# Is it true that for every prime number $p >3$, the next one will be less than $p+p/2$?

I know that for every prime $p$, the next prime is less than $2p$. Can we improve this statement? Can it be less than $(3/2)p$? What is the best function of $p$ for which this is true (for every prime, not just infinitively many)? (assume $p>3$, or $p>$ than a fixed $p$ if necessary).

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For every positive $\epsilon$, there is an $N=N(\epsilon)$ such that if $p\gt N$ there is always a prime between $p$ and $(1+\epsilon)p$. An $N(\epsilon)$ that works can be specified explicitly.

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It is a consequence of the prime number theorem, and, at the computing $N(\epsilon)$ level, a consequence of theorems of Pierre Dusart (and others) about the size of prime gaps. – André Nicolas May 29 '13 at 23:33

Its true for every prime greater then or equal to $11$

Much stronger results are also known also, for example there exists primes between

$n$ and $(1+\frac{1}{5})n$  $\forall n\ge25$

$n$ and $(1+\frac{1}{16597})n$  $\forall n\ge 2010760$

$n$ and $(1+\frac{1}{\ln(n)^2})n$  $\forall n\ge463$

$n$ and $(1+\frac{1}{2\ln(n)^2})n$  $\forall n\ge3275$

$n$ and $(1+\frac{1}{25\ln(n)^2})n$  $\forall n\ge 396738$

It is important to note that results of this sort should not be taken as good estimates of the distribution of the primes

These are useful in that they assert the existence of primes over intervals for relatively small numbers, in general if you order the primes from decreasing to increasing value, then the $n$th prime will be asymptotic to $n\ln(n)$, this suggests that the primes are distributed roughly logarithmically,

That is if your given a prime number $m$ your on average going to see the next prime at around $m+\ln(m)$

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Do you know the proof? Can this be improved (even if it means making p > than a prime greater than 11)? Thanks! – user79303 May 29 '13 at 22:52
Yes, there are some very elementary proofs of this – Ethan May 29 '13 at 22:53
Do you have a link? – user79303 May 29 '13 at 22:55
@user79303 Yes, see (en.wikipedia.org/wiki/Bertrand's_postulate), Paul Erdős gave a very elementry proof of the fact that there exists a prime between $n$ and $2n$ the clever idea behind it is that the central binomial coeiffient $\binom{2n}{n}$ is always an integer and is always divisible by all the primes between $n$ and $2n$. – Ethan May 29 '13 at 23:26

There is no best function $f(p)$ for which this is true, since $g(p)$ with $g(p_n)=(f(p_n)+p_{n+1})/2$ would be better. If you replace "less than" by "less than or equal to", there is a unique best function $f(p)$, namely $f(p_n)=p_{n+1}$.

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Doesn't work for $p=3$. I seem to recall that $p+\ln p$ was a good upper bound on the prime successor of $p$, in general (or at least for $p$ large enough) but I'm not finding any references at the moment.

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Just edited the question. I am interested in large p, small primes can be handled separately. – user79303 May 29 '13 at 22:49
This bound will fail infinitely often; this follows, e.g., from Rankin's theorem. – Charles May 31 '13 at 17:26

The wikipedia article on "prime gap" has a good deal of useful information.

In particular, it give the history of the result that $p_{n+1} < p_n+p_n^c$ for various values of $c$ (all less than 1) for sufficiently large $n$.

The best so far is $c < 3/4+\epsilon$ for any $\epsilon > 0$.

Another result is that there is always a prime between $n^3$ and $(n+1)^3$ for sufficiently large $n$.

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