Q:

Let $a_1=1, a_n=(n-1)a_{n-1}+1,n\ge 2.$ Find $n$ such that $n|a_n.$

Let $a_1=1, a_n=(n-1)a_{n-1}+1,n\ge 2.$ Find $n$ such that  $n|a_n.$
My progress: Given recurrence can be rewritten as
$\frac{a_n}{(n-1)!}-\frac{a_{n-1}}{(n-2)!}=\frac{1}{(n-1)!}$
$\implies a_n=(n-1)!\sum_{k=0}^{n-1}\frac{1}{k!}$

A:

The sequence is OEIS A064383.  It begins $$ 1, 2, 4, 5, 10, 13, 20, 26, 37, 52, 65, 74, 130, 148, 185, 260, 370, 463, 481, 740, 926, 962, 1852, 1924, 2315, 2405, 4630, 4810, 6019, 9260, 9620, 12038, 17131, 24076, 30095, 34262, 60190, 68524, 85655, 120380, 171310, 222703, 342620$$
An amusing remark is that up to $600\ 000$ it is all the divisors of $2^2\cdot 5\cdot 13\cdot 37\cdot 463=4\ 454\ 060$

