[seqfan] Re: ceiling inequality
Robert G. Wilson v
rgwv at rgwv.com
Tue Nov 4 04:15:49 CET 2014
k = 2; mx = 0; While[k < 10000001, c = k^4/((3 k^2 - 5) Ceiling[k/Sqrt[3]]^2); If[c > mx, mx = c; AppendTo[lst, k]; Print[k]]; k++]; lst
{2, 3, 5, 19, 45, 71, 168, 265, 627, 989, 2340, 3691, 8733, 13775, 32592, 51409, 121635, 191861, 453948, 716035, 1694157, 2672279, 6322680, 9973081}
FindLinearRecurrence[lst] -> {0, 4, 0, -1, 0, 0, 0}
-----Original Message-----
From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Charles Greathouse
Sent: Monday, November 03, 2014 10:37 AM
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: ceiling inequality
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/
>
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