Bertrand's postulate says:

For every $n>1$ there is always at least one prime $p$ such that $n<p<2n$.

Is the following statement:

For every $n>3$ there is always at least one prime $p$ such that $F_n<p<F_{n+1}$ ($F_n$ is $n$-th Fibonacci number).

also valid?

If it is invalid, is there a finite or infinite number of $n$s such that there is no prime between $F_n$ and $F_{n+1}$?

This question is inspired by another question. I feel intuitively that it may be interesting, but don't have enough number theory background to tackle it.

  • $\begingroup$ Part I: determine the smallest positive integer $a$ such that $2F_n\le F_{n+a}$... $\endgroup$ – abiessu May 29 '15 at 22:12
  • $\begingroup$ a=2..... @abiessu $\endgroup$ – VividD May 29 '15 at 22:14
  • 1
    $\begingroup$ In fact the prime number theorem implies that for any $\varepsilon > 0$ and sufficiently large $n$, there is a prime between $n$ and $(1 + \varepsilon) n$. For stronger results than this see en.wikipedia.org/wiki/Legendre%27s_conjecture. $\endgroup$ – Qiaochu Yuan May 29 '15 at 23:13

The conjecture is true.

From a result of Nagura, for $n\ge 25$, there is always a prime between $n$ and $1.2n$. Note that $F_9=34$, $F_{10}=55$, and their ratio is $1.617>1.2$. In fact, for all $n\ge 9$, we can prove that $\frac{F_{n+1}}{F_n}>1.2$ (*). Hence, combining these results gives a prime between $F_n$ and $F_{n+1}$ for all $n\ge 9$. It remains merely to check the smaller Fibonacci numbers.

(*) The ratio approaches $\phi=\frac{1+\sqrt{5}}{2}\approx 1.6$ as a limit, and rather quickly.


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