A Family Of Permutations of the Positive Integers

Leroy Quet qq-quet at mindspring.com
Sat Dec 20 04:18:29 CET 2003


[also  posted to sci.math] 

Let

a(1) = 1;
a(2) = n, n = integer >= 2;

a(m+2)  = the |a(m+1) -a(m)|th highest yet-unpicked positive integer.


By "yet-unpicked", 
I mean an integer that is not among {a(1),a(2),...,a(m+1)}.

So, for each n >=2, we get a permutation of the + integers. 

Now, the n =2 case is uninteresting, just giving the + integers in their 
own order.

But it seems that for all sufficiently large n's, the terms of the 
sequence fall into a specific pattern.

And, for all sufficiently high n's, every 3rd term forms a sequence of 
constants.

So, if a(n,m) = the m_th term of the sequence with a(n,2) = n (and a(n,1) 
= 1),

then, as n -> oo, we can get the sequence:

A(m) = a(n,3m-2),

which begins (perhaps...):

A(m)  ->  1, 2, 5, 6, 9(?),...   

Generally, the a-sequence, for all sufficiently high n's,

is(??)..

1, n, n+1, 2, n+3, n+6, 5, n+8, n+11, 6, ...


Anything interesting anyone can add to the knowledge of these 
permutations??


thanks,
Leroy
  Quet





More information about the SeqFan mailing list