OFFSET
1,2
COMMENTS
Inverse weigh transform of A001970.
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..2000
N. J. A. Sloane, Transforms
FORMULA
Product_{n>=1} (1 + x^n)^a(n) = Product_{k>=1} 1/(1 - x^k)^p(k), where p(k) = number of partitions of k (A000041).
EXAMPLE
(1 + x) * (1 + x^2)^3 * (1 + x^3)^3 * (1 + x^4)^8 * (1 + x^5)^7 * ... * (1 + x^n)^a(n) * ... = 1/((1 - x) * (1 - x^2)^2 * (1 - x^3)^3 * (1 - x^4)^5 * (1 - x^5)^7 * ... * (1 - x^k)^p(k) * ...).
MATHEMATICA
nn = 40; f[x_] := Product[(1 + x^n)^a[n], {n, 1, nn}]; sol = SolveAlways[0 == Series[f[x] - Product[1/(1 - x^k)^PartitionsP[k], {k, 1, nn}], {x, 0, nn}], x]; Table[a[n], {n, 1, nn}] /. sol // Flatten
CROSSREFS
KEYWORD
nonn
AUTHOR
Ilya Gutkovskiy, Jun 11 2018
STATUS
approved