# [seqfan] Re: A sequence

Neil Sloane njasloane at gmail.com
Wed Nov 26 18:17:01 CET 2014

```Robert Israel's simplified version seems pretty nice,
and captures the idea behind the original problem.
Maybe submit it as a joint submission?

Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Email: njasloane at gmail.com

On Wed, Nov 26, 2014 at 11:07 AM, <israel at math.ubc.ca> wrote:

> Too many arbitrary parameters? How about a(k) = the largest k such that
> 2^k <= k*n (for n >= 2)? This would be floor((-LambertW(-1,-ln(2)/n)/ln(2))):
> first few values are 2,3,4,4,4,5,5,5,5,6,6,6,6,6,6,
> 6,6,7,7,7,7,7,7,7,7,7,7,7,7,7,8.
>
> Cheers,
> Robert
>
>
> On Nov 26 2014, Bob Selcoe wrote:
>
>  Hi Antreas & Seqfans,
>>
>> The question boils down to what is the max value of n where 2^(n-4) < 10n
>> - 15 (i.e. 10). BTW - it should be x_6 = 142, x_7 = 194
>>
>> I don't know if the sequence is interesting enough on its own, but
>> perhaps a sequence where some variant of the equation is used, say
>> something like 2^(n-k) < 10n - 3k, and generate a sequence where a(n) is
>> the max value as k increases?
>>
>> Cheers,
>> Bob Selcoe
>>
>> --------------------------------------------------
>> From: "Antreas Hatzipolakis" <anopolis72 at gmail.com>
>> Sent: Wednesday, November 26, 2014 6:07 AM
>> To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
>> Subject: [seqfan] A sequence
>>
>>  Following is a sequence problem proposed in the Greek Math. magazine
>>> EUCLID (November 1970).
>>>
>>> Probably the sequence is not interesting for inclunding in OEIS, but
>>> anyway
>>> here it is....
>>>
>>> Translation of the problem:
>>>
>>> We consider the sequence x_1, x_2, x_3,... x_n,....:
>>>
>>> x_1 = 11, x_2 = 32, x_3 = 54, x_4= 78, x_5 = 106, x_6  = 194,....
>>>
>>> To find the greatest of the terms of the sequence each one of them is
>>> less that the sum of the two previous terms.
>>>
>>> Since the word-by-word translation doesn't make much sense, I explain:
>>>
>>> Find the greatest index m such that:
>>>
>>> x_m < x_(m-1) + x_(m-2)
>>>
>>> Well.... the formula of the sequence is
>>>
>>> x_n =  2^(n-1) + 20n - 10
>>>
>>> The greatest term in question is x_10 = 702:
>>>
>>> x_10 < x_8 + x_9 ie 702  < 278 + 428  = 706
>>>
>>> For n >10, we have that x_n > x_(n-1) + x_(n-2).
>>>
>>>
>>> APH
>>>
>>> _______________________________________________
>>>
>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>
>>>
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>>
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```