A184786 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).

1, 1, 2, 8, 35, 147, 654, 3009, 14219, 68605, 336623, 1674517, 8425573, 42806200, 219285459, 1131431170, 5874504011, 30670279153, 160916320637, 847994498527, 4486473924741, 23821682237692, 126897559943046, 677992017255423

Offset

0,3

Comments

Limit a(n+1)/a(n) = phi^(phi+2) = (phi+1)^(phi+1)/phi^phi = 5.7032759...

Links

Table of n, a(n) for n=0..23.

Formula

a(n) = floor( binomial(phi^2*n, n)/(phi*n+1) ) where phi = (sqrt(5)+1)/2.

Example

G.f.: A(x) = 1 + x + c2*x^2 + c3*x^3 + c3*x^4 + c5*x^5 +...
A(x)^(phi^2) = 1 + c2*x + c3*x^2 + c4*x^3 + c5*x^4 + c6*x^5 +...
where the coefficients begin:
c2 = 2.6180339887..., c3 = 8.9721359549..., c4 = 35.015865823...,
c5 = 147.58190992..., c6 = 654.49854850..., c7 = 3009.5978243...,
c8 = 14219.000049..., c9 = 68605.600329..., c10 = 336623.1131..., ...;
the floor of the coefficients of A(x) forms this sequence.

Prog

(PARI) {a(n)=local(phi=(1+sqrt(5))/2); if(n<0, 0, floor(binomial(phi^2*n, n)/(phi*n+1)))}

Crossrefs

Cf. A184785.

Sequence in context: A037723 A037618 A326294 * A082759 A243204 A279013

Adjacent sequences: A184783 A184784 A184785 * A184787 A184788 A184789

Keyword

nonn

Author

Paul D. Hanna, Jan 21 2011

Status

approved