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(3)x^3)...

1, 1, 1, 2, 2, 3, 4, 8, 8, 14, 18, 29, 40, 68, 88, 174, 210, 344, 492, 852, 1144, 1962, 2786, 4601, 6704, 11240, 16096, 27738, 39650, 64936, 97108, 168408, 236880, 397110, 589298, 979496, 1459960, 2421132, 3604880, 6086790

Offset: 1

Author

Neil Fernandez, Nov 07 2008

Keywords

• nonn

Examples

			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.
		

Crossrefs

Cf. A000045, A147558

Cf. A147542. [From R. J. Mathar, Mar 12 2009]

Extensions

More terms from R. J. Mathar, Mar 12 2009