Bounded by a constant?

Question

What exactly is meant by "constant" when it is said that Legendre's conjecture implies that the upper bound on the prime gap above n could be bounded by the product of a constant and the square root of n? For example, if $2\sqrt n$ were the upper bound for the prime gap above n, would the 2 here be this constant?

Answer 1

You are correct. Often the Landau "Big O" notation is used. So, in some books, you might see it said that the prime gap is $O(\sqrt{n})$ (read as "order of $\sqrt{n}$" or "big oh of $\sqrt{n}$" if the conjecture is true. There are several equivalent definitions of what this means, but the idea is that a function $g(x)$ is said to be $O(f(x))$ if for there is a constant $c>0$ and point $x_0$ such $\vert g(x) \vert \leq cf(x)$ as soon as $x>x_0.$