A127202 a(1)=1, a(2)=2; a(n) = the smallest positive integer not occurring earlier in the sequence such that gcd(a(n), a(n-1)) does not equal gcd(a(n-1), a(n-2)).

1, 2, 4, 3, 6, 5, 10, 7, 14, 8, 9, 12, 11, 22, 13, 26, 15, 18, 16, 17, 34, 19, 38, 20, 21, 24, 23, 46, 25, 30, 27, 28, 32, 29, 58, 31, 62, 33, 36, 35, 40, 37, 74, 39, 42, 41, 82, 43, 86, 44, 45, 48, 47, 94, 49, 56, 50, 51, 54, 52, 53, 106, 55, 60, 57, 59, 118, 61, 122, 63, 66

Offset: 1

Author

Leroy Quet, Jan 08 2007

Keywords

• nonn

Comments

This sequence appears to be a permutation of the positive integers. - Leroy Quet, Jan 08 2007

From N. J. A. Sloane, Jan 26 2017: (Start)

Theorem: This is a permutation of the positive integers.

Proof: (Outline. For details see the link.)

1. Sequence is infinite.

2. For all m, either m is in the sequence or there exists an n_0 such that for n >= n_0, a(n) > m.

3. For all primes p, there is a term divisible by p.

4. For all primes p, there are infinitely many multiples of p in the sequence.

5. Every prime appears in the sequence.

6. For any number m, there are infinitely many multiples of m in the sequence.

7. Every number m appears in the sequence.

(End)

Comment from N. J. A. Sloane, Feb 28 2017: (Start)

There are several short cycles and at least one apparently infinite orbit:

[1], [2], [3, 4], [5, 6], [7, 10, 8],

[9, 14, 22, 19, 16, 26, 24, 20, 17, 15, 13, 11],

[21, 34, 29, 25],

and the first apparently infinite orbit is, in the forward direction,

[23, 38, 33, 32, 28, 46, 41, 40, 35, 58, 51, 45, 42, 37, 62, 106, ...] (see A282712), and in the reverse direction

[23, 27, 31, 36, 39, 44, 50, 57, 65, 73, 82, 47, 53, 61, 68, 77, ...] (see A282713). (End)

Conjecture: The two lines in the graph are (apart from small local deviations) defined by the same equations as the two lines in the graph of A283312. - N. J. A. Sloane, Mar 12 2017

Examples

			gcd(a(7), a(8)) = gcd(10,7) = 1. So a(9) is the smallest positive integer which does not occur earlier in the sequence and which is such that gcd(a(9), 7) is not 1. So a(9) = 14, since gcd(14,7) = 7.
		

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..75000 (First 10000 terms from Rémy Sigrist)
• Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang, Finding structure in sequences of real numbers via graph theory: a problem list, arXiv:2012.04625, Dec 08, 2020
• N. J. A. Sloane, Proof that A127202 is a permutation.

Crossrefs

Cf. A127203, A283312.

Agrees with A280985 for first 719 terms.

For fixed points see A281353. See also A282712, A282713.

Programs

Mathematica

f[l_List] := Block[{k = 1, c = GCD[l[[ -1]], l[[ -2]]]},While[MemberQ[l, k] || GCD[k, l[[ -1]]] == c, k++ ];Append[l, k]];Nest[f, {1, 2}, 69] (* Ray Chandler, Jan 16 2007 *)

PARI

\\ based on Rémy Sigrist's program for A280985
{ seen = 0; p = 1; g = 2;
        for (n=1, 10000,
                a = 1;
while (bittest(seen, a) || (n>2 && gcd(p,a)==g), a++; );
                print (n " " a);
                g = gcd(p,a);
                p = a;
                seen += 2^a;
        )
}

Extensions

Extended by Ray Chandler, Jan 16 2007