# [seqfan] Re: The Right Arm Reach Sequence

Thu Aug 17 19:31:08 CEST 2023

``` Hi Allan,
Thank you, again and again, for your continuous encouragement. I really appreciate your support and your insight.
I have added the sequence to the OEIS. It's A363584.
Best,
Ali

On Wednesday, August 16, 2023 at 07:16:27 PM GMT+1, Allan Wechsler <acwacw at gmail.com> wrote:

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.
Neil, there is something vaguely Recamanian about this idea. You like this 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.
“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

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