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Suppose x is an arbitrarily large prime number. Is there a well defined maximum gap before the next prime number y shows up. I am looking for maximum limit of (y - x) in terms of x. Not in terms of n where x may be nth prime number. So, the answer I am looking for is in terms of magnitude of the prime number x, not in terms of its index on the prime number line. I saw in this answer - Upper bound for the prime gap above $n$ that there is no elementary proof that the prime gap after p can be at most 2√p. Is there any other (proven or conjectured) max gap that is smaller than 2√p?

kpv
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    There is lots of information at https://en.m.wikipedia.org/wiki/Prime_gap – Empy2 Dec 26 '21 at 01:49
  • The formula $c\log n\log\log n\log\log\log\log n/\log\log\log n$ in that refence, that the gap gets above infinitely often. You can also replace $n$ with $x$ in that formula because $\log x\approx \log n +\log\log n$, and $\log x/\log n\to 1$ – Empy2 Dec 26 '21 at 02:03
  • This appears approximation answer. I am looking for exact limit. Anyway, do you know if how does this approximation compare with < 2√p? Is 2√p smaller or the approximation you are referring to? – kpv Dec 26 '21 at 03:31
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    The same reference gives the gap below $x^{0.525}$ (which is bigger than $2\sqrt x$), but that is only when $x$ is large enough. Sorry I don't know of absolute bounds. – Empy2 Dec 26 '21 at 03:58

2 Answers2

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Bertrand's Postulate states that there's always a prime between $x$ and $2x$ giving a bound of $y-x\leq x$.

Alex R.
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Well the Bertrand–Chebyshev theorem states there is always a prime between $n$ and $2n$ for any $n>1$. Hence, given a prime, $p$, there is guaranteed to be a prime within $2p$. I am sure that can be reduced, as you have noted there is probably a prime within $2\sqrt{n}$, but this appears to be conjecture at the moment.

In practice, primes appear far far far more frequently than we understand them. We know that the probability a random number $n$ is prime is

$$\frac{1}{\log(n)}$$

This is the prime number theorem. Hence, the frequency of primes is so common that even the reciprocals of prime numbers still diverge (this is known as density, "the primes are very dense within the integers"),

$$\sum_{n=1}^{\infty}\frac{1}{p_n} = \infty$$

Proved by Euler. Something like the squares have converging reciprocals and are therefore not very dense and extremely sparse. Based on the prime number theorem, there is a 50% you will find a prime within $[n,n+\log(n)]$. Continuing this binomial distribution, for example, if $n$ is a 1000 digit number, then the probability of finding a prime within 1,000,000 (which is nothing compared to the size of $n$) of $n$ is $1-2.225e^-189$ (a very near certainty).

Primes are extremely more common than $2\sqrt{n}$. No one really knows why.

Bobby Ocean
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  • Do you happen to know what is source of 2 sqrt(n) conjecture? – kpv Dec 26 '21 at 01:52
  • @kpv Sorry, I don't. The post you provided says it is called Legendre's conjecture. The post goes on to say, "if true". I took that to mean it hasn't be shown to be true yet. Again, 1000 digit number has a square root that is 500 digits long, which is still an insanely large number. Probabalisitcally a prime will certainty show up in less than adding only a 1,000,000 which us insanely smaller than the square root. – Bobby Ocean Dec 26 '21 at 02:30
  • Bobby Ocean, thank you for the info, however, I am looking for best exact gap limit proven, or conjectured, not approximations, or probabilistic ones. – kpv Dec 26 '21 at 03:35
  • @kpv Gotch ya. If your question is more towards specific latest arxiv.org research you might want to try the question on mathoverflow instead. – Bobby Ocean Dec 26 '21 at 04:50