cp's OEIS Frontend

Let P = the set of distinct prime divisors of j = a(n-1), and let Q = the set of distinct prime divisors of k = a(n). Let g = gcd(j, k) > 1 and let G = {P intersect Q}. Noncoprime j and k implies |G| > 0. This sequence is such that |G| > 0, |P| > |G|, and |Q| == |G|, or vice versa.

Coprimality and equality are forbidden, forcing primes into divisibility. Because of this, a(i) = mp -> a(i+1) = p -> a(i+2) = sp, with composite m and s not powers of p > 2. For n > i, multiples of p including powers may appear in the sequence. Consequently, odd primes enter the sequence late.

Numbers m that immediately precede and follow prime p have omega(m) > 1.

For any adjacent pair of terms with n > 1, if one is prime then the other cannot be a power of that prime, since such a pair would have the same number of distinct prime divisors.

This sequence requires an asymmetric version of the relation of j and k seen in A337687. In that sequence, we have P != Q, |P| > |G| and |Q| > |G|, therefore we have symmetry in that there is at least 1 prime p | j that does not divide k, and at least 1 prime q | k that does not divide j. That sequence occurs among composites m with omega(m) > 1, but this sequence admits primes, since, for |G| = 1, we must have |P| or |Q| equal to 1, and there is no prohibition for multiplicity to exceed 1.

A353917 is a version of this sequence that prohibits divisibility, hence primes do not appear, but composite prime powers do.

Open question: do the primes appear in order? (They do for n <= 2^16).

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.