[seqfan] Re: A sequence

[Charles Greathouse]

I think that would be nice. Antreas, you might consider submitting the
original as well with a pointer to the simplified/generalized sequence, in
case someone looks for that particular problem. (Or not, as you prefer.)

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Wed, Nov 26, 2014 at 12:17 PM, Neil Sloane <njasloane at gmail.com> wrote:

> 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.
> Phone: 732 828 6098; home page: http://NeilSloane.com
> 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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