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A366909
Lexicographically earliest infinite sequence of distinct positive integers such that, for n > 2, a(n) shares a factor with a(n-1) but not with n.
1
1, 5, 10, 15, 21, 35, 20, 25, 55, 33, 30, 65, 40, 85, 17, 51, 39, 13, 26, 91, 52, 117, 42, 7, 14, 49, 70, 45, 57, 19, 38, 95, 50, 75, 66, 11, 22, 77, 28, 63, 60, 115, 23, 69, 161, 105, 56, 119, 34, 187, 44, 99, 78, 143, 104, 169, 130, 125, 110, 121, 88, 165, 80, 135, 84, 133, 76, 171, 152, 209
OFFSET
1,2
COMMENTS
To ensure the sequence is infinite a(n) must be chosen so that it has at least one distinct prime factor that is not a factor of n+1. The first time this rule is required is when determining a(5); see the examples below. It also does not allow a(2) to equal 3 as that would then share its only prime factor with n = 3. As 2 and 4 share a factor with n = 2, this leaves a(2) = 5 as the first valid value.
One can easily show that no 3-smooth number, see A003586, can be a term; these are all blocked by the requirement that a(n) shares no factor with n, else are blocked as such a choice would violate this condition when choosing a(n+1).
For the terms studied beyond the prime a(855) = 277 all subsequent primes appear in their natural order. The earlier primes 7, 11, 13, 17, 19, 197, 199, 211, 223, 277, 281 are either out of order or reversed. The behavior of prime ordering for larger values of n is unknown.
LINKS
Scott R. Shannon, Image of the first 100000 terms. The green line is a(n) = n.
EXAMPLE
a(4) = 15 as 15 does not share a factor with 4 while sharing the factor 5 with a(3) = 10.
a(5) = 21 as 21 does not share a factor with 5 while sharing the factor 3 with a(4) = 15. Note that 3 is unused and satisfies these requirements but as 5 + 1 = 6 = 2*3 contains 3 as a prime factor, a(5) cannot contain 3 as its only distinct prime factor else a(6) would not exist. Likewise a(5) cannot equal 6, 9, 12 or 18.
KEYWORD
nonn
AUTHOR
Scott R. Shannon, Oct 27 2023
STATUS
approved