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

1,2

Comments

Inverse weigh transform of A001147.

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

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 - ...)))))).

Example

(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 + ...

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

Crossrefs

Cf. A001147, A168246, A305868, A305869.

Sequence in context: A124191 A065087 A366733 * A342395 A058337 A025503

Adjacent sequences: A305867 A305868 A305869 * A305871 A305872 A305873

Keyword

nonn

Author

Ilya Gutkovskiy, Jun 12 2018

Status

approved