Reminder: The OEIS is hiring a new managing editor, and the application deadline is January 26.
%I #9 Mar 30 2012 18:37:25
%S 1,1,2,8,35,147,654,3009,14219,68605,336623,1674517,8425573,42806200,
%T 219285459,1131431170,5874504011,30670279153,160916320637,
%U 847994498527,4486473924741,23821682237692,126897559943046,677992017255423
%N Let A(x) satisfy: A(x) = 1 + x*A(x)^(phi^2) where phi = (sqrt(5)+1)/2, then this sequence equals the integer part of the coefficients of A(x).
%C Limit a(n+1)/a(n) = phi^(phi+2) = (phi+1)^(phi+1)/phi^phi = 5.7032759...
%F a(n) = floor( binomial(phi^2*n, n)/(phi*n+1) ) where phi = (sqrt(5)+1)/2.
%e G.f.: A(x) = 1 + x + c2*x^2 + c3*x^3 + c3*x^4 + c5*x^5 +...
%e A(x)^(phi^2) = 1 + c2*x + c3*x^2 + c4*x^3 + c5*x^4 + c6*x^5 +...
%e where the coefficients begin:
%e c2 = 2.6180339887..., c3 = 8.9721359549..., c4 = 35.015865823...,
%e c5 = 147.58190992..., c6 = 654.49854850..., c7 = 3009.5978243...,
%e c8 = 14219.000049..., c9 = 68605.600329..., c10 = 336623.1131..., ...;
%e the floor of the coefficients of A(x) forms this sequence.
%o (PARI) {a(n)=local(phi=(1+sqrt(5))/2);if(n<0,0,floor(binomial(phi^2*n,n)/(phi*n+1)))}
%Y Cf. A184785.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Jan 21 2011