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Lexicographically earliest infinite sequence of distinct positive integers such that, for n > 2, a(n) shares a factor with a(n-2)+a(n-1) but not with |a(n-2)-a(n-1)|.
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%I #15 Sep 19 2024 11:01:36

%S 1,5,9,21,25,23,3,39,7,69,19,11,15,13,35,27,31,29,33,93,49,71,45,87,

%T 55,213,67,63,75,115,57,43,65,51,145,77,37,81,59,85,99,207,17,91,123,

%U 107,95,101,119,121,105,113,109,111,125,177,151,41,117,79,147,565,89,129,327,133,135,201,161

%N Lexicographically earliest infinite sequence of distinct positive integers such that, for n > 2, a(n) shares a factor with a(n-2)+a(n-1) but not with |a(n-2)-a(n-1)|.

%C To ensure the sequence is infinite a(n) must be chosen so that a(n-1)+a(n) has one or more prime factors that are not factors of |a(n-1)-a(n)|. This criteria is first encountered when finding a(3). The smallest number that shares a factor with a(1)+a(2) = 6 while not sharing one with |a(1)-a(2)| = 4 is 3, but choosing 3 would halt the sequence since then a(4) would need to share a factor with 8 but not with 2.

%C The sequence starts with a(2) = 5 as it is naturally a sequence of odd numbers - starting with an even number will quickly result in two consecutive odd numbers, after which all terms must be odd so they do not share a factor with the difference of the previous two terms, which will always be even. Choosing a(2) = 3 however is not possible due to a similar reason to that given above.

%C The terms are slightly concentrated along many straight lines, with a higher concentration along a line slightly above a(n) = 2n. Terms below this line appear to be all prime values; see the attaching image of the first 50000 terms.

%C The primes do not occur in their natural order, and the only fixed point in the first 100000 terms is a(n) = 37; it is unknown if more exist.

%H Scott R. Shannon, <a href="/A376288/b376288.txt">Table of n, a(n) for n = 1..10000</a>

%H Scott R. Shannon, <a href="/A376288/a376288_1.png">Image of the first 50000 terms</a>. Prime values are shown in yellow. The green line is a(n) = n.

%F a(4) = 21 as a(2) + a(3) = 5 + 9 = 14 while |a(2) - a(3)| = |5 - 9| = 4, and 21 has a common factor with 14 while not sharing one with 4. Note that 7 also satisfies these criteria but it cannot be chosen else a(5) would not exist.

%Y Cf. A336957, A098550, A005408, A064413.

%K nonn

%O 1,2

%A _Scott R. Shannon_, Sep 19 2024