OFFSET
0,1
COMMENTS
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 5,3,4. For example, starting with 6 we get 6,8,10,24,7,25,15,9,12,5,3,4, stop (A235599).
By beginning with 3 or 4, we make sure that the 5,3,4 dead-end is never available.
If infinite, is it a permutation of the integers >= 3? This seems likely. Proving it doesn't seem easy though.
Comment from Jim Nastos, Dec 30 2013: Your question about whether the sequence can 'paint itself into a corner' is essentially asking if the Pythagorean graph has a Hamiltonian path. As far as I know, the questions in the Cooper-Poirel paper (see link) are still unanswered. They ask 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 integers >=3).
Lars Blomberg has computed the sequence out to 3 million terms without finding a dead end.
Position of k>2: 0, 1, 2, 7, 10, 6, 4, 8, 35, 3, 67, 30, 5, 13, 89, 15, 143, 12, 22, 118, 385, 9, 11, ..., see A236243. - Robert G. Wilson v, Jan 17 2014
LINKS
Robert G. Wilson v, Table of n, a(n) for n = 0..10000
J. Cooper and C. Poirel, Note on the Pythagorean Triple System
MATHEMATICA
f[s_List] := Block[{n = s[[-1]]}, sol = Solve[ x^2 + y^2 == z^2 && x > 0 && y > 0 && z > 0 && (x == n || z == n), {x, y, z}, Integers]; Append[s, Min[ Complement[ Union[ Extract[sol, Position[ sol, _Integer]]], s]]]]; lst = Nest[f, {3}, 250] (* Robert G. Wilson v, Jan 17 2014 *)
CROSSREFS
KEYWORD
nonn,nice
AUTHOR
Jack Brennen, Dec 26 2013
STATUS
approved