16
$\begingroup$

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 $$ \underbrace{1!+2!}_{\color{blue}{3}}+\color{blue}{3}! \text{ and } \underbrace{1!+2!+\cdots+10!}_{3^2\times \color{red}{11}\times 40787}+\color{red}{11}! $$

$\endgroup$
  • 7
    $\begingroup$ 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. $\endgroup$ – David E Speyer Mar 8 '12 at 16:18
  • 2
    $\begingroup$ By the way, I just checked up to the 500th prime, 3571, without finding another example. $\endgroup$ – David E Speyer Mar 8 '12 at 16:20
  • 4
    $\begingroup$ Checked for other examples for primes up to 15000000; none found. $\endgroup$ – Douglas S. Stones Dec 29 '12 at 22:27
  • 3
    $\begingroup$ 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 $p\mid 1!+2!+\cdots+(p-1)!$. $\endgroup$ – Paolo Leonetti Feb 26 '18 at 16:48
  • 1
    $\begingroup$ 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). $\endgroup$ – Peter Košinár Feb 26 '18 at 19:41
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.