, 2250, 2474, 2718
Offset: 1
Keywords
Examples
G.f. = q + q^2 + 2*q^3 + 2*q^4 + 3*q^5 + 3*q^6 + 4*q^7 + 5*q^8 + 6*q^9 + 7*q^10 + ...
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..10000
- Michael Somos, A Remarkable eta-product Identity
Programs
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PARI
{a(n) = my(A); if( n<1, 0, n--; A = x * O(x^n); polcoeff( eta(x^3 + A) * eta(x^4 + A) * eta(x^5 + A) * eta(x^60 + A) / (eta(x + A) * eta(x^12 + A) * eta(x^15 + A) * eta(x^20 + A)), n))};
Formula
Expansion of F(q) * F(q^2) in powers of q^3 where F(q) is the g.f. for A103263.
Euler transform of a period 60 sequence.
G.f. is a period 1 Fourier series which satisfies f(-1 / (60 t)) = g(t) where q = exp(2 Pi i t) and g() is the g.f. for A143751.
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = (u^2 + v^2) * (1 + u + v) * (u + v + u*v) - u*v * (1+ 2*u + 2*v + u*v)^2.
G.f.: x * Product_{k>0} P(30, x^k) * P(60, x^k) where P(n, x) is the n-th cyclotomic polynomial.
G.f.: -1 + Product_{k>0} (1 + x^k) * (1 + x^(15*k)) / ((1 + x^(6*k)) * (1 + x^(10*k))). - Seiichi Manyama, May 04 2017
a(n) ~ exp(2*Pi*sqrt(n/15)) / (2 * 15^(1/4) * n^(3/4)). - Vaclav Kotesovec, Jun 03 2018