[seqfan] Re: A sequence

Bob Selcoe rselcoe at entouchonline.net
Wed Nov 26 15:56:27 CET 2014

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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