$n$ is a natural number $>1$, $\varphi(n)$ denotes the Euler's totient function, $P_n$ is the $n^\text{th}$ prime number and $\sigma(n)$ is the sum of the divisors of $n$. Consider the expression:
$$F(n)=\varphi(|P_{n+2}-\sigma(n)|)+1$$
Conjecture: when $F(n) \equiv 3 \pmod {20}$ then this number is a prime or not. When the number is not a prime it can be a power of prime by calculating $|P_{n+2}-\sigma(n)|=p^k$ ($p$ prime, $k$ a natural number${}> 1$). Vice-versa we introduce the following expression:
$$G(n)=|P_{n+2}-\sigma(n)|$$
When $G(n) \equiv 3 \mod 20$ then this number is a prime or not. When the number is not a prime it can be a power of prime by calculating $\varphi(|P_{n+2}-\sigma(n)|)+1$
Examples:
$n=10\,270\,001\,113$, we have:
$$F(10\,270\,001\,113) =\varphi(|P_{10\,270\,001\,115}-\sigma(10\,270\,001\,113)|)+1 =\varphi(259\,189\,944\,599-10\,468\,624\,896)+1=248\,721\,319\,703$$ which is prime because it ends by $03.$
A counterexample is found with $n=680$:
$$F(680)=\varphi(|P_{682}-\sigma(680)|)+1 = \varphi(5101 - 1620)+1=3423$$ which is not prime but we have $P_{n+2}-\sigma(n)=p^2$, more precisely it is the square of $59.$
Interestingly for $n \leq 526\,388\,126$ (calculations with PARI/GP) all counterexamples are the power of prime.
Another example is found for $k=6$, this is $n=526\,388\,126$. In this case, we have:
$$F(n)=10\,549\,870\,323$$
which is not prime and $|P_{n+2}-\sigma(n)|=47^6$ (here $k=6$).
Another example is with $n=154$ we have:
$G(154)=|P_{156}-\sigma(154)|=623=7\cdot89$ which is not a prime and not a power of prime. So we calculate $\varphi(|P_{156}-\sigma(154)|)+1=529=23^2$ which is a prime power
The question is: "Are there only these two solutions?
- A power of prime if the result is not a prime
- Or the result is prime
$F(n) \equiv 3 \mod (20)$ and $P_{n+2}$ is odd. Moreover $F(n)$ is odd if and only if $P_{n+2}-\sigma(n)$ is positive and odd. Consequently $F(n)$ is prime or the power of a prime.
