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Consider the function $$f(a,b):=\sum_{j=0}^a (bj)!=1+b!+(2b)!+\cdots +(ab)!$$

Given a positive integer $b$ , are there always only a finite number of positive integers $a$ such that $f(a,b)$ is prime ?

Idea :

To show that we only have finite many prime numbers, we only have to find a prime $p$ and a positive integer $a$ such that $p<ab$ and $p\mid f(a,b)$. In this case we have the same prime factor $p$ for every larger $a$ hence only finite many primes.

However, for some positive integers $b$ , I could not find such a prime $p$ and a suitable $a$, for example for $b=15$ and $b=19$. Can we expect only finite many primes also in those cases ? Or does even a pair $(a,p)$ exist for every positive integer $b$ ?

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    $\begingroup$ What is the source of this question, please? $\endgroup$
    – Gerry Myerson
    Commented Jul 16, 2022 at 10:46
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    $\begingroup$ Well, definitely there are values of $b$ for which $f(a,b)$ is not a prime for any $a$. Indeed, if $p>3$ is a prime and $b=p-1$, by Wilson's Theorem we know that $$(p-1)!\equiv -1\pmod{p}$$ thus $p\mid f(a,b)$ for any $a\geq 1$. $\endgroup$
    – Federico Clerici
    Commented Jul 16, 2022 at 10:53
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    $\begingroup$ WHAT IS THE SOURCE OF THIS QUESTION, PLEASE? $\endgroup$
    – Gerry Myerson
    Commented Jul 17, 2022 at 21:58
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    $\begingroup$ This function is my invention. I never saw it before. It generalizes the factorial numbers $n!+1$ which have been checked for primality upto a very high level. $\endgroup$
    – Peter
    Commented Jul 18, 2022 at 8:11

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