[seqfan] Re: Simple sequence based on Pythagorean triples
nastos at gmail.com
Tue Dec 31 04:30:56 CET 2013
Your questions about whether the sequence can 'paint intself in a corner'
is essentially asking if the pyth graph has a hamiltonian path... as far as
I know, the questions in this paper are still unanswered:
where it is asked whether the graph is k-colorable with a finite k, or
whether it is even connected (sort of equivalent to your question of
whether it is a permutation of the ints>=3).
If anyone knows whether those questions in the paper have been resolved,
I would be very interested.
On Thu, Dec 26, 2013 at 10:48 AM, Jack Brennen <jfb at brennen.net> wrote:
> This sequence seems simple enough and yet is not in the OEIS:
> Begin with a(0) = 3.
> Let a(n) for n > 0 be the smallest positive integer not yet
> in the sequence which forms part of a Pythagorean triple
> when paired with a(n-1).
> I believe that the sequence begins:
> (Any typos are part of my cut-and-paste...)
> Two questions:
> Is the sequence infinite? Can it "paint itself into
> a corner" at any point? Note that picking any starting
> point >= 5 seems to lead to a finite sequence ending in
> 6,8,10,24,7,25,15,9,12,5,3,4 stop
> By beginning with 3 or 4, you make sure that the 5,3,4
> dead-end is never available.
> If infinite, is it a permutation of the integers >= 3?
> It seems likely. Proving it doesn't seem easy though.
> Seqfan Mailing list - http://list.seqfan.eu/
More information about the SeqFan