OFFSET
0,3
COMMENTS
It follows from Atanassov et al. that a(n) << sqrt(phi)^n, which matches the trivial a(n) >> sqrt(phi)^n up to a constant factor. - Charles R Greathouse IV, Feb 06 2013
Conjecture: Fibonacci(m)*Fibonacci(n) with 2 < m < n is a perfect power only for (m,n) = (3,6). This has been verified for 2 < m < n <= 900. - Zhi-Wei Sun, Jan 02 2025
LINKS
Charles R Greathouse IV, Table of n, a(n) for n = 0..10000
K. T. Atanassov, Ron Knott, Kiyota Ozeki, A. G. Shannon, and László Szalay, Inequalities among related pairs of Fibonacci numbers, Fibonacci Quarterly 41:1 (2003), pp. 20-22.
Clark Kimberling, Orderings of products of Fibonacci numbers, Fibonacci Quarterly 42:1 (2004), pp. 28-35.
Zhi-Wei Sun, Perfect powers as products of two Fibonacci or Lucas numbersQuestion 485037 at MathOverflow, December 30, 2024.
EXAMPLE
25 is in the sequence since it is the product of two, not necessarily distinct, Fibonacci numbers, 5 and 5.
26 is in the sequence since it is the product of two Fibonacci numbers, 2 and 13.
27 is not in the sequence because there is no way whatsoever to represent it as the product of exactly two Fibonacci numbers.
MATHEMATICA
Take[ Union@Flatten@Table[ Fibonacci[i]Fibonacci[j], {i, 0, 16}, {j, 0, i}], 61] (* Robert G. Wilson v, Dec 14 2005 *)
PROG
(PARI) list(lim)=my(phi=(1+sqrt(5))/2, v=vector(log(lim*sqrt(5))\log(phi), i, fibonacci(i+1)), u=List([0]), t); for(i=1, #v, for(j=i, #v, t=v[i]*v[j]; if(t>lim, break, listput(u, t)))); vecsort(Vec(u), , 8) \\ Charles R Greathouse IV, Feb 05 2013
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved