Re: Guess the Permutation

From: Leroy Quet (qqquet_at_mindspring.com)
Date: 01/13/05 

Date: 13 Jan 2005 11:02:00 -0800

I am posting the solution below.

It would be interesting if Jyrki Lahtonen's hypotheses
(especially that involving sqrt(3))
turned out to be correct.

Leroy Quet wrote:
> Here are the first terms of a sequence which
> forms a permutation of the positive integers.
>
> 1,2,3,6,4,9,5,12,7,16,8,19,10,23,11,26,
> 13,30,14,33,15,36,17,...
>
> Try to guess the rule I used to generate this permutation.
> (Not in EIS.)
>
> I know there are an infinite number of rules
> which generate permutations which start with
> the same terms as above.
> But try to guess the rule anyway.
>
> I will give the answer in a few days if no one else gives it sooner.
> thanks,
> Leroy Quet

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a(2n-1) = the lowest positive integer not occurring earlier in the
sequence.

a(2n) = the a(2n-1)th lowest positive integer not occurring earlier in
the sequence.

So, for example, the positive integers not among the first 5 terms form
the sequence
{b_5(k)}: 5,7,8,9,10,11,...

a(5) is 4; so the 4th term of {b_5(k))} is 9, which is a(6).
thanks,
Leroy Quet

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