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Earliest sequence of integers > 1 such that gcd(a(n),a(n+k)) = 1, where k = 1..a(n-1), with a(1) = 1 and a(2) = 2.
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%I #11 Sep 11 2021 03:58:05

%S 1,2,3,2,5,3,7,2,5,11,7,3,5,13,17,2,5,11,19,3,5,13,7,11,5,23,29,13,17,

%T 31,19,11,23,2,37,41,5,7,37,3,43,47,17,7,23,43,37,31,53,59,29,11,23,

%U 61,67,71,73,2,13,79,83,7,13,19,23,89,97,101,103,43,13,107,109,41,79,113,127,3,5

%N Earliest sequence of integers > 1 such that gcd(a(n),a(n+k)) = 1, where k = 1..a(n-1), with a(1) = 1 and a(2) = 2.

%C As the sequence always takes the earliest number satisfying the restriction gcd(a(n),a(n+k)) = 1, all the terms beyond a(1) will be prime.

%e a(3) = 3, as a(1) = 1, a(2) = 2, so the next one term after a(2) cannot share a divisor with 2, and the smallest such number is 3.

%e a(4) = 2 and a(5) = 5, as a(2) = 2, a(3) = 3, so the next two terms after a(3) cannot share a divisor with 3. The first such term is 2. But now a(3) = 3 and a(4) = 2, so the next three terms after a(4) cannot share a divisor with 2. The smallest number which satisfies both of these restrictions is 5.

%Y Cf. A000040, A027749.

%K nonn

%O 1,2

%A _Scott R. Shannon_, Sep 09 2021