A121593 Expansion of (eta(q^7) / eta(q))^4 in powers of q.

1, 4, 14, 40, 105, 252, 574, 1236, 2564, 5124, 9948, 18788, 34685, 62664, 111132, 193672, 332325, 561996, 937958, 1546132, 2519825, 4062888, 6486008, 10257324, 16079389, 24996636, 38555216, 59025820, 89728900, 135486960, 203274344

Offset: 1

Author

Michael Somos, Aug 09 2006

Keywords

• nonn

Comments

G.f. A(q) is denoted by tau(q) / 49 in Klein and Fricke 1890.

Examples

			G.f. = q + 4*q^2 + 14*q^3 + 40*q^4 + 105*q^5 + 252*q^6 + 574*q^7 + ...
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000
• Kevin Acres, David Broadhurst, Eta quotients and Rademacher sums, arXiv:1810.07478 [math.NT], 2018. See Table 1 p. 10.
• F. Klein and R. Fricke, Vorlesungen über die Theorie der elliptischen Modulfunctionen, Teubner, Leipzig, 1890, Vol. 1, see p. 745, Eq. (3).

Crossrefs

Cf. A030181.

Programs

Mathematica

a[ n_] := SeriesCoefficient[ q (QPochhammer[ q^7] / QPochhammer[ q])^4, {q, 0, n}]; (* Michael Somos, Jan 02 2015 *)
nmax = 40; Rest[CoefficientList[Series[x * Product[((1 - x^(7*k)) / (1 - x^k))^4, {k, 1, nmax}], {x, 0, nmax}], x]] (* Vaclav Kotesovec, Sep 07 2015 *)
eta[q_]:=q^(1/6) QPochhammer[q]; a[n_]:=SeriesCoefficient[(eta[q^7] / eta[q])^4, {q, 0, n}]; Table[a[n], {n, 4, 35}] (* Vincenzo Librandi, Oct 18 2018 *)

PARI

{a(n) = my(A); if( n<1, 0, n--; A = x * O(x^n); polcoeff( (eta(x^7 + A) / eta(x + A))^4, n))};

Formula

Euler transform of period 7 sequence [4, 4, 4, 4, 4, 4, 0, ...].

G.f.: x * (Product_{k>0} (1 - x^(7*k)) / (1 - x^k))^4.

G.f. A(q) satisfies 0 = f(A(q), A(q^2)) where f(u, v) = (u + v) * (u - v)^2 - u*v * (1 + 7*u) * (1 + 7*v).

G.f. is a period 1 Fourier series which satisfies f(-1 / (7 t)) = 7^-2 g(t) where q = exp(2 Pi i t) and g() is the g.f. for A030181. - Michael Somos, Jan 02 2015

G.f. A(q) satisfies j(q) = f(49 * A(q)) where f(x) := (x^2 + 13*x + 49) * (x^2 + 5*x + 1)^3 / x. - Michael Somos, Jan 02 2015

Convolution inverse of A030181. - Michael Somos, Jan 02 2015

a(n) ~ exp(4*Pi*sqrt(n/7)) / (49 * sqrt(2) * 7^(1/4) * n^(3/4)). - Vaclav Kotesovec, Sep 07 2015

a(1) = 1, a(n) = (4/(n-1))*Sum_{k=1..n-1} A113957(k)*a(n-k) for n > 1. - Seiichi Manyama, Apr 01 2017