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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?

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    $\begingroup$ 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. $\endgroup$
    – ViHdzP
    Jan 5, 2020 at 21:23
  • $\begingroup$ Yes but C is a multiple of 4 @ URL $\endgroup$
    – mick
    Jan 5, 2020 at 21:24
  • $\begingroup$ Also 6 is not a prime !! @ URL $\endgroup$
    – mick
    Jan 5, 2020 at 21:25
  • $\begingroup$ 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 $\endgroup$
    – clathratus
    Jan 5, 2020 at 21:31
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    $\begingroup$ @mick : The Question has "every $n$" and "$n > 4$", not "every prime $n$", so the Question as written is required to work for $n = 6$. $\endgroup$ Jan 5, 2020 at 21:31

1 Answer 1

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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.

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    $\begingroup$ Props to Wolfram|Alpha for helping me find this. $\endgroup$
    – ViHdzP
    Jan 5, 2020 at 21:29
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    $\begingroup$ 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. $\endgroup$
    – mick
    Jan 5, 2020 at 21:36
  • $\begingroup$ 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 $n$. $\endgroup$
    – ViHdzP
    Jan 5, 2020 at 21:37
  • $\begingroup$ It fails for 103 too. There we have 3 prime factors. Not many factors though $\endgroup$
    – mick
    Jan 5, 2020 at 21:45
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    $\begingroup$ @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 $k$. $\endgroup$ Jan 5, 2020 at 21:53

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