A367575 Lexicographically earliest infinite sequence of distinct positive numbers such that, when all terms are written as a product of their prime factors with specific primes as the first and last factor, the sum of the two primes adjacent to the commas between the terms equals the magnitude of the difference between the terms.

2, 6, 12, 5, 15, 9, 14, 24, 11, 33, 25, 18, 28, 13, 39, 34, 54, 48, 23, 69, 55, 45, 35, 27, 21, 7, 16, 20, 30, 26, 42, 38, 60, 29, 87, 77, 63, 49, 40, 19, 57, 51, 17, 36, 52, 56, 66, 50, 46, 72, 65, 80, 70, 74, 114, 76, 37, 111, 105, 91, 75, 68, 64, 31, 93, 85, 104, 78, 82, 126, 102, 88

Offset: 1

Author

Scott R. Shannon, Nov 23 2023

Keywords

• nonn

Comments

The sequence is a prime factorization version of the 'Commas sequence', A121805. Although for many terms a following number can be chosen that is smaller than the term given in the sequence and meets the term difference requirements, all such choices ultimately lead to the sequence halting as a number is eventually reached for which no unused next number exists. See the examples for the specific factorization order for the terms.

Giving a term that can be written as p*k, where p is prime and k is either prime or composite, then p*k + 2*Gpf(k) is always a next possible term, where Gpf(k) is the greatest prime dividing k. Alternatively if a term p is prime then 3*p is a next possible term. This implies the sequence is infinite as these rules could be used to find all subsequent terms once a term larger than any previous term appears in the sequence.

Numerous values can never appear in the sequence as their only possible predecessor or successor have already appeared and did not produce the term in question. For example 3,4 and 8 can never appear as their only possible preceding terms are 3,4,8,9,12 or 15, and as these later three terms appear early in the sequence and do not produce 3,4 or 8, then these can never appear. However, unlike A367465, primes and prime powers can appear as terms, the first being 5, 9, 11, 25, 13, 23,... .

Examples

			The prime factorization of the terms, with the required prime factors in the first and last position, begins: 2, 2*3, 3*2*2, 5, 5*3, 3*3, 2*7, 3*2*2*2, 11, 11*3, 5*5, 3*2*3, 7*2*2, 13, 13*3, 2*17, 3*3*2*3, 3*2*2*2*2, 23, 23*3, 11*5, 5*3*3, 7*5, 3*3*3, 3*7, 7, 2*2*2*2, 2*2*5, 5*3*2, 2*13, 3*7*2, 2*19, 3*5*2*2, 29, 29*3, 7*11, 3*3*7, 7*7, 2*5*2*2, 19, 19*3, 3*17,...
a(7) = 14 as a(6) = 9 which is written as 3*3, and 14 = 2*7, so the two primes adjacent to the term separating comma are 3 and 2, and 3+2 = 5, which equals |14 - 9|. Note that after a(6) = 9 there are three possible numbers that would meet the difference requirement for a(7) : 3, 4, 14. Choosing 3 forces the following term to be 8, which forces the following term to be 4, but 4's only successors are 8 and 9, both of which have already been used. Likewise choosing 4 leads to a similar dead-end. This leaves 14 as the smallest choice.
		

Links

• Scott R. Shannon, Table of n, a(n) for n = 1..2000

Crossrefs

Cf. A367465 (multiplication), A367504, A121805, A027746, A006530, A020639.