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

The question was stimulated by this one. Here it comes:

When you look at the sum $$\sum\limits_{k=1}^N k!$$ for $$N\geq 10$$, you'll always find $$3$$ and $$11$$ among the prime factors, due to the fact that $$\sum\limits_{k=1}^{10}k!=3^2\times 11\times 40787.$$ Increasing $$N$$ will give rise to factors $$3$$ resp. $$11$$.

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

I think, one has to show, that $$\sum\limits_{k=1}^{N}k!$$ has a factor of $$N+1$$, because the upcoming sum will always share the $$N+1$$ factor as well. This happens for $$\underbrace{1!+2!}_{\color{blue}{3}}+\color{blue}{3}! \text{ and } \underbrace{1!+2!+\cdots+10!}_{3^2\times \color{red}{11}\times 40787}+\color{red}{11}!$$

• 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. Mar 8, 2012 at 16:18
• By the way, I just checked up to the 500th prime, 3571, without finding another example. Mar 8, 2012 at 16:20
• Checked for other examples for primes up to 15000000; none found. Dec 29, 2012 at 22:27
• There is really little known about the distribution of factorials modulo a prime $p$. For instance, it is conjectured that the sequence $(n!)$ attains $\approx (1-\frac{1}{e})p$ residues modulo $p$ and it is known unconditionally that it is at least $$\frac{p\log\log p}{\log\log\log p}.$$ However, this does not help for your question. If I should guess an answer, I'd bet (given the "random" behaviour of the primes) that there are infinitely many primes $p$ such that $p\mid 1!+2!+\cdots+(p-1)!$. Feb 26, 2018 at 16:48
• It might be a good idea to make the problem statement unambiguous by explicitly pointing out you're looking for prime factors $p$ shared by all the sums with $N\geq (p-1)$, rather than using $N\geq 10$ (which would make the question trivial). Feb 26, 2018 at 19:41

As pointed out in the comments, the case is trivial( at least if you know some theorems) if you fix $$n>10$$, instead of $$n>p-1$$ for prime $$p$$.
It follows from Wilson's Theorem, that if you have a multiple of $$p$$ at index $$n= p-2$$ you won't at index $$p-1$$ because it will decrease out of being one for that index. $$p>12$$ implies at least $$1$$ index where it is NOT a multiple if it worked at index $$11$$. That leaves us with $$p<12$$ which would have to be factors of the sum up to $$p-1$$, $$2$$ is out as the sum is odd, $$5$$, needs $$24+1+2+6=33$$ to be a multiple of $$5$$, it isn't. Lastly $$7$$ needs $$1+2+6+24+120+720=873$$ to be a multiple which would force $$33$$ to be a multiple of $$7$$ which it isn't.