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

[Neil Fernandez]

Hi all,

In message <CABdyEv8BABsr-yaxgn0YUtWr5nAZA8k29JFsFTurkGjXXX9udA at mail.gma
il.com>, Vladimir Reshetnikov <v.reshetnikov at gmail.com> writes

>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.

For n>2k-1, k>0, floor(n!/((2k+1)e)) seems to be a candidate, but I
haven't checked for very high n.

Neil

-- 
Neil Fernandez