A352394 a(n) = n for n <= 3; let i = a(n-2) and j = a(n-1); a(n+1) = least k not already in the sequence such that (j, k) = 1 and (i, k) = m > 1 and only one of either omega(i) or omega(k) exceed omega(m), where omega = A001221, and either i | k or k | i.

1, 2, 3, 10, 21, 5, 7, 15, 14, 165, 182, 11, 13, 22, 39, 110, 273, 55, 91, 220, 819, 4, 9, 20, 63, 260, 693, 26, 33, 130, 231, 65, 77, 195, 154, 3315, 2926, 17, 19, 34, 57, 170, 399, 85, 133, 255, 266, 51, 38, 357, 190, 119, 95, 238, 285, 2618, 3705, 187, 247

Offset: 1

Author

Michael De Vlieger, Jun 23 2022

Keywords

• nonn

Comments

Theorem: i | k implies i < k, otherwise k | i implies i > k, a consequence of definition.

Theorem: Prime i implies i < k, since prime i is forced into i | k. Conversely, prime k implies i > k.

Theorem: even terms cannot be adjacent. Proof: If prime p | j, then p cannot divide k as well, because then (j, k) >= p and by definition of "prime", p > 1, which contradicts the axiom (j, k) = 1. Since 2 is prime, consecutive even terms are prohibited.

A restriction on the Yellowstone sequence A098550 analogous to A113552 regarding its relationship to A064413.

Conjecture: sequence is not a permutation of natural numbers. Proof sketch: Since either i | k or k | i, and defining m as the smaller of the 2 terms, as n increases, it becomes harder to reach all numbers through multiplication or division by m. Therefore it would seem that there is strong tendency for the sequence to fall into multiplicative recurrence as does A113552.

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..5000
• Michael De Vlieger, Annotated log-log scatterplot of a(n), n = 1..2^12, labeling records in red and local minima in blue, highlighting primes in green, composite prime powers in cyan, with fixed points in gold.

Crossrefs

Cf. A001221, A098550, A113552.

Programs

Mathematica

nn = 120; c[_] = False; a[1] = 1; i = a[2] = 2; j = a[3] = 3; u = 4; c[1] = c[2] = True; facs = {2}; Do[k = u; While[Nand[! c[k], Xor[And[Length@ Complement[facs, #] > 0, Divisible[i, k]], And[Length@ Complement[#, facs] > 0, Divisible[k, i]]] &[FactorInteger[k][[All, 1]]], CoprimeQ[j, k]], k++]; Set[{a[n], c[k]}, {k, True}]; i = j; j = k; facs = FactorInteger[i][[All, 1]]; If[k == u, While[c[u], u++]], {n, 4, nn}]; Array[a, nn]