A305870 Product_{n>=1} (1 + x^n)^a(n) = g.f. of A001147 (double factorial of odd numbers).

1, 3, 12, 90, 816, 9206, 122028, 1859550, 32002076, 613891800, 12989299596, 300556868286, 7550646317520, 204687481411974, 5955892982437120, 185158929517924710, 6125200081143892800, 214837724609534836158, 7963817560398871790604, 311101285877490394183800, 12773912991134665452205048

Offset: 1

Author

Ilya Gutkovskiy, Jun 12 2018

Keywords

• nonn

Comments

Inverse weigh transform of A001147.

Examples

			(1 + x) * (1 + x^2)^3 * (1 + x^3)^12 * (1 + x^4)^90 * (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

• Alois P. Heinz, 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

Crossrefs

Cf. A001147, A168246, A305868, A305869.

Programs

Maple

b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
      add(binomial(a(i), j)*b(n-i*j, i-1), j=0..n/i)))
    end:
a:= proc(n) option remember; doublefactorial(2*n-1)-b(n, n-1) end:
seq(a(n), n=1..23);  # Alois P. Heinz, Jun 13 2018

Mathematica

nn = 21; f[x_] := Product[(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

Formula

Product_{n>=1} (1 + x^n)^a(n) = 1/(1 - x/(1 - 2*x/(1 - 3*x/(1 - 4*x/(1 - 5*x/(1 - ...)))))).