|
%I #33 Oct 18 2018 03:06:16
%S 1,6,27,98,315,912,2456,6210,14937,34390,76317,163896,342062,695736,
%T 1382880,2691586,5139906,9644622,17808040,32393370,58113312,102914152,
%U 180062622,311488920,533124225,903324372,1516110165,2521780688,4158863310,6803237280,11043320922,17794350786
%N Expansion of (eta(q^5) / eta(q))^6 in powers of q.
%H Seiichi Manyama, <a href="/A121591/b121591.txt">Table of n, a(n) for n = 1..10000</a>
%H Kevin Acres, David Broadhurst, <a href="https://arxiv.org/abs/1810.07478">Eta quotients and Rademacher sums</a>, arXiv:1810.07478 [math.NT], 2018. See Table 1 p. 10.
%F Euler transform of period 5 sequence [6, 6, 6, 6, 0, ...].
%F G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = u*v * (1 + 125 * u*v) - (u+v) * (u^2 - 13 * u*v + v^2). - _Michael Somos_, May 22 2013
%F G.f. is a period 1 Fourier series which satisfies f(-1 / (5 t)) = 1/125 g(t) where q = exp(2 Pi i t) and g() is the g.f. for A106248. - _Michael Somos_, May 22 2013
%F G.f.: x * (Product_{k>0} (1 - x^(5*k)) / (1 - x^k))^6.
%F Convolution inverse of A106248, 6th power of A035959. - _Michael Somos_, Aug 09 2015
%F a(n) ~ exp(4*Pi*sqrt(n/5)) / (125 * sqrt(2) * 5^(1/4) * n^(3/4)). - _Vaclav Kotesovec_, Sep 07 2015
%F a(1) = 1, a(n) = (6/(n-1))*Sum_{k=1..n-1} A116073(k)*a(n-k) for n > 1. - _Seiichi Manyama_, Mar 31 2017
%e G.f. = q + 6*q^2 + 27*q^3 + 98*q^4 + 315*q^5 + 912*q^6 + 2456*q^7 + 6210*q^8 + ...
%t a[ n_] := SeriesCoefficient[ q (QPochhammer[ q^5] / QPochhammer[ q])^6, {q, 0, n}]; (* _Michael Somos_, May 22 2013 *)
%t nmax = 40; Rest[CoefficientList[Series[x * Product[((1 - x^(5*k)) / (1 - x^k))^6, {k, 1, nmax}], {x, 0, nmax}], x]] (* _Vaclav Kotesovec_, Sep 07 2015 *)
%o (PARI) {a(n) = my(A); if( n<1, 0, n--; A = x * O(x^n); polcoeff( (eta(x^5 + A) / eta(x + A))^6, n))};
%Y Cf. A035959, A106248.
%K nonn
%O 1,2
%A _Michael Somos_, Aug 09 2006
|
