A358267 a(1) = 1, a(2) = 2. Thereafter:(i). If no prime divisor of a(n-1) divides a(n-2), a(n) is the least novel multiple of the squarefree kernel of a(n-1). (ii). If some (but not all) prime divisors of a(n-1) do not divide a(n-2), a(n) is the least of the least novel multiples of all such primes. (iii). If every prime divisor of a(n-1) also divides a(n-2), a(n) = u, the least unused number.

1, 2, 4, 3, 6, 8, 5, 10, 12, 9, 7, 14, 16, 11, 22, 18, 15, 20, 24, 21, 28, 26, 13, 17, 34, 30, 25, 19, 38, 32, 23, 46, 36, 27, 29, 58, 40, 35, 42, 33, 44, 48, 39, 52, 50, 45, 51, 68, 54, 57, 76, 56, 49, 31, 62, 60, 55, 66, 63, 70, 64, 37, 74, 72, 69, 92, 78, 65

Offset: 1

Author

David James Sycamore, Nov 06 2022

Keywords

• nonn

Comments

Let a(n-2) = i, a(n-1) = j. The sequence is generated from divisor relationships j->i, ranging from coprime: gcd(i,j) =1, to partial: 1 < gcd(i,j) < j, to total: gcd(i,j) = j, using conditions described in the definition.

The first 26 terms are the same as those of A280864 and A280866.

A prime cannot occur consequent to condition (i). a(n) = prime p either because p|a(n-1) but not a(n-2); see (ii), or because every prime divisor of a(n-1) also divides a(n-2), as when for example a(n-1) is a prime power q^k and q|a(n-2), which forces a(n) = u prime, see (iii).

If a(n) = u = p from condition (iii), a(n+1) = 2*p. If p|a(n-1)-> a(n) = p we see m*p->p->u (and u may of course be prime, as in ...,13,17,...). 13 is the first prime to appear consequent to condition (ii), see Example. Consecutive primes appear often: (13,17); (53,59); (61,67); ... Sequence is conjectured to be a permutation of the positive integers with primes appearing slowest, and in natural order.

Local minima consist of 1 and the primes p, while 4p dominates the maxima as n increases. - Michael De Vlieger, Nov 06 2022

Examples

			a(1) = 1, a(2) = 2 and since 2|a(2) but not a(1), and no other primes are involved, a(3) = 4, the least novel multiple of 2, the squarefree kernel of 2 (by (i)).
Every prime divisor of a(3) = 4 also divides a(2) = 2, thus a(4) = 3, the least unused number (by (iii)).
a(23) = 13 because 13|a(22) = 26, but does not divide a(21) = 28 (by (ii)). Then since every prime divisor of a(23) also divides a(22), a(24) = 17, the least unused term (by (iii)). This is the first occasion of consecutive primes.
a(25) = 34, a(26) = 30 and there are two primes (3,5) which divide 30 but not 34. At this point the least novel multiples of 3 and 5 are 27 and 25 respectively, so a(27) = 25 (by (ii)). This is the first departure from A280864/A280866, which both have a(27) = 45.
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Michael De Vlieger, Scatterplot of a(n), n = 1..2^20
• Michael De Vlieger, Annotated scatterplot of a(n) n = 1..200, showing primes p in red, 2p in light blue, 3p in green, and 4p in dark blue.
• Michael De Vlieger, Log-log scatterplot of a(n) n = 1..2^14, showing maxima in red, local minima in blue, highlighting primes p in green and other prime powers in gold.

Crossrefs

Cf. A280864, A280866, A352187, A357942, A357963, A357992, A357994.

Programs

Mathematica

Block[{a, c, f, g, k, m, q, u, nn}, nn = 120; c[] = False; q[] = 1; Array[Set[{a[#], c[#]}, {#, True}] &, 3]; q[2] = 2; u = 3; Do[m = FactorInteger[a[n - 1]][[All, 1]]; f = Select[m, CoprimeQ[#, a[n - 2]] &]; If[AllTrue[m, Mod[a[n - 2], #] == 0 &], k = u, Set[{k, q[#1]}, {#2, #2/#1}] & @@ First@ MinimalBy[Map[{#, Set[g, q[#]]; While[c[g #], g++]; # g} &, f], Last] ]; Set[{a[n], c[k]}, {k, True}]; If[k == u, While[c[u], u++]], {n, 3, nn}]; Array[a, nn] ] (* Michael De Vlieger, Nov 06 2022 *)

Extensions

More terms from Michael De Vlieger, Nov 06 2022