This is inspired from this post: Prime numbers which end with $19, 39, 59, 79$ or $99$
Here I found a new formula:
$n$ is a natural number $>1$, $\varphi(n)$ denotes the Euler's totient function, $P_n$ is the $n^{th}$ prime number and $\sigma(n)$ is the sum of the divisors of n. Consider the expression:
$$F(n)=\varphi(|\sigma(n)-P_{n+2}|)+1$$
Conjecture: if this ends with 03, 23, 43, 63 or 83 then this number is always prime.
Example with $n=10270001113$, we have:
$$F(10270001113)=\varphi(\sigma(10270001113)-P_{10270001115})+1=\varphi(10468624896-259189944599)+1=248721319703$$ which is prime because it ends by 03.