A290633 - OEIS

%I #19 Aug 18 2017 14:06:52

%S 2,2,4,4,2,6,3,3,6,2,8,4,6,6,4,8,2,10,4,12,2,14,4,10,2,12,3,9,6,8,8,6,

%T 9,3,12,4,14,2,16,4,18,2,20,4,16,2,18,3,15,5,5,10,6,12,8,10,5,15,3,18,

%U 4,20,2,22,4,24,2,26,4,22,2,24,3,21,6,10,8,12

%N Lexicographically earliest sequence of positive integers such that, for any m and n > 0, gcd(a(n), a(n+1)) > 1 and a(n) != a(n+2), and if m < n then a(m) != a(n) or a(m+1) != a(n+1).

%C a(n) > 1 for any n > 0.

%C If we drop the constraint "a(n) != a(n+2)", then we obtain the positive even numbers interspersed with 2's: 2, 2, 4, 2, 6, ...

%C Conjecturally, (a(n), a(n+1)) runs over all pairs of noncoprime positive integers; in this sense, this sequence is opposite to sequences like Stern's diatomic series (A002487).

%C This sequence has connections with A067992: here we avoid duplicate ordered pairs of consecutive terms, there unordered pairs, here we deal with noncoprime consecutive terms, there we (conjecturally) have coprime consecutive terms; also, the scatterplots of these sequences have similarities.

%C For any prime p, the sequence contains a multiple of p: by contradiction:

%C - let p be the least prime whose multiples are missing from the sequence (note that p > 2),

%C - there is only a finite number of pairs of noncoprime (p-1)-smooth numbers < p^2,

%C - so eventually we must have a term, say a(m), > p^2,

%C - if q is the least prime factor of a(m-1), then p*q would have been a better choice for a(m), hence the contradiction.

%C Also, if p is an odd prime, then the first multiple of p appearing in the sequence is a semiprime p*q with q < p.

%C If p < q are prime, then the first multiple of p appears before the first multiple of q.

%C For any prime p, the first occurrence of p in the sequence is immediately followed by a second occurrence of p.

%C For any prime p > 3:

%C - there is a semiprime p*q with q < p in the sequence,

%C - if q = 2, then this first p*q is followed by a 4,

%C - if q > 2, then this first p*q is followed by a 2,

%C - so there are infinitely many 2's or 4's in the sequence,

%C - if there are infinitely many 2's in the sequence, then the n-th occurrence of 2 is followed by 2*(n+e) with |e| <= 1, and every even

%C   number appears in the sequence,

%C - the same conclusion applies if there are infinitely many 4's,

%C - hence every even number appear in the sequence.

%C For any n > 1, the first occurrence of n in the sequence must be either preceded or followed by the least prime factor of n (A020639).

%H Rémy Sigrist, <a href="/A290633/b290633.txt">Table of n, a(n) for n = 1..10000</a>

%H Rémy Sigrist, <a href="/A290633/a290633.gp.txt">PARI program for A290633</a>

%H Rémy Sigrist, <a href="/A290633/a290633.png">Scatterplot of the first 10000 pairs of consecutive terms</a>

%H Rémy Sigrist, <a href="/A290633/a290633_1.png">Colorized scatterplot of the first 100000 pairs of consecutive terms</a>

%e a(1) = 2 is suitable.

%e a(2) = 2 is suitable.

%e a(3) cannot be either 2 (=a(1)) or 3 (gcd(2,3)=1).

%e a(3) = 4 is suitable.

%e a(4) cannot be either 2 (=a(2)) or 3 (gcd(4,3)=1).

%e a(4) = 4 is suitable.

%e a(5) = 2 is suitable.

%e a(6) cannot be 2 (pair (2,2) already seen), 3 (gcd(2,3)=1), 4 (pair (2,4) already seen) or 5 (gcd(2,5)=1).

%e a(6) = 6 is suitable.

%o (PARI) See Links section.

%Y Cf. A002487, A020639, A067992.

%K nonn,look

%O 1,1

%A _Rémy Sigrist_, Aug 08 2017