[seqfan] Re: FW: A100083

[William Keith]

Instead of summing factorials, one may consider only the constant term (in
n) of the sum

(n+1)! + n! + ... + 2!
= (...(n+1)+1)(n) + 1)(n-1) +1)(n-2)+1)(n-3)+1) ... )(n-(n-2)).

The sequence of these constant terms is

1, 0, -1, 0, -3, 8, -45, 264, -1855, 14832, ...

which, after the first 1 (there because the sum is of (m+1)! instead of
m!), are the sums of reversed negative factorials, e.g.

264
= -6 + (-6)(-5) + (-6)(-5)(-4) + (-6)(-5)(-4)(-3) + (-6)(-5)(-4)(-3)(-2) +
(-6)(-5)(-4)(-3)(-2)(-1)
= - 6 + 30 - 120 + 360 - 720 + 720.

Dropping also the signs and the first 0 as well, these are the rencontres
numbers, A000240.  The question becomes when n divides the (n-2)-th term of
this sequence, where by convention we consider the -1st and 0th terms 0, so
n=1 and n=2 qualify.

Cordially,
William Keith