A367504
a(1) = 2; for n > 1, a(n) = a(n-1) + 2*gpf(a(n-1)), where gpf(k) = A006530(k) = greatest prime dividing k.
Original entry on oeis.org
2, 6, 12, 18, 24, 30, 40, 50, 60, 70, 84, 98, 112, 126, 140, 154, 176, 198, 220, 242, 264, 286, 312, 338, 364, 390, 416, 442, 476, 510, 544, 578, 612, 646, 684, 722, 760, 798, 836, 874, 920, 966, 1012, 1058, 1104, 1150, 1196, 1242, 1288, 1334, 1392, 1450, 1508, 1566, 1624, 1682, 1740, 1798
Offset: 1
a(7) = 40 as a(6) = 30 = 2*3*5, thus A006530(30) = 5 and a(7) = a(6) + 2*5 = 30 + 2*5 = 40.
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NestList[#+2FactorInteger[#][[-1,1]]&,2,100] (* Paolo Xausa, Dec 31 2023 *)
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.
Original entry on oeis.org
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
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.
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