[seqfan] Re: ceiling inequality
apovolot at gmail.com
apovolot at gmail.com
Tue Nov 4 06:00:34 CET 2014
http://www.wolframalpha.com/input/?i=%7B2%2C+3%2C+5%2C+19%2C+45%2C+71%2C+168%2C+265%2C+627%2C+989%2C+2340%2C+3691%2C+8733%2C+13775%2C+32592%2C+51409%2C+121635%2C+191861%2C+453948%2C+716035%2C+1694157%2C+2672279%2C+6322680%2C+9973081%2C...%7D
> On Nov 3, 2014, at 10:15 PM, Robert G. Wilson v <rgwv at rgwv.com> wrote:
>
> 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
>>
>>
>>
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>>
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