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McKay-Thompson series of class 90b for the Monster group.
3

%I #24 Jun 29 2018 04:29:54

%S 1,-1,0,0,0,0,0,-1,1,0,0,-1,1,-1,1,0,0,-1,2,-2,2,-1,0,-1,3,-2,1,-1,1,

%T -2,4,-5,3,-1,1,-2,4,-6,4,-2,3,-5,6,-7,6,-2,1,-6,10,-10,9,-6,4,-7,12,

%U -12,9,-7,6,-10,18,-20,13,-8,9,-12,18,-24,20,-12,13,-21,27,-29,24,-14,13,-25,36,-38,35,-25,19,-30,46,-46,36

%N McKay-Thompson series of class 90b for the Monster group.

%H G. C. Greubel, <a href="/A112215/b112215.txt">Table of n, a(n) for n = 0..1000</a>

%H D. Ford, J. McKay and S. P. Norton, <a href="http://dx.doi.org/10.1080/00927879408825127">More on replicable functions</a>, Commun. Algebra 22, No. 13, 5175-5193 (1994).

%H <a href="/index/Mat#McKay_Thompson">Index entries for McKay-Thompson series for Monster simple group</a>

%F Expansion of q^(1/3)*(eta(q)*eta(q^6)*eta(q^10)*eta(q^15))/(eta(q^2) *eta(q^3)*eta(q^5)*eta(q^30)) in powers of q. - _G. C. Greubel_, Jun 06 2018

%F a(n) ~ (-1)^n * exp(2*Pi*sqrt(n/5)/3) / (2 * sqrt(3) * 5^(1/4) * n^(3/4)). - _Vaclav Kotesovec_, Jun 29 2018

%e T90b = 1/q - q^2 - q^20 + q^23 - q^32 + q^35 - q^38 + q^41 - q^50 + ...

%t eta[q_]:= q^(1/24)*QPochhammer[q]; a[n_]:= SeriesCoefficient[q^(1/3)* (eta[q]*eta[q^6]*eta[q^10]*eta[q^15])/(eta[q^2]*eta[q^3]*eta[q^5]* eta[q^30]), {q, 0, n}]; Table[a[n], {n, 0, 50}] (* _G. C. Greubel_, Mar 04 2018 *)

%t nmax = 100; CoefficientList[Series[Product[(1 - x^(2*k - 1))*(1 - x^(30*k - 15))/((1 - x^(6*k - 3))*(1 - x^(10*k - 5))), {k, 1, nmax}], {x, 0, nmax}], x] (* _Vaclav Kotesovec_, Jun 29 2018 *)

%o (PARI) q='q+O('q^80); F=(eta(q)*eta(q^6)*eta(q^10)*eta(q^15))/(eta(q^2) *eta(q^3)*eta(q^5)*eta(q^30)); Vec(F) \\ _G. C. Greubel_, Jun 06 2018

%K sign

%O 0,19

%A _Michael Somos_, Aug 28 2005