[seqfan] Naturals permutated in triplets a, b, c "with no common factor"

[William Marshall]

> From: "Eric Angelini"<Eric.Angelini at kntv.be>
> Subject: [seqfan]  Naturals permutated in triplets a, b,	c "with no
> 	common factor"
>
> Hello SeqFans,
>
> (mail also available here:
> http://www.cetteadressecomportecinquantesignes.com/TripletsPermut.htm)
>
> S = 1,2,3,4,5,9,6,7,13,8,11,19,10,17,27,12,23,35,14,15,29,16,21,37,18,25,43,...
>
> See S as a succession of triplets [a,b,c]:
>
> 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,
> 18,25,43,
> ...
>
> Rule 1) a+b=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)
> Rule 3) S is a permutation of the Naturals
>
> To build S is easy:
> - write N
> - start from the left and:
>     ->  put a "+" on top of two yet unmarked integer which will obey
>        rules (1) and (2) (always start with the smallest unmarked integer)
>     ->  put a "=" on top of the result taking the same rules into account
>
> We have:
>
> 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 ...
>
>      + + =
> 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 ...
> giving the first triplet [1,2,3]
> (used integers will be marked with a dot "." from now on)
>
>      . . . + +       =
> 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 ...
> giving the second triplet [4,5,9]
>
>      . . . . . + +   .          =
> 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 ...
> giving the third triplet [6,7,13]
>
>      . . . . . . . + .    +     .                 =
> 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 ...
> giving the third triplet [8,11,19]
>
>      . . . . . . . . . +  .     .           +     .                       =
> 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 ...
> giving the fourth triplet [10,17,27]
>
>      . . . . . . . . . .  .  +  .           .     .           +            .                      =
> 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 ...
> giving the sixth triplet [12,23,35]
>
> etc.
>
> I'm not 100% sure that S is infinite...
> Could someone please compute a hundred or so terms (if of interest)
> The "c" terms succession migth also be of interest.
> Best,
> ?.

Nobody else seems to have commented on it in the list since it was 
posted last month, although R. J. Mathar gave more terms. (The sequence 
is now A171100.) It's definitely infinite though.

If x and y are coprime then x+y is pairwise coprime to both x and y. For 
the smallest unused positive integer x, there are infinitely many larger 
integers coprime to it, but only a finite number of integers have been 
eliminated by previous triplets.

So it's always possible to construct a new triplet from the smallest 
integer x by selecting the smallest unused candidate y (coprime to x) 
such that x+y is also unused.

It also appears that the ratios a(n-2):a(n-1):a(n), n a multiple of 3, 
tends to 1:phi:phi^2 as n tends to infinity, where phi is the golden 
ratio. Is there a simple proof of this?