A112193 Coefficients of replicable function number "54b".
1, 1, 1, 2, 2, 3, 4, 5, 6, 7, 9, 11, 13, 16, 19, 23, 27, 32, 38, 44, 52, 61, 71, 82, 95, 110, 127, 145, 167, 191, 218, 249, 283, 322, 365, 414, 469, 529, 597, 673, 757, 851, 955, 1071, 1199, 1341, 1499, 1673, 1865, 2078, 2313, 2572, 2857, 3171, 3517, 3897
Offset: 0
Keywords
Examples
G.f. = 1 + x + x^2 + 2*x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 5*x^7 + 6*x^8 + ... _Michael Somos_, Oct 06 2019 G.f. = q^-1 + q^2 + q^5 + 2*q^8 + 2*q^11 + 3*q^14 + 4*q^17 + 5*q^20 + ...
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..10000
- D. Ford, J. McKay and S. P. Norton, More on replicable functions, Comm. Algebra 22, No. 13, 5175-5193 (1994).
- Index entries for McKay-Thompson series for Monster simple group
Crossrefs
Programs
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Maple
b:= proc(n, i) option remember; local r; `if`(2*n>i*(i+1)-(j-> 9*j*(j+1))(iquo(i, 9, 'r')), 0, `if`(n=0, 1, b(n, i-1)+`if`(i>n or r=0, 0, b(n-i, i-1)))) end: a:= n-> b(n$2): seq(a(n), n=0..80); # Alois P. Heinz, Aug 31 2015
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Mathematica
nmax = 50; CoefficientList[Series[Product[(1 + x^k) / (1 + x^(9*k)), {k, 1, nmax}], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 31 2015 *) b[n_, i_] := b[n, i] = Module[{q, r}, {q, r} = QuotientRemainder[i, 9]; If[2*n > i*(i+1) - 9*q*(q+1), 0, If[n == 0, 1, b[n, i-1] + If[i>n || r == 0, 0, b[n-i, i-1]]]]]; a[n_] := b[n, n]; Table[a[n], {n, 0, 80}] (* Jean-François Alcover, Oct 07 2016, after Alois P. Heinz *) a[ n_] := SeriesCoefficient[ QPochhammer[ q^2] QPochhammer[ q^9] / (QPochhammer[ q] QPochhammer[ q^18]), {q, 0, n}]; (* Michael Somos, Oct 06 2019 *)
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PARI
{a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( eta(x^2 + A) * eta(x^9 + A) / (eta(x + A) * eta(x^18 + A)), n))}; /* Michael Somos, Oct 06 2019 */
Formula
a(n) ~ exp(2*Pi*sqrt(2*n/3)/3) / (6^(3/4) * n^(3/4)) * (1 - (9*sqrt(3)/ (16*Pi*sqrt(2)) + sqrt(2)*Pi/(9*sqrt(3))) / sqrt(n)). - Vaclav Kotesovec, Aug 31 2015, extended Jan 21 2017
From Michael Somos, Oct 06 2019: (Start)
Expansion of q^(1/3) * eta(q^2) * eta(q^9) / (eta(q) * eta(q^18)) in powers of q.
Euler transform of period 18 sequence [1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, ...].
G.f. is a period 1 Fourier Series which satisifies f(-1 / (18 t)) = f(t) where q = exp(2 Pi i t).
Given g.f. A(x), then B(q) = A(q^3) / q satisfies 0 = f(B(q), B(q^2)) where f(u, v) = (1 + u*v) * (u^3 + v^3) - u*v * (1 + u^2*v^2).
Given g.f. A(x), then B(q) = A(q^3) / q satisfies 0 = f(B(q), B(q^2), B(q^4)) where f(u, v, w) = (w^2 - v) * (u^2 - v) - 2*u*v*w.
Convolution inverse of A261733.
(End)
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