

A365984


Starting with a(1) = 2, the lexicographically earliest infinite sequence of distinct positive integers such that a(n)  a(n1) is a divisor of a(n), and where a(n)  a(n1) > 1.


2



2, 4, 6, 8, 10, 12, 9, 18, 15, 20, 16, 14, 21, 24, 22, 33, 30, 25, 50, 40, 32, 28, 26, 39, 36, 27, 54, 45, 42, 35, 70, 56, 48, 44, 46, 69, 66, 55, 60, 57, 38, 76, 72, 63, 84, 77, 88, 80, 64, 62, 93, 90, 75, 78, 52, 65, 130, 104, 91, 98, 96, 92, 94, 141, 138, 115, 110, 99, 102, 51, 34, 68, 85, 170
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OFFSET

1,1


COMMENTS

For the sequence to be infinite no term can be a prime except for a(1) = 2. One can easily show that if a(n) is a prime p, then the only possible value for a(n1) or a(n+1) is 2p. If a(n) = p was a term then the difference between it and the previous term must also be p, implying the previous term is a multiple of p, so it must be 2p. As 2p has now already appeared the term after p would not exist, thus terminating the sequence.
The first term that is not a prime power that cannot be used even though it satisfies being divisible by the difference between it and the previous term is 175, which appears to be a valid value for a(214) since a(213) = 350. However the next term after 175 would have to be one of 140, 150, 168, 170, 180, 182, 200, 210, 350, but all of those values have already appeared as previous terms, so 175 can never appear else it would terminate the sequence.


LINKS

a(4) = 8 as 8  a(3) = 8  6 = 2, and 2 is a divisor of 8. Note that 3 would also satisfy this requirement, but as shown above a prime will terminate the sequence so is not permitted.


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



