[seqfan] Re: The Right Arm Reach Sequence

Allan Wechsler acwacw at gmail.com
Wed Aug 16 20:16:15 CEST 2023

```I waited a few days until I could remember Ali Sada's definition but not
the entries, and then I regenerated 1,3,2,7,15,4,31,63,127,255,5...

I confirm that this isn't in OEIS but I haven't tried Superseeker. It has a
very weird dynamic, and while an inductive argument quickly convinced me
that every number would appear, the sequence looks like it will spend most
of its time putting out astronomically high terms, but will occasionally
relent and let the next small number slip through. The exponential
excursion between 4 and 5 is probably typical.

The definition makes sense and I don't have a strong objection to it. I
would be tempted to look for a denser statement, like, "... subject to the
constraint that if k <= a(m), then a(m+k) > 2*a(m)." But I think as soon as
somebody writes code, the code may suggest a more graceful statement.

It may be very hard to predict exactly when even moderate values will first
appear. When will 10 show up? How about 100? There may be a fast way to
figure these questions out, but at the moment I can't see it.

sort of thing, so I think this is the sort of thing you will like :)

On Sun, Jul 30, 2023 at 3:39 AM Ali Sada via SeqFan <seqfan at list.seqfan.eu>
wrote:

> Hi everyone,
>
> Each positive integer k has k “right arm reach”. And numbers between k+1
> and 2k cannot be within k's right arm reach.
> Now, 2 cannot follow directly because the second position is within the
> right arm reach of 1. So, a(2) = 3.
> This step will send 4,5, and 6 to the “bullpen” waiting for their turns,
> while 2 can go back to the field since it’s out of 1’s right arm reach now.
> a(3) = 2.
> After that, the least positive integer that’s not in the bullpen is 7, so,
> a(4) = 7.  This step will send all the numbers between 8 and 14 to the
> bullpen. Since we are now in the 3’s and 7’s right arm reaches, the least
> available positive integer is 15. So, a(5) = 15. And now we are outside of
> the 3’s right arm reach, so, 4 comes out of the bullpen and a(6) = 4. And
> so on.
> of this sequence?
> “Starting at a(1) = 1, a(n) is the least positive integer, not already in
> the sequence, that satisfies the following condition for all existed terms:
> numbers between a(m)+1 and 2a(m) cannot be positioned at any location
> between m+1 and m + a(m).”
>
> Best,
>
> Ali
>
>
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>
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