Permutation? Floor Of Ratio Of Adjacent Terms = 2

[franktaw at netscape.net]

I would want to include cross-refs to A050000 and A114183.

(For that matter, A050000 and A114183 ought to cross-ref
each other.)

To answer your first question, the sequence is infinite.  When
m occurs in the sequence, at least one of 3m-1 and 3m-2
will not have appeared -- because if either appears without m
having appeared, it will immediately be followed by m, so the
other cannot have appeared yet.

I don't currently see how to prove that the sequence does
include every positive integer, but I'm pretty sure it does.
The set of larger numbers such that m can be the next
term is [2m..3m-1].  If m has not occurred, numbers in this
range must be followed by a number between ceiling(2m/3)
and m; so if m does not occur, at least
m - ceiling(2m/3) = floor(m/3) values in this range must also
not occur.  This implies that the set of skipped values has
finite density, and strongly suggests that it has density one.

Franklin T. Adams-Watters

-----Original Message-----
From: Leroy Quet <q1qq2qqq3qqqq at yahoo.com>

First, thanks to everyone who contributed to the
last discussion I started.

Now a new topic.

I just submitted this:

%S A139080 1,2,4,8,3,6,12,5,10,20,7,14,28,11,22,9
%N A139080 a(1)=1. a(n) = the smallest positive
integer not occurring earlier in the sequence
such that
floor(max(a(n),a(n-1))/min(a(n),a(n-1))) = 2.
%C A139080 Is there always an unused positive
integer, a(n), such that
floor(max(a(n),a(n-1))/min(a(n),a(n-1))) = 2, or
does the sequence terminate? If the sequence is
infinite, is it a permutation of the positive
integers?
%O A139080 1
%K A139080 ,more,nonn,

...