A354853 a(1) = 4, a(2) = 9; let i = a(n-2) and j = a(n-1); a(n+1) = k 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 neither i | k nor k | i.

4, 9, 10, 21, 8, 27, 14, 15, 16, 25, 6, 35, 32, 49, 12, 77, 30, 121, 18, 55, 42, 125, 24, 65, 64, 169, 20, 39, 70, 81, 28, 33, 128, 243, 22, 45, 256, 105, 26, 63, 512, 231, 34, 99, 289, 66, 85, 36, 625, 78, 95, 48, 361, 60, 133, 40, 343, 90, 91, 50, 2197, 110

Offset: 1

Author

Michael De Vlieger, Jun 23 2022

Keywords

• nonn

Comments

A restriction on the Yellowstone sequence A098550 analogous to A353917 regarding its relationship to A064413. This sequence exhibits phases similar to those in A353917, except between every other term instead of adjacent terms.

Let P = the set of distinct prime divisors of i = a(n-2), and let Q = the set of distinct prime divisors of k = a(n). Let g = gcd(i, k) > 1 and let G = {P intersect Q}. Noncoprime i and k implies |G| > 0. This sequence is such that |G| > 0, |P| > |G|, and |Q| == |G|, or vice versa, yet neither i | k nor k | i.

Theorem: terms are composite. Proof: since divisibility and coprimality between i and k is prohibited and since primes must either divide or be coprime to other numbers, no primes appear in the sequence.

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. Hence we start the sequence with {4, 9}.

Theorem: squarefree semiprimes i = pq are followed by k = p^2 or k = q^2. Proof: since omega(i) = |P| = 2 and is squarefree, we have 2 cases pertaining to successor k, both with gcd(i, k) > 1.

1.) |P| == |G| implies |Q| > |G| and |Q| > |P|.

2.) |Q| == |G| implies |P| > |G| and |P| > |Q|.

The first case implies some prime r | k yet gcd(i, r) = 1. But this would require i | k, which is prohibited. The second case suggests either p | k or q | k, but so as to satisfy non-divisibility axiom, we are forced into either k = p^e, e > 1, or k = q^m, m > 1.

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..10^4, with records in red, local minima in blue, highlighting composite prime powers in magenta, squarefree semiprimes in gold, and other squarefree numbers in green.

Crossrefs

Cf. A001221, A098550, A353917.

Programs

Mathematica

nn = 120; s = {4, 9}; state = {3, 7}; u = 4; c[] = 0; p[] = 2; p[2] = p[3] = 3; 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[#], c[s[[#]]]}, {s[[#]], #}] &, Length[s]]; While[Nand[c[u] == 0, CompositeQ[u]], u++]; Set[{i, j}, s[[-2 ;; -1]]]; Do[k = u; If[PrimeNu[i] == PrimeOmega[i] == 2, k = Min[Map[#^p[#] &, FactorInteger[i][[All, 1]]]], 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[PrimePowerQ@ k, p[FactorInteger[k][[1, 1]]]++]; If[k == u, While[Nand[c[u] == 0, CompositeQ[u]], u++]], {n, Length[s] + 1, nn}]; Array[a, nn]