%I A147558
%S A147558 1,1,1,2,2,3,4,8,8,14,18,29,40,68,88,174,210,344,492,852,1144,1962,2786,
%T A147558 4601,6704,11240,16096,27738,39650,64936,97108,168408,236880,397110,
%U A147558 589298,979496,1459960,2421132,3604880,6086790
%N A147558 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(3x^3)...
%e A147558 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 A147558 Cf. A000045, A147558
%Y A147558 Cf. A147542. [From R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 12 
               2009]
%Y A147558 Sequence in context: A011784 A032252 A112708 this_sequence A032243 A153922 
               A153943
%Y A147558 Adjacent sequences: A147555 A147556 A147557 this_sequence A147559 A147560 
               A147561
%K A147558 nonn
%O A147558 1,4
%A A147558 N. Fernandez (primeness(AT)borve.org), Nov 07 2008
%E A147558 More terms from R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Mar 12 2009