

A171100


Natural numbers arranged into triples a, b, c with no common factor.


2



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, 20, 31, 51, 22, 39, 61, 24, 41, 65, 26, 33, 59, 28, 45, 73, 30, 47, 77, 32, 49, 81, 34, 53, 87, 36, 55, 91, 38, 63, 101, 40, 57, 97, 42, 67, 109, 44, 69, 113, 46, 71, 117, 48, 79
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OFFSET

1,2


COMMENTS

Regard the sequence S as a succession of triple [a,b,c]:
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 natural numbers.
To build S is easy:
 write down N
 start from the left and:
> put a "+" on top of two as yet unmarked integers which will satisfy 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 triple [1,2,3]
(used integers will be marked with a circle "o" from now on)
....o.o.o.+.+.......=
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 triple [4,5,9]
....o.o.o.o.o.+.+...o..........=
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 triple [6,7,13]
....o.o.o.o.o.o.o.+.o....+.....o.................=
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 triple [8,11,19]
....o.o.o.o.o.o.o.o.o.+..o.....o...........+.....o.......................=
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 fifth triple [10,17,27]
....o.o.o.o.o.o.o.o.o.o..o..+..o...........o.....o...........+............o......................=
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 triple [12,23,35]
etc.
Is the sequence infinite?
Contribution from William Rex Marshall, Nov 07 2010: (Start)
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").
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)


LINKS

W.R. Marshall, Table of n, a(n) for n=1..10000
Eric Angelini, Naturals permuted in triple a, b, c "with no common factor"
E. Angelini, Naturals permuted in triple a, b, c "with no common factor" [Cached copy, with permission]


CROSSREFS

Cf. A171101 (values of c).
Sequence in context: A227778 A166276 A101544 * A240062 A255129 A119586
Adjacent sequences: A171097 A171098 A171099 * A171101 A171102 A171103


KEYWORD

nonn,tabf


AUTHOR

N. J. A. Sloane, Sep 24 2010, based on a posting by Eric Angelini to the Sequence Fans Mailing List, Sep 16 2010


EXTENSIONS

More terms from R. J. Mathar, Sep 25 2010


STATUS

approved



