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Product_{n>=1} 1/(1 - x^n)^a(n) = g.f. of A001147 (double factorial of odd numbers).
5

%I #12 Aug 09 2019 11:00:02

%S 1,2,12,87,816,9194,122028,1859460,32002076,613890984,12989299596,

%T 300556859080,7550646317520,204687481289946,5955892982437120,

%U 185158929516065160,6125200081143892800,214837724609502834082,7963817560398871790604,311101285877489780292000,12773912991134665452205048

%N Product_{n>=1} 1/(1 - x^n)^a(n) = g.f. of A001147 (double factorial of odd numbers).

%C Inverse Euler transform of A001147.

%H Seiichi Manyama, <a href="/A305868/b305868.txt">Table of n, a(n) for n = 1..404</a>

%H N. J. A. Sloane, <a href="/transforms.txt">Transforms</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/DoubleFactorial.html">Double Factorial</a>

%H <a href="/index/Fa#factorial">Index entries for sequences related to factorial numbers</a>

%F 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 - ...)))))).

%F a(n) ~ 2^(n + 1/2) * n^n / exp(n). - _Vaclav Kotesovec_, Aug 09 2019

%e 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 + ...

%t 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

%t 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 *)

%Y Cf. A001147, A112354, A305867, A305870.

%K nonn

%O 1,2

%A _Ilya Gutkovskiy_, Jun 12 2018

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Last modified September 21 07:45 EDT 2024. Contains 376083 sequences. (Running on oeis4.)