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

[Robert Gerbicz]

Divide the formula for a(n) by 2*(n^3-12*n^2+41*n-45), and examine the same
recursion for the terms of b(n)=n!/e=n!*sum(k=0,inf,(-1)^k/k!) (here
abs(6*a(n)-b(n))<=6).

The multiplier of the (-1)^k/k!*(n-4)! term on the right side of the
equation is
(2*(n-1)*(n^3-11*n^2+30*n-21)*(n-1)*(n-2)*(n-3)+12*(n-2)^2*(n-2)*(n-3)-2*(n^3-6*n^2+17*n-21)*(n-3)+2*(n-3)*(n^3-9*n^2+20*n-15))/(2*(n^3-12*n^2+41*n-45)),
but that is exactly n*(n-1)*(n-2)*(n-3), so the sum of these 4 terms is
(-1)^k/k!*n!, what exactly appear in the formula for b(n). So this
recursion is true for b(n)=n!/e if you drop out the
(n-3)*(n^3-10*n^2+26*n-26)/(2*(n^3-12*n^2+41*n-45)) error term. (but for
n<4 you should return n!/e instead of b(n)=0). Of course you need the
closely related a(n)=floor(n!/(3*e))/2, for this you can use very good
approx. for the tail sum of b(n). (note that in the original conjectured
formula there is also a small term to possible correct the floor operation:
a term of roughly n/2.)


2015-11-02 20:34 GMT+01:00 Vladimir Reshetnikov <v.reshetnikov at gmail.com>:

> 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
>
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