login

Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.

Result of using the positive integers 1,2,3,... as coefficients in an infinite polynomial series in x and then expressing this series as Product_{k>=1} (1+a(k)x^k).
10

%I #17 Jun 24 2018 11:31:48

%S 1,2,1,3,0,-2,0,9,0,-6,0,4,0,-18,0,93,0,-54,0,72,0,-186,0,232,0,-630,

%T 0,1020,0,-2106,0,10881,0,-7710,0,13824,0,-27594,0,49440,0,-97902,0,

%U 191844,0,-364722,0,590800,0,-1340622,0,2656920,0,-4918482,0,9791784,0,-18512790

%N Result of using the positive integers 1,2,3,... as coefficients in an infinite polynomial series in x and then expressing this series as Product_{k>=1} (1+a(k)x^k).

%H Seiichi Manyama, <a href="/A147654/b147654.txt">Table of n, a(n) for n = 1..500</a>

%F Product_{k>=1} (1+a(k)*x^k) = 1 + Sum_{k>=1} k*x^k. - _Seiichi Manyama_, Jun 24 2018

%e From the positive integers 1,2,3,..., construct the series 1+x+2x^2+3x^3+4x^4+... a(1) is always the coefficient of x, here 1. Divide by (1+a(1)x), i.e. here (1+x), to get the quotient (1+a(2)x^2+...), which here gives a(2)=2. Then divide this quotient by (1+a(2)x^2), i.e. here (1+2x^2), to get (1+a(3)x^3+...), giving a(3)=1.

%Y Cf. A028310, A147541, A147559.

%K sign

%O 1,2

%A _Neil Fernandez_, Nov 09 2008

%E More terms from _Seiichi Manyama_, Jun 23 2018