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A367465
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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 product of the two primes adjacent to the commas between the terms equals the magnitude of the difference between the terms.
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3
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2, 6, 12, 16, 20, 10, 14, 28, 24, 15, 30, 26, 52, 48, 39, 78, 72, 63, 42, 21, 70, 35, 60, 45, 36, 27, 18, 22, 44, 40, 50, 46, 92, 88, 66, 33, 154, 77, 126, 105, 80, 76, 38, 34, 68, 102, 51, 340, 255, 170, 85, 110, 55, 176, 172, 86, 82, 164, 160, 135, 120, 111, 222, 148, 74, 1443, 962, 481, 312
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OFFSET
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1,1
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COMMENTS
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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. The sequence is infinite as at any time an even number is encountered that is larger than any previous term, one could choose all subsequent terms to be a(n) = a(n-1) + 2*Gpf(a(n-1)), where Gpf(a(n-1)) is the greatest prime factor of a(n-1) and where that prime is placed last in the factorization ordering while 2 is placed first. This guarantees an infinite sequence that follows the required difference rule. See A367504.
One can show that no prime p, other than a(1) = 2, can be a term as its preceding term must be p*(p+1), but the only term following p must also be p*(p+1), which has already appeared. Prime powers can appear but are rare; in the first 2500 terms the prime powers are a(4) = 16, a(26) = 27, a(694) = 729, a(1425) = 1331, a(2251) = 2197. The later four are all odd cubes. In the same range the only fixed points are 100 and 1899, although more likely exist as the terms appear to spike up to large values only to decrease again to below the line a(n) = n.
See A367504 for the conjectured sequence when an additional requirement is added that the primes in the factorization of each term must be in order.
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LINKS
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EXAMPLE
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The prime factorization of the terms, with the required prime factors in the first and last position, begins: 2, 2*3, 2*3*2, 2*2*2*2, 2*5*2, 5*2, 2*7, 2*7*2, 2*2*2*3, 3*5, 3*5*2, 2*13, 2*13*2, 2*2*2*2*3, 3*13, 2*13*3, 2*2*2*3*3, 3*3*7, 3*2*7, 3*7, 7*2*5, 7*5, 5*2*2*3, 5*3*3, 3*2*2*3, 3*3*3, 3*3*2, 2*11, 2*11*2, 2*2*5*2, 5*5*2, 2*23, 2*23*2,... .
a(4) = 16 as a(3) = 12 which is written as 2*3*2, and 16 = 2*2*2*2, so the two primes adjacent to the term separating comma are 2 and 2, and 2*2 = 4, which equals |16 - 12|. Note that after a(3) = 12 there are five possible numbers that would meet the difference requirement : 3, 8, 16, 18, 21. The first is a prime so can be discarded, while choosing 8 forces the following number to be 4, which then has no available choices so would halt the sequence. This leaves 16 as the smallest choice.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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