[seqfan] Re: A sequence

israel at math.ubc.ca israel at math.ubc.ca
Wed Nov 26 17:07:14 CET 2014


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
>>
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>>
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>> 
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