Conjecture: $ x\leqslant f(x)\leqslant x+x^{\log_{113}13} \{x|1\leqslant x\leqslant+\infty,x\in \mathrm{positive~integer}\}$?

Question

Function f(x) is the most closest prime number not less than $x$.

$f(5)=5\qquad f(9)=11$

Conjecture: $ x\leqslant f(x)\leqslant x+x^{\log_{113}13} \{x|1\leqslant x\leqslant+\infty,x\in \mathrm{positive~integer}\}$?

States that there is a prime number between $x$ and $x+x^{\log_{113}13}$ for every positive integer $x$?

For all positive integer $n$, where $p_n$ is the $n$th prime number.

$p_{n+1}-1-p_n\leqslant {\left(p_n\right)}^{\log_{113}13} \qquad$ ?

Answer 1

This result is true for all sufficiently large $x$ due to a theorem of R.C. Baker, G. Harman, and J. Pintz (see: http://www.cs.umd.edu/~gasarch/BLOGPAPERS/BakerHarmanPintz.pdf, or the official page in the journal: https://academic.oup.com/plms/article-abstract/83/3/532/1479119). In particular, they proved the following statement:

There exists a positive real number $x_0$ such that for all $x > x_0$, the interval $[x, x+ x^{0.525}]$ contains a prime number.

Answer 2

Let us assume the Riemann Hypothesis. Carneiro, Milinovich, and Soundararajan proved in 2017 that for any $x>4$, there exists a prime in $[x,x+\frac{22}{25}\sqrt{x}\log x]$. Using this result, it is straightforward to prove the OP's conjecture for $x\geq 10^{47}$ (under the Riemann Hypothesis).

See also my response to this related MO question.