A229900 G.f. satisfies: A(x) = x*exp( Sum_{n>=1} A(Lucas(n)*x^n) / n ).

1, 1, 3, 7, 22, 54, 192, 496, 1722, 4799, 17013, 48362, 169458, 498733, 1776570, 5331037, 18608396, 57109292, 201213331, 625146398, 2176243724, 6873338754, 24052502138, 76626395556, 265774736523, 856909651252, 2979519037203, 9675077715466, 33455355338926, 109591055905415

Offset

1,3

Comments

Compare to: G(x) = x*exp( Sum_{n>=1} G(x^n)/n ), which is the g.f. of A000081, the number of rooted trees with n nodes.

Compare to: exp( Sum_{n>=1} Lucas(n)*x^n/n ) = 1/(1-x-x^2), which is the g.f. of the Fibonacci numbers.

The limit a(n+1)/a(n) seems to exist (near 3.5...); is this true?

Links

Table of n, a(n) for n=1..30.

Example

G.f.: A(x) = x + x^2 + 3*x^3 + 7*x^4 + 22*x^5 + 54*x^6 + 192*x^7 + 496*x^8 +...
where
A(x) = x*exp(A(x) + A(3*x^2)/2 + A(4*x^3)/3 + A(7*x^4)/4 + A(11*x^5)/5 + A(18*x^6)/6 + A(29*x^7)/7 + A(47*x^8)/8 + A(76*x^9)/9 + A(123*x^10)/10 +...).

Prog

(PARI) {Lucas(n)=fibonacci(n-1)+fibonacci(n+1)}
{a(n)=local(A=x); for(i=1, n, A=x*exp(sum(k=1, n, subst(A, x, Lucas(k)*x^k +x*O(x^n))/k))); polcoeff(A, n)}
for(n=1, 30, print1(a(n), ", "))

Crossrefs

Cf. A000032 (Lucas), A229807, A229901.

Sequence in context: A182174 A080882 A229807 * A079120 A092566 A036719

Adjacent sequences: A229897 A229898 A229899 * A229901 A229902 A229903

Keyword

nonn

Author

Paul D. Hanna, Oct 02 2013

Status

approved