[seqfan] Re: ceiling inequality

[Bob Selcoe]

Except for the omission of 12, Robert's sequence actually appears to be the 
numbers "closest" to an integer; however, omitting a(1)=2, adding a(3)=12 
and bisecting the sequence generates two sequences:  1) Charles' sequence 
and 2) recurrence equation same as Charles' with different seeds: a(m) = 
4*a(m-1) - a(m-2); a(1)=3, a(2)=12.

The second sequence therefore has the same relevant properties as discussed 
in my prior post from today (apparently asymptotic positive (k'+1)/k'); so 
the previously-discussed propositions still hold (provided the two sequences 
are proven "closest").

Cheers,
Bob Selcoe
(PS Note - error in previous posting: please disregard a(0)=1; should have 
read a(0)=-1 but this is not pertinent).

--------------------------------------------------
From: <apovolot at gmail.com>
Sent: Monday, November 03, 2014 11:00 PM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Cc: "Charles Greathouse" <crgreathouse at gmail.com>
Subject: [seqfan] Re: ceiling inequality

> 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
>>>
>>>
>>>
>>> _______________________________________________
>>>
>>> Seqfan Mailing list - http://list.seqfan.eu/
>>
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