[seqfan] Re: ceiling inequality

[Charles Greathouse]

We (potentially) run into trouble where n/sqrt(3) falls just short of an
integer. Using the continued fraction of sqrt(3) I'd check 5, 19, 71, 265,
... (where each term is 4 times the previous minus the one before that). Up
to 15,000 digits these hold, so it seems likely that this is true. Maybe
some effective form of Roth's theorem can be used to prove it.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Mon, Nov 3, 2014 at 10:03 AM, Kimberling, Clark <ck6 at evansville.edu>
wrote:

>
> Hello SeqFans,
>
> It appears that  (3n^2 - 5)*ceiling(n/sqrt(3))^2 > n^4      for all n > 1.
>
> Can someone prove this are cite something?
>
> Thanks.
>
> Clark Kimberling
>
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>