# Are $3$ and $11$ the only common prime factors for $\sum_{k=1}^N k!$ for $N\geq 10$?

The question was stimulated by this one:

When you look at values of $\sum_{k=1}^N k!$ for $N\geq 10$, you'll always find $3$ and $11$ among the prime factors, due to the fact that $$\sum_{k=1}^{10}k!=3^2\times 11\times 40787.$$ Every futher addend also shares a $3$ resp. $11$. Are $3$ and $11$ the only common prime factors for $\sum_{k=1}^N k!$ for $N\geq 10$?

I think, one has to show, that $\sum_{k=1}^{N}k!$ has a factor of $N+1$, because the upcoming sum will always share the $N+1$ factor as well. Therefore all sums with $N\geq 2$ share a prime factor of $3$.

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Any common factor of all divides the gcd of the first two $\rm = gcd(99\cdot 40787,99\cdot 443987) = 99$ Conversely, $99$ divides them all. So $99$ is the gcd of all of them. – Bill Dubuque Mar 8 '12 at 15:46
Let $a_p = \sum_{k=1}^p k!$. If we assume that $a_p \equiv 0 \mod p$ with "probability" $1/p$, then the expected number of times this happens for $p<N$ is $\log \log N$. This quantity is not larger then $3$ until $N$ is about 500,000,000. So the fact that you have only found two examples with a short search does not convince me that there are no more. – David Speyer Mar 8 '12 at 16:18
Let $\rm\:s(n) = n!+\cdots+1!.\:$ Note $\rm\: (s(n+1),s(n)) = ((n+1)!+s(n),s(n)) = ((n+1)!,s(n)).\:$ Hence it suffices to show that the latter gcd $= 99$. – Bill Dubuque Mar 8 '12 at 16:20
By the way, I just checked up to the 500th prime, 3571, without finding another example. – David Speyer Mar 8 '12 at 16:20
$\rm (a,b)\:$ denotes $\rm\:gcd(a,b),\:$ (standard number theory notation). – Bill Dubuque Mar 8 '12 at 18:59