A224415 G.f.: exp( Sum_{n>=1} 5*L(n)*x^n/n ) where L(n) = Fibonacci(n-1)^2 + Fibonacci(n+1)^2 = A069921(n-1).

1, 5, 25, 100, 380, 1348, 4610, 15250, 49250, 155860, 485228, 1489780, 4520475, 13577775, 40423155, 119413496, 350336200, 1021523000, 2962214500, 8547193700, 24551057380, 70231278200, 200150437000, 568435763000, 1609247086325, 4542394525369, 12786764813645

Offset: 0

Author

Paul D. Hanna, Apr 05 2013

Keywords

• nonn

Comments

Given g.f. A(x), note that A(x)^(1/5) does not yield an integer series.

Compare to: exp( Sum_{n>=1} Lucas(n)*x^n/n ) = 1/(1-x-x^2) where Lucas(n) = Fibonacci(n-1) + Fibonacci(n+1).

Examples

			G.f.: 1 + 5*x + 25*x^2 + 100*x^3 + 380*x^4 + 1348*x^5 + 4610*x^6 +...
where
log(A(x))/5 = x + 5*x^2/2 + 10*x^3/3 + 29*x^4/4 + 73*x^5/5 + 194*x^6/6 + 505*x^7/7 + 1325*x^8/8 +...+ A069921(n-1)*x^n/n +...
		

Links

• Index entries for linear recurrences with constant coefficients, signature (5,0,-25,5,48,5,-25,0,5,-1).

Crossrefs

Cf. A069921.

Programs

PARI

{L(n)=fibonacci(n-1)^2+fibonacci(n+1)^2}
{a(n)=polcoeff(exp(sum(m=1, n, 5*L(m)*x^m/m)+x*O(x^n)), n)}
for(n=0, 30, print1((a(n)), ", "))

PARI

{a(n)=polcoeff(1/((1+x)^4*(1-3*x+x^2)^3+x*O(x^n)),n)}
for(n=0, 30, print1((a(n)), ", "))

Formula

G.f.: 1 / ( (1+x)^4 * (1 - 3*x + x^2)^3 ).