Let $n$ be 113. Use $n+C\log^2n$ to find the next consecutive prime or at least approximately how far away it is. Will you show me how to work this out step by step to show me how to use this formula? What is the constant $C$? What is the reason why $(log(n))^2$ is used in this conjecture?
The gap from 113 to 127 is 14. So $C\log^2$113 should be equal to 14 or slightly greater than 14, but less than 22.
$(ln(113))^2$ is approximately 22.348.
Is the conjecture that there is ways a prime between $n$ and $n+C\log^2n$ the same as Cramér's conjecture $O((\log n)^2)$?
Is this conjecture somehow derived from the prime counting function?
$n+C\log^2n$ is not a good approximation of the next prime after $n$. It's much more likely to be around $n+\log n.$ Rather, the largest prime gap between any two primes up to $n$ is in the neighborhood of $C\log^2n.$ It's like the difference between the time it would take to run 26 miles (a day?) vs. the world record for a marathon (~2:03). Similarly, the prime after 10^18 is 10^18 + 3 but the record gap up to 10^18 is 1442.
It's believed that, for any $C<2/e^\gamma\approx1.229,$ the largest gap up to $x$ will be greater than $C\log^2x$ infinitely often. I don't know of anyone brave enough to conjecture what would be good enough, but it wouldn't be surprising if for any $C>2/e^\gamma$ there were only finitely many (even 0?) gaps $q-p$ where $q-p>C\log^2q.$
Cramér's conjecture is precisely the assertion that there exists some $C$ such that for all large enough $x$ (say, $x\ge3$) there is always a prime between $x$ and $x+C\log^2x$. (That's what the big-O statement means.)