# Is $\sum_{k=1}^{n}\frac{(n-1)!}{(k-1)!}$ composite for $n\geq 4$?

Define $$a_n$$ as follows:
$$a_1=1,\ \ a_{n+1}=na_n+1\$$ At this time, the sequence $$a_n$$ is as follows: $$a_n=\sum_{k=1}^{n}\frac{(n-1)!}{(k-1)!}$$ I made some discoveries about this sequence.
The first:$$a_k\equiv 0\pmod{m}\Rightarrow a_{k+Nm}\equiv 0\pmod{m}~~~~\forall k,m,N\in\mathbb{N}$$ The second:$$n\geq 4\,\Rightarrow\,a_n ~\mathrm{is~composite}$$ I was able to prove the first, but not the second. My expectation is that the second is correct, but I'm not sure it can be proved. My friend used computer and check $$a_n$$ is composite for $$4\leq n\leq 48$$. After $$a_{49}$$, it is too large number to check on his computer. Please let me know if you come up with a proof method. Any help is welcome!

(I am a Japanese college student. I'm sorry for my poor English.)

• This is oeis.org/A000522 – Fredrik Johansson Jul 27 '20 at 6:58
• $a_i$ is odd resp. even when $i$ is odd resp. even. So $a_i$ is certainly composite for all even $i$. – Ben Smith Jul 27 '20 at 8:14
• The series $a_{n+1}=\sum_{k=0}^{n} \frac{n!}{k!}$ can be written in another representation. That is $$a_{n+1}=2^n+\sum_{k=2}^{n} \binom{n}{k}2^{n-k}D_k$$. Where, $D_k$ is the number of derangements. – Alapan Das Jul 27 '20 at 8:29

$$a_n$$ is composite for $$4 \le n \le 2016$$.
$$a_{2017}$$ appears to be prime (it passes a strong pseudoprime test). I have not tried to certify that it is prime (this would take a while as the number has 5789 digits).
• A pity this wasn't discovered $3$ years ago. – Robert Israel Jul 27 '20 at 14:19