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%I #14 Jun 20 2018 06:55:08
%S 1,1,2,1,3,4,5,6,7,11,15,17,22,24,34,40,48,56,69,84,104,118,144,164,
%T 200,234,273,318,372,436,511,582,681,775,906,1036,1192,1362,1562,1784,
%U 2046,2315,2647,2988,3409,3860,4371,4936,5573,6288,7104,7967,8979,10052
%N McKay-Thompson series of class 45A for the Monster group with a(0) = 1.
%H G. C. Greubel, <a href="/A226054/b226054.txt">Table of n, a(n) for n = -1..1000</a>
%H Vaclav Kotesovec, <a href="https://arxiv.org/abs/1509.08708">A method of finding the asymptotics of q-series based on the convolution of generating functions</a>, arXiv:1509.08708 [math.CO], 2015-2016.
%F Expansion of (eta(q^3) * eta(q^15))^2 / (eta(q) * eta(q^5) * eta(q^9) * eta(q^45)) in powers of q.
%F Euler transform of period 45 sequence [ 1, 1, -1, 1, 2, -1, 1, 1, 0, 2, 1, -1, 1, 1, -2, 1, 1, 0, 1, 2, -1, 1, 1, -1, 2, 1, 0, 1, 1, -2, 1, 1, -1, 1, 2, 0, 1, 1, -1, 2, 1, -1, 1, 1, 0, ...].
%F G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = u*v * (u*v + 3) - (u+v) * (u^2 + u*v + v^2).
%F G.f. is a period 1 Fourier series which satisfies f(-1 / (45 t)) = f(t) where q = exp(2 Pi i t).
%F a(n) = A058684(n) unless n=0.
%F a(n) ~ exp(4*Pi*sqrt(n/5)/3) / (5^(1/4) * sqrt(6) * n^(3/4)). - _Vaclav Kotesovec_, Oct 14 2015
%e 1/q + 1 + 2*q + q^2 + 3*q^3 + 4*q^4 + 5*q^5 + 6*q^6 + 7*q^7 + 11*q^8 + ...
%t nmax=60; CoefficientList[Series[Product[(1-x^(3*k))^2 * (1-x^(15*k))^2 / ((1-x^k) * (1-x^(5*k)) * (1-x^(9*k)) * (1-x^(45*k))),{k,1,nmax}],{x,0,nmax}],x] (* _Vaclav Kotesovec_, Oct 14 2015 *)
%t eta[q_] := q^(1/24)*QPochhammer[q]; a:= CoefficientList[Series[q*(eta[q^3]*eta[q^15])^2/(eta[q]*eta[q^5]*eta[q^9]*eta[q^45]), {q, 0, 60}], q]; Table[a[[n]], {n, 1, 50}] (* _G. C. Greubel_, Jun 20 2018 *)
%o (PARI) {a(n) = local(A); if( n<-1, 0, n++; A = x * O(x^n); polcoeff( (eta(x^3 + A) * eta(x^15 + A))^2 / (eta(x + A) * eta(x^5 + A) * eta(x^9 + A) * eta(x^45 + A)), n))}
%Y Cf. A058684.
%K nonn
%O -1,3
%A _Michael Somos_, May 24 2013
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