Calcule $\gcd(0!+1!+\ldots+n!, (n+1)!)$

Question

I have to compute $d=\gcd(0!+1!+\ldots+n!, (n+1)!)$, so let $a=0!+1!+\ldots+n!$ and $b=(n+1)!$. Then:

$a=0!+1!+\ldots+n!=3!+0!+1!+2!+4!+...+n!=6+4+4!+...+n! \equiv 2 \mod 4$

Thus, $a$ and $b$ are even, but $4$ does not divide $a$, so $2 | d$. Now, if no other prime $p>2$ divides $a$ and $b$ we will have $d=2$. It is enough then to check the case where $n+1=p$, obviously, $p|(n+1)!$, so I have to check that $p$ does not divide $a=0!+1!+\ldots+n!=0!+1!+\ldots+(p-1)!\equiv 1!+\ldots+(p-2)! \mod p$

But I don't know how to continue. I'm sure that this is the way, but I don't find the tools to prove it.

If you're allowed to use internet sources, have you tried to compute the first handful of terms by hand and typing them into the OEIS search box? — hmakholm left over Monica, Jun 01 '16 at 08:02
First time in my life I've heard about OEIS! But will this page answer my question? — Xavier Tapia Gonzalez, Jun 01 '16 at 08:07
x @Xavier: I don't know -- but it's always worth a try for this kind of musings. — hmakholm left over Monica, Jun 01 '16 at 08:10
I found some interesting things, but not exactlly what I need. I'm sure I'll find the anwer some day — Xavier Tapia Gonzalez, Jun 01 '16 at 08:43

score 4 · Accepted Answer · edited Apr 13 '17 at 12:58

Let me first summarize the observations you already made yourself, using the following notations: $S_n=1+1+2!+\ldots + n!$, $d_n=(S_n,(n+1)!)$

Then $2|d_n$, $2^2\nmid d_n$; the second property follows $2^2\nmid S_3$, but $2^2|(n+1)!$ for $n\ge 3$ and $S_{n+1}=S_n+(n+1)!$ so $2^2\nmid S_n$ for all $n$ by induction. Also, $d_n=2$ initially; a new prime $p$ would occur in $d$ precisely if $S_{p-1}\equiv 0 \bmod p$. We suspect that this happens for no $p$, which would imply that $d_n=2$ for all $n$.

After some googling, I found that the conjecture that $S_{p-1}\not\equiv 0\bmod p$ is known as Kurepa's conjecture, and it seems that its status is unclear (see also the comment at the linked question).

Ok, so is harder than I thought, it is indeed an open problem. But this closes my question for sure. Thank you very much Christian! — Xavier Tapia Gonzalez, Jun 02 '16 at 06:48

Calcule $\gcd(0!+1!+\ldots+n!, (n+1)!)$

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