[seqfan] floor(n!/(3e))/2

[Vladimir Reshetnikov]

Dear SeqFans,

A great question was recently asked at Math.SE:
http://math.stackexchange.com/q/1508821/19661

It's known that floor(n!/e) is always even (see https://oeis.org/A014508).
The question asks for other irrational numbers z such that floor(n!/z) is
always even. I found two candidates: floor(n!/(3e)) and floor(n!/(11e)).
The former has been proved to be always even in an answer to the question.

A natural question arises: what is the sequence floor(n!/(3e))/2?
It starts 0, 0, 0, 0, 1, 7, 44, 309, 2472, 22249, 222493, ...

Mathematica found a conjectured recurrence that successfully checks for
several thousands terms in the sequence:

a(0) = 0, a(1) = 0, a(2) = 0, a(3) = 0, 2 (n^3 - 12 n^2 + 41 n - 45) a(n) =
2 (n - 1) (n^3 - 11 n^2 + 30 n - 21) a(n-1) + 12 (n - 2)^2 a(n-2) - 2 (n^3
- 6 n^2 + 17 n - 21) a(n-3) + 2 (n - 3) (n^3 - 9 n^2 + 20 n - 15) a(n-4) +
(n - 3) (n^3 - 10 n^2 + 26 n - 26).

Can we prove that this recurrence indeed holds for all a(n)? Is there an
explicit formula or a generating function for a(n)? Does it have
connections to other sequences in OEIS? Can we find a similar recurrence
for floor(n!/(11e))/2?

--
Thanks
Vladimir Reshetnikov