login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A207969
G.f.: exp( Sum_{n>=1} 5*Fibonacci(n)^4 * x^n/n ).
7
1, 5, 15, 60, 295, 1625, 9430, 56465, 345010, 2139595, 13419500, 84926105, 541398665, 3472389210, 22385362895, 144945232375, 942089445030, 6143582084115, 40181143112035, 263482860974570, 1731780213622125, 11406235045261205, 75268685723935940
OFFSET
0,2
COMMENTS
Conjecture: exp( Sum_{n>=1} 5*Fibonacci(n)^(2*k) * x^n/n ) is an integer series for integers k>=0.
Note that exp( Sum_{n>=1} 5*Fibonacci(n)^(2*k+1) * x^n/n ) is not an integer series for integers k.
Note that exp( Sum_{n>=1} Fibonacci(n)^(2*k) * x^n/n ) is not an integer series for integers k.
FORMULA
The o.g.f. A(x) = 1 + 5*x + 15*x^2 + 60*x^3 + ... is an algebraic function: A(x)^5 = (1 + 3*x + x^2)^4/( (1 - 7*x + x^2)*(1 - 2*x + x^2)^3 ). Cf. A203804. - Peter Bala, Apr 03 2014
a(n) ~ 2^(4/5) * 5^(1/10) * phi^(4*n) / (Gamma(1/5) * 3^(1/5) * n^(4/5)), where phi = A001622 = (1+sqrt(5))/2 is the golden ratio. - Vaclav Kotesovec, Oct 18 2020
EXAMPLE
G.f.: A(x) = 1 + 5*x + 15*x^2 + 60*x^3 + 295*x^4 + 1625*x^5 + 9430*x^6 +...
such that
log(A(x))/5 = x + x^2/2 + 2^4*x^3/3 + 3^4*x^4/4 + 5^4*x^5/5 + 8^4*x^6/6 + 13^4*x^7/7 +...+ Fibonacci(n)^4*x^n/n +...
PROG
(PARI) {a(n)=polcoeff(exp(sum(k=1, n, 5*fibonacci(k)^4*x^k/k)+x*O(x^n)), n)}
for(n=0, 25, print1(a(n), ", "))
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Feb 22 2012
STATUS
approved