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In this post , the function $$f(n):=\sum_{j=1}^n j!^2$$ is mentioned. $f(n)$ seems to be squarefree for every positive integer $n$.

Do we have $n+1\mid f(n)$ for some positive integer $n$ ? The conjecture is that the answer to this question is "no" : For no positive integer $n\le 10^5$ , this divisibility holds.

Motivation : I want to show that the smallest prime factor of $f(n)$ must exceed $n$. If $p$ is a prime number with $p\le n$ and if we have $p\mid f(n)$ , then we can easily conclude $p\mid f(p-1)$ , so this has to be ruled out. It would be enough for this purpose to show the above conjecture for the case that $n+1$ is prime , but I think , if there is a proof then the proof covers all positive integers.

Any ideas ?

Peter
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1 Answers1

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First of all, we can reduce our search to the $n$ such that $n+1$ is prime.

The proof of this can be done using induction (as you sorta already did):

Let $n+1$ be the first number such that $n+1 | f(n)$.

If $n+1$ is composite we can write $n+1=k \cdot p$ with $k \in \mathbb N$, $k > 1$ and $p < n$. We have $$n+1 | f(n) \implies k\cdot p | f(n) \implies p | f(n) \implies p|f(p-1) \,.$$ Since $p < n+1$ we have found a smaller number that satisfies the relation. Therefore the first $n+1$ such that $n+1|f(n)$ must be a prime.

Using this I wrote a program that searches for the first $p$ such that $p | f(p-1)$.

I found $p=1248829$ giving $n=1248828$, and indeed $n+1|f(n)$, disproving the conjecture.

However, I have not found any composite integer satisfying the relation yet.

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    Apparently, this number was already found and is listed here, here and here. – jorisperrenet Jun 07 '23 at 19:43
  • For $n=1248828$, do we also have $(n+1)^2\mid f(n)$? – mr_e_man Jul 10 '23 at 14:46
  • No we do not! My program says $f(n)\equiv869856854002\bmod(n+1)^2$. So we might be able to answer the linked question: $f(x)$ is not a perfect power for $x\geq n$, as it has only one factor of $n+1$. It remains to check that $f(x)$ is not a perfect power for $x<n$. – mr_e_man Jul 10 '23 at 14:58