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