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

[Vladimir Reshetnikov]

I found an explicit solution for the conjectured recurrence:

floor(n!/(3e))/2 = !n/6 + cos(Pi*n/3)/9 + sin(Pi*n/3)/(3*sqrt(3)) -
(-1)^n/36 - 1/4, where !n = A000166(n) is subfactorial.

In general, I conjecture that for every positive integer m the sequence
floor(n!/(m*e)) = !n/m + (some periodic sequence).

--
Vladimir Reshetnikov

On Mon, Nov 2, 2015 at 11:34 AM, Vladimir Reshetnikov <
v.reshetnikov at gmail.com> wrote:

> 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
>
>