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A128525
McKay-Thompson series of class 11A for the Monster Group with a(0) = 6.
5
1, 6, 17, 46, 116, 252, 533, 1034, 1961, 3540, 6253, 10654, 17897, 29284, 47265, 74868, 117158, 180608, 275562, 415300, 620210, 916860, 1344251, 1953974, 2819664, 4038300, 5746031, 8122072, 11413112, 15943576, 22153909, 30620666
OFFSET
-1,2
FORMULA
Expansion of (1 + 3*F)^2 * (1/F + 1 + 3*F) where F = eta(q^3) * eta(q^33) / (eta(q) * eta(q^11)) in powers of q.
G.f. A(x) satisfies 0 = f(A(x), A(x^2)) where f(u, v) = u^2 + v^2 - u^2*v^2 + 12*u*v*(u+v) - 20*(u^2+v^2) - 53*u*v + 56*(u+v) - 44.
G.f. A(x) satisfies 0 = f(A(x), A(x^2), A(x^4)) where f(u, v, w) = u^2 + w^2 + u*w - v^2*(u+w) + 12*v^2 + 12*v*(u+w) - 20*(u+w) - 53*v + 56.
G.f. is a period 1 Fourier series which satisfies f(-1 / (11 t)) = f(t) where q = exp(2 Pi i t).
a(n) = A058205(n) unless n = 0.
Convolution of A028160 and A032442. - Michael Somos, Apr 21 2015
a(n) ~ exp(4*Pi*sqrt(n/11)) / (sqrt(2)*11^(1/4)*n^(3/4)). - Vaclav Kotesovec, Dec 04 2015
EXAMPLE
G.f. = 1/q + 6 + 17*q + 46*q^2 + 116*q^3 + 252*q^4 + 533*q^5 + 1034*q^6 + ...
MATHEMATICA
a[ n_] := SeriesCoefficient[ With[{F = q QPochhammer[ q^3] QPochhammer[ q^33] / (QPochhammer[ q] QPochhammer[ q^11])}, (1 + 3 F)^2 (1/F + 1 + 3 F)], {q, 0, n}]; (* Michael Somos, Apr 21 2015 *)
PROG
(PARI) {a(n) = my(A); if( n<-1, 0, n++; A = x * O(x^n); A = x * eta(x^3 + A) * eta(x^33 + A) / (eta(x + A) * eta(x^11 + A)); polcoeff( (1 + 3*A)^2 * (1/A + 1 + 3*A), n-1))};
CROSSREFS
KEYWORD
nonn
AUTHOR
Michael Somos, Mar 07 2007
STATUS
approved