[seqfan] Re: A269783

[Neil Sloane]

The present version of the definition of A269783 is still unclear

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 Mon, Mar 7, 2016 at 10:12 AM, Max Alekseyev <maxale at gmail.com> wrote:

> Oh, I probably see what you mean.
> Should not it be "Sequence of integers such that for each n, the
> subsequence { a(k+1) | a(k)=n } is ..." ?
> Please confirm.
> Max
>
>
> On Mon, Mar 7, 2016 at 10:03 AM, Max Alekseyev <maxale at gmail.com> wrote:
> > Hi Franklin,
> > The sequence definition is not clear to me. It says
> > "Subsequence following the instances of n in the sequence is n, n-1,
> > n+1, n-2, n+2, ... for n > 0 and n, n+1, n-1, n+2, n-2, ... for n <=
> > 0."
> > There is exactly one instance of n (the first term) in each of the
> > cases, so what is exactly "subsequence following" it?
> > Regards,
> > Max
> >
> > On Mon, Mar 7, 2016 at 3:10 AM, Frank Adams-Watters
> > <franktaw at netscape.net> wrote:
> >> Can we find a proof of the comment in https://oeis.org/draft/A269783
> that every ordered pair occurs? It seems obvious, and the numerical
> evidence is strong, but I have not been able to prove it.
> >>
> >> It is trivially easy to show that the conjecture is true if every
> integer occurs infinitely often, and still quite easy to show that this
> follows if there are  infinitely many integers occurring infinitely often.
> >>
> >> One approach to proving the latter would be to show that a(n) = o(n),
> perhaps O(sqrt(n)), but I don't see any way towards that result.
> >>
> >> Alternatively, one could get the desired final result from careful
> analysis of the patterns in the sequence; but again, I don't see  how to
> get there.
> >>
> >> Note that sequences with definitions like this can easily fail to
> include all integer pairs. For example, if the numbers following each n are
> 0, 1, -1, 2, -2, ..., independent of n we get the sequence 0, 0, 1, 0, -1,
> 0, 2, 0, -2, 0, .... If we always go up first (n, n+1, n-1, n+2, ...), we
> will get 0, 0, 1, 1, 2, 2, 3, 3, 4, 4, .... Such problems usually manifest
> themselves right away - but that isn't a proof.
> >>
> >> Franklin T. Adams-Watters
> >>
> >> P.S. If people take this offline, please include me in the discussion.
> >>
> >> --
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>
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