A354720 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.

1, 2, 3, 10, 21, 4, 7, 6, 35, 8, 5, 12, 55, 9, 11, 15, 22, 25, 16, 45, 14, 27, 32, 33, 20, 81, 64, 39, 28, 13, 42, 65, 18, 125, 66, 85, 24, 17, 30, 119, 36, 49, 60, 77, 40, 121, 70, 99, 50, 231, 128, 63, 26, 105, 169, 75, 52, 165, 182, 135, 56, 195, 154, 117, 44

Offset: 1

Author

Michael De Vlieger, Jun 23 2022

Keywords

• nonn

Comments

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 A353916 regarding its relationship to A064413.

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..10000
• Michael De Vlieger, Annotated log-log scatterplot of a(n), n = 1..2^14, showing records in red, local minima in blue, highlighting fixed points in gold, primes in green, and composite prime powers in light green.

Crossrefs

Cf. A001221, A098550, A353916.

Programs

Mathematica

nn = 120; s = Range[3]; state = {2, 3, 4, 7}; u = 1; c[] = 0; f[j, k_] := Which[j == k, 5, GCD[j, k] == 1, 0, True, 1 + FromDigits[Map[Which[Mod[##] == 0, 1, PowerMod[#1, #2, #2] == 0, 2, True, 0] & @@ # &, Permutations[{k, j}]], 3]]; Array[Set[{a[s[[#]]], c[#]}, {#, s[[#]]}] &, Length[s]]; While[c[u] > 0, u++]; Set[{i, j}, s[[-2 ;; -1]]]; Do[k = u; While[Nand[c[k] == 0, MemberQ[state, f[i, k]], CoprimeQ[j, k]], k++]; Set[{a[n], c[k], i, j}, {k, n, j, k}]; If[k == u, While[c[u] > 0, u++]], {n, Length[s] + 1, nn}]; Array[a, nn]