# Is this sequence based on the Fibonacci numbers a prime generator?

Consider the Fibonacci sequence $$\text{fibo}(n)$$ and the fractions

$$A(n) = \frac{\text{fibo}\left(n^2\right)}{\text{fibo}(n)^2} = \frac{b(n)}{c(n)},$$

where the fractions $$\frac{b(n)}{c(n)}$$ are in reduced form.

Now, it appears that for all $$n > 4$$:

1. If $$c(n) = 1$$, then $$b(n)$$ is of the form $$2^q p$$ where $$p$$ is a prime.

2. If $$c(n) > 1$$, then either $$c(n)$$ or $$b(n)$$ is of the form $$2^q p$$ where $$p$$ is a prime.

So this formula generates a prime for every $$n$$.

A weaker conjecture is that this holds for the cases when $$n$$ is a prime.

Are these conjectures true? How can we prove them?

• For $n=6$, $c(n)=4$, $b(n)=3^3\times17\times19\times107$. If I understood correctly, this should be a counterexample to your stronger conjecture. Jan 5, 2020 at 21:23
• Yes but C is a multiple of 4 @ URL
– mick
Jan 5, 2020 at 21:24
• Also 6 is not a prime !! @ URL
– mick
Jan 5, 2020 at 21:25
• note that we have the explicit formula for the $n$-th Fibonacci number $$F_n=\frac{\phi^n-(-1)^n\phi^{-n}}{\sqrt5},$$ where $\phi=\frac{1+\sqrt5}{2}$. This may help Jan 5, 2020 at 21:31
• @mick : The Question has "every $n$" and "$n > 4$", not "every prime $n$", so the Question as written is required to work for $n = 6$. Jan 5, 2020 at 21:31

For $$n=19$$, a prime, we have $$\frac{F_{n^2}}{\left(F_n\right)^2}=\frac{297695973435970582594631907579321477163892921001085193295076858332955181}{4181}.$$

The numerator and denominator are odd, the former is divisible by $$6567762529$$, and the latter is divisible by $$37$$. So, both of your conjectures are false.

• Props to Wolfram|Alpha for helping me find this. Jan 5, 2020 at 21:29
• Hmm thank you. It is weird how often this works though. For n = 23 and n = 29 it holds again. Perhaps you can modify the conjecture ? Maybe it fails miserably for very large n ... I will test some more numbers. Thank you.
– mick
Jan 5, 2020 at 21:36
• I don't believe we should expect any pattern of primes to pop up here. And "most" large numbers are not prime. So, I conjecture that in fact, this fails for infinitely many $n$. Jan 5, 2020 at 21:37
• It fails for 103 too. There we have 3 prime factors. Not many factors though
– mick
Jan 5, 2020 at 21:45
• @mick I think this is less weird than you might think; $F_n\mid F_{n^2}$, which immediately shrinks the size of the numbers involved. When you start talking about restricting to prime $n$, you eliminate several other possible factors from either side, because 'most' small factors can't divide $F_{kp}$ for any $k$. Jan 5, 2020 at 21:53