[seqfan] Re: Is the definition of this sequence correct?

[M. F. Hasler]

Dear Ali et al.,
Now the example is clear and also "the other direction" had been clarified
in the second proposal of name.
This clarification confirms that you can/must actually require a minimum
distance of a(n)+1 (because of the "other direction" you clarified using
the index m2).

Then I would suggest:

a(n) is the least positive integer not already in the sequence such that,
if a(m) = a(n)+1, then |m - n| > a(n).

( a(1)=1 follows from the definition.
If we feel an urge to specify it, I'd suggest to put this to the end so
that the main idea comes as early as possible in the "name", considering
esp. the truncation of names in the "pop-up titles".)

Or maybe better: (all numbers are "already in the sequence"...)

Lexicographically first permutation of the positive integers such that, if
a(m) = a(n)+1, then |m - n| > a(n), for all indices m and n.

(We can use "permutation" as shortcut for "...not occurring earlier"
because any integer will indeed occur, as soon a possible: the condition on
the distance can only "delay" its occurrence a little bit.)

- Maximilian

On Sun, Jul 3, 2022, 21:43 Ali Sada via SeqFan <seqfan at list.seqfan.eu>
wrote:

>
> Thank you, Tom. I really appreciate your response. It will take me some
> time to write a VBA program to find the terms. Manually, I will make
> mistakes!
>
> Would any of these two versions work?
>
> a(1)=1; a(n) is the least positive integer not already in the sequence
> such that the absolute distance between a(n) and a(n)+1 is >= a(n).
>
> or
>
> a(1)=1; a(n) is the least positive integer not already in the sequence
> such that |n-m1|>= a(n) and |n-m2|>= a(n)-1, where m1 and m2 are the
> indices of a(n)+1 and a(n)-1 respectively.
>
> Best,
>
> Ali
>
>
>
>
>     On Sunday, July 3, 2022 at 10:45:47 PM GMT+1, Tom Duff <
> eigenvectors at gmail.com> wrote:
>
>  No, sorry, my definition is bogus. The sequence is more complicated than I
> made it out to be.
>
> On Sun, Jul 3, 2022 at 15:01 Tom Duff <eigenvectors at gmail.com> wrote:
>
> > I think this should read:
> > a(1)=1; a(n+1) is the smallest positive integer, distinct from all a(m),
> > m<=n, with |a(n+1)-a(n)|>=a(n).
> >
> > Sequences, not their entries, are “lexicographically earliest”. The way a
> > sequence gets to be lexicographically earliest is by picking the smallest
> > eligible entry at each step. “Distance … in both directions” is best
> > expressed by explicitly saying that it’s the absolute difference.
> > All that said, I’m surprised that this sequence is not already in the
> > OEIS. Compute a bunch of terms (it’s easy, you shouldn’t need help) and
> > search for it. If it’s not there, add it.
> >
> > On Sun, Jul 3, 2022 at 04:18 Ali Sada via SeqFan <seqfan at list.seqfan.eu>
> > wrote:
> >
> >> Hi everyone,
> >>
> >> Please check this definition
> >>
> >> a(1) =1; a(n) is the lexicographically earliest positive integer such
> >> that the distance between a(n) and a(n)+1 is >= a(n) in both directions.
> >> (The distance between a(n) and a(m) is |n-m|)
> >>
> >> a(1) = 1
> >> a(2) = 2
> >> Now, a(3) cannot be 3, so a(3) = 4.
> >> a(4) cannot be 3 nor 5, so a(4) = 6.
> >> a(5) cannot be 3 nor 5 nor 7, so a(5) = 8.
> >> Now, we can use 3 for a(6) (the distance with 4 is 3).
> >> And so on.
> >>
> >> I would appreciate your help with the correct definition and terms.
> >>
> >> Best,
> >>
> >> Ali
> >>
> >> --
> >> Seqfan Mailing list - http://list.seqfan.eu/
> >>
> >
>
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