%I #6 Jan 09 2024 16:29:42
%S 1,1,1,2,2,3,4,8,8,14,18,29,40,68,88,174,210,344,492,852,1144,1962,
%T 2786,4601,6704,11240,16096,27738,39650,64936,97108,168408,236880,
%U 397110,589298,979496,1459960,2421132,3604880,6086790
%N Result of using the Fibonacci numbers as coefficients in an infinite polynomial series in x and then expressing this series as (1+a(1)x)(1+a(2)x^2)(1+a(3)x^3)...
%e From the Fibonacci numbers, beginning 1,1, construct the series 1+x+x^2+2x^3+3x^4+5x^5+... 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)=1. Then divide this quotient by (1+a(2)x^2), i.e. here (1+x^2), to get (1+a(3)x^3+...), giving a(3)=1.
%Y Cf. A000045, A147558
%Y Cf. A147542. [From _R. J. Mathar_, Mar 12 2009]
%K nonn
%O 1,4
%A _Neil Fernandez_, Nov 07 2008
%E More terms from _R. J. Mathar_, Mar 12 2009