%I
%S 1,2,3,4,5,9,6,7,13,8,11,19,10,17,27,12,23,35,14,15,29,16,21,37,18,25,
%T 43,20,31,51,22,39,61,24,41,65,26,33,59,28,45,73,30,47,77,32,49,81,34,
%U 53,87,36,55,91,38,63,101,40,57,97,42,67,109,44,69,113,46,71,117,48,79
%N Natural numbers arranged into triples a, b, c with no common factor.
%C Regard the sequence S as a succession of triple [a,b,c]:
%C 1,2,3,
%C 4,5,9,
%C 6,7,13,
%C 8,11,19,
%C 10,17,27,
%C 12,23,35,
%C 14,15,29,
%C 16,21,37,
%C 18,25,43,
%C ...
%C Rule 1) a+b=c
%C Rule 2) "a" and "b" share no common factor (except 1), "b" and "c" share no common factor (except 1), "c" and "a" share no common factor (except 1)
%C Rule 3) S is a permutation of the natural numbers.
%C To build S is easy:
%C  write down N
%C  start from the left and:
%C > put a "+" on top of two as yet unmarked integers which will satisfy rules (1) and (2) (always start with the smallest unmarked integer)
%C > put a "=" on top of the result taking the same rules into account
%C We have:
%C N = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 ...
%C ... + + =
%C N = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 ...
%C giving the first triple [1,2,3]
%C (used integers will be marked with a circle "o" from now on)
%C ....o.o.o.+.+.......=
%C N = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 ...
%C giving the second triple [4,5,9]
%C ....o.o.o.o.o.+.+...o..........=
%C N = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 ...
%C giving the third triple [6,7,13]
%C ....o.o.o.o.o.o.o.+.o....+.....o.................=
%C N = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 ...
%C giving the fourth triple [8,11,19]
%C ....o.o.o.o.o.o.o.o.o.+..o.....o...........+.....o.......................=
%C N = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 ...
%C giving the fifth triple [10,17,27]
%C ....o.o.o.o.o.o.o.o.o.o..o..+..o...........o.....o...........+............o......................=
%C N = 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 ...
%C giving the sixth triple [12,23,35]
%C etc.
%C Is the sequence infinite?
%C Contribution from _William Rex Marshall_, Nov 07 2010: (Start)
%C Sequence is infinite. A new triple is always possible as there are always infinitely many candidate "b" values coprime to the smallest unused integer "a", and previous triples can only rule out a finite number of them and their sums (which are necessarily pairwise coprime to "a" and "b").
%C It appears that the ratios a(n2):a(n1):a(n), when n is a multiple of 3, tend to 1:phi:phi^2 as n tends to infinity, where phi is the golden ratio (A001622). Is there a simple proof of this? (End)
%H W.R. Marshall, <a href="/A171100/b171100.txt">Table of n, a(n) for n=1..10000</a>
%H Eric Angelini, <a href="http://www.cetteadressecomportecinquantesignes.com/TripletsPermut.htm">Naturals permuted in triple a, b, c "with no common factor"</a>
%H E. Angelini, <a href="/A171100/a171100.pdf">Naturals permuted in triple a, b, c "with no common factor"</a> [Cached copy, with permission]
%Y Cf. A171101 (values of c).
%K nonn,tabf
%O 1,2
%A _N. J. A. Sloane_, Sep 24 2010, based on a posting by Eric Angelini to the Sequence Fans Mailing List, Sep 16 2010
%E More terms from _R. J. Mathar_, Sep 25 2010
