OFFSET
0,2
COMMENTS
Moss and Ward prove that this is an integral sequence. - Peter Bala, Nov 28 2022
Let A(x) be the g.f. for this sequence. Note that the expansion of A(x)^(1/5) = exp( Sum_{n>=1} Fibonacci(n^2) * x^n/n ) does not have integer coefficients.
LINKS
Patrick Moss and Tom Ward, Fibonacci along even powers is (almost) realizable, arXiv:2011.13068 [math.NT], 2020; Fibonacci Quart. 60 (2022), no. 1, 40-47.
EXAMPLE
G.f.: A(x) = 1 + 5*x + 20*x^2 + 115*x^3 + 1665*x^4 + 82650*x^5 + ...
such that
log(A(x))/5 = x + 3*x^2/2 + 34*x^3/3 + 987*x^4/4 + 75025*x^5/5 + 14930352*x^6/6 + 7778742049*x^7/7 + ... + Fibonacci(n^2)*x^n/n + ...
PROG
(PARI) {a(n)=polcoeff(exp(sum(k=1, n, 5*fibonacci(k^2)*x^k/k)+x*O(x^n)), n)}
for(n=0, 16, print1(a(n), ", "))
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul D. Hanna, Feb 22 2012
STATUS
approved