Imagine each natural number as a point in space along a path on which one can stand and walk. Imagine standing at any one point and looking forward toward the next prime. If we stand at $1$ and look forward, we see that the distance to the next prime is $1$. If we stand at $2$ and look forward, we see that the distance to the next prime is $1$. Let $n$ be any natural number. Let $m$ be the distance from $n$ to the next prime. According to Bertrand's postulate, $m<n$. I have strong reason to believe that $m<2\sqrt n$. We can use the squares of the primes as framework for navigating along the natural number line. I divided the natural number line into sections. I call $1$ and $2$ section $0$ because when $n$ is $1$ or $2$, the number of prime numbers which can affect the distance to the next prime is $0$. At $1$, $m=1$. At $2$, $m=1$.
Is there a mathematical expression for this? I use m because it is the first letter of the word maximum. I say, at $1$, $m=1$. At $2$, $m=1$. Is there already a symbol in use for the maximum distance from any given number n to the next prime? Should this be expressed as a function?
$m(1)=1$ $m$ of one equals one. $m(2)=1$ $m$ of two equals one. $m(3)=2$ $m(4)=2$ From $4$, you have to go no farther than $2$ to get to the next prime. $m(5)=2$ $m(6)=2$ $m(7)=4$
Legendre's conjecture says that for every $n$ there is a prime number between $n^2$ and $(n+1)^2$, which is very close to what you surmise (since $(n+1)^2-n^2=2n+1=2\sqrt{n^2}+1$). If true it would imply that the gap after a prime $p$ could be bounded by a constant times$~\sqrt p$; however neither the conjecture nor that consequence have been proved. Given the amount of effort that no doubt has gone into trying to prove this, you can be sure that (1) no counterexamples have been found, but also that (2) there exists no elementary proof that the prime gap after $p$ can be at most $2\sqrt p$. There is more information at the linked article