A121589 Series expansion of (eta(q^9) / eta(q))^3 in powers of q.
1, 3, 9, 22, 51, 108, 221, 429, 810, 1476, 2631, 4572, 7802, 13056, 21519, 34918, 55935, 88452, 138332, 213990, 327852, 497592, 748833, 1117692, 1655719, 2434938, 3556791, 5161808, 7445631, 10677096, 15226658, 21599469, 30485268, 42817788
Offset: 1
Keywords
Examples
G.f. = q + 3*q^2 + 9*q^3 + 22*q^4 + 51*q^5 + 108*q^6 + 221*q^7 + 429*q^8 + ...
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.
Programs
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Maple
N:= 100: # to get a(1)..a(N) S:= series(q*Product(1-q^(9*k),k=1..N/9)/Product((1-q^k)^3, k=1..N),q,N+1): seq(coeff(S,q,n),n=1..N); # Robert Israel, Nov 02 2017
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Mathematica
nmax = 40; CoefficientList[Series[Product[((1-x^(9*k))/(1-x^k))^3, {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Sep 07 2015 *) QP = QPochhammer; s = (QP[q^9]/QP[q])^3 + O[q]^40; CoefficientList[s, q] (* Jean-François Alcover, Nov 30 2015, adapted from PARI *) a[ n_] := SeriesCoefficient[ q (QPochhammer[ q^9] / QPochhammer[ q])^3, {q, 0, n}]; (* Michael Somos, Nov 02 2017 *)
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PARI
{a(n) = my(A); if( n<1, 0, n--; A = x * O(x^n); polcoeff( (eta(x^9 + A) / eta(x + A))^3, n))};
Formula
Euler transform of period 9 sequence [3, 3, 3, 3, 3, 3, 3, 3, 0, ...].
G.f.: x * (Product_{k>0} (1 - x^(9*k) / (1 - x^k))^3.
Expansion of c(q^3) / (3 * b(q)) = (c(q) / (3 * b(q^3))^3 in powers of q where b(), c() are cubic AGM functions.
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = (u - v^2) * (u^2 - v) - 2 * u * v * ( 3 * (u + v) + 13 * u * v ).
G.f. A(x) satisfies 0 = f(A(x), A(x^3)) where f(u, v) = u^3 - v * (1 + 9 * v + 27 * v^2) * (1 + 9 * u + 27 * u^2).
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^3), A(x^6)) where f(u1, u2, u3, u6) = u1 * u2 - (1 + 3 * (u1 + u2)) * (u3 + u6 + 9 * u3 * u6).
G.f. is a period 1 Fourier series which satisfies f(-1 / (9 t)) = (1/27) g(t) where q = exp(2 Pi i t) and g() is the g.f. of A131986.
a(n) ~ exp(4*Pi*sqrt(n)/3) / (27 * sqrt(6) * n^(3/4)). - Vaclav Kotesovec, Sep 07 2015
a(1) = 1, a(n) = (3/(n-1))*Sum_{k=1..n-1} A116607(k)*a(n-k) for n > 1. - Seiichi Manyama, Apr 01 2017
Extensions
Second formula corrected by Vaclav Kotesovec, Sep 07 2015
Comments