# Prove that $F_{x^{n+1}} \sim 5^{\frac{x-1}{2}}F_{x^n}^x \forall x,n \geq 1$, holding either variable constant while the other goes to infinity

I noticed from looking at the prime factorizations of some Fibonacci numbers that all those with an index equal to a power of 5 divided that power of five, a property not guaranteed by the strong divisibility property alone. It was while searching for the approximate value of the characteristic factors of these numbers that I noticed $\frac{F_{5^{n+1}}}{F_{5^n}^{5}}\sim 5^2$ as n goes to infinity. Recognizing 5 as 2*2+1, I found that the above could be generalized to a special case of the asymptotic equality that I'm looking to prove in which x is odd. The difference between the two sides of the identity in question grows quite rapidly in absolute value as n increases, but this value is telescoping. For particular x, can we somehow use the set of approximations given by repeated application of the main identity to recursively identify the value of the Fibonacci number whose index is the following power of x?

Since $F_n \sim 5^{-\frac 12} \phi^n$ for large $n$, (where $\phi = \frac {1+ \sqrt 5}2$)
For fixed $x$ and large $n$, we have $F_{x^{n+1}} \sim 5^{-\frac 12} \phi^{x^{n+1}} = 5^{\frac {x-1}2}(5^{-\frac 12} \phi^{x^n})^x \sim 5^\frac{x-1}2 F_{x^n}^x$
• to prove the equivalence for fixed $n$ and large $x$ you need more precision on how $F_n$ behaves and a more involved calculation. – mercio Aug 6 '13 at 9:21
• @mercio I'm not sure you'd need more information on the behaviour of $F_n$ to analyse the "fixed $n$, large $x$" case. The value $x^n$ is going to be large if $x$ is large and the asymptotics for $F_k\sim 5^{-1/2} \phi^k$ is pretty tight enough even for small values of $k$ (since the other root, $\frac{1-\sqrt{5}}{2}$, is very small in absolute value). – Peter Košinár Aug 6 '13 at 10:38