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Let $f(n)$ be an integer function such that $$ f(n) = \sum\limits_{i=1}^{n} \sum\limits_{j=1}^{i} \sum\limits_{k=1}^{j} [(3ijk - (ij + ik + jk)) = n]. $$

Here square bracket denotes Iverson bracket.

I conjecture that if $f(n) = 0$, then $2n+1$ is always prime.

Note that these primes are just a subsequence of all primes and the sequence is not in the OEIS.

Is there a way to prove it?

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    $\begingroup$ similar appearance: a positive integer $n$ is idoneal if and only if it cannot be written as $ij + ik + jk$ for distinct positive integers $i,j,k$ oeis.org/A000926 $\endgroup$ Commented Mar 24 at 0:12

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Suppose $2n+1$ is composite, then it factors as $(2a+1)(2b+1)$ for some $a\geq b\geq 1$. Setting $i=a+1,j=b+1,k=1$ we get $3ijk-(ij+ik+jk)=2ab+a+b=n$, so $f(n)>0$.

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