A305868 Product_{n>=1} 1/(1 - x^n)^a(n) = g.f. of A001147 (double factorial of odd numbers).
1, 2, 12, 87, 816, 9194, 122028, 1859460, 32002076, 613890984, 12989299596, 300556859080, 7550646317520, 204687481289946, 5955892982437120, 185158929516065160, 6125200081143892800, 214837724609502834082, 7963817560398871790604, 311101285877489780292000, 12773912991134665452205048
Offset: 1
Keywords
Examples
1/((1 - x) * (1 - x^2)^2 * (1 - x^3)^12 * (1 - x^4)^87 * (1 - x^5)^816 * ... * (1 - x^n)^a(n) * ...) = 1 + 1*x + 1*3*x^2 + 1*3*5*x^3 + 1*3*5*7*x^4 + ... + A001147(k)*x^k + ...
Links
- Seiichi Manyama, Table of n, a(n) for n = 1..404
- N. J. A. Sloane, Transforms
- Eric Weisstein's World of Mathematics, Double Factorial
- Index entries for sequences related to factorial numbers
Programs
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Mathematica
nn = 21; f[x_] := Product[1/(1 - x^n)^a[n], {n, 1, nn}]; sol = SolveAlways[0 == Series[f[x] - 1/(1 + ContinuedFractionK[-k x, 1, {k, 1, nn}]), {x, 0, nn}], x]; Table[a[n], {n, 1, nn}] /. sol // Flatten nmax = 20; s = ConstantArray[0, nmax]; Do[s[[j]] = j*(2*j - 1)!! - Sum[s[[d]]*(2*j - 2*d - 1)!!, {d, 1, j - 1}], {j, 1, nmax}]; Table[Sum[MoebiusMu[k/d]*s[[d]], {d, Divisors[k]}]/k, {k, 1, nmax}] (* Vaclav Kotesovec, Aug 09 2019 *)
Formula
Product_{n>=1} 1/(1 - x^n)^a(n) = 1/(1 - x/(1 - 2*x/(1 - 3*x/(1 - 4*x/(1 - 5*x/(1 - ...)))))).
a(n) ~ 2^(n + 1/2) * n^n / exp(n). - Vaclav Kotesovec, Aug 09 2019
Comments