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%I #24 Apr 13 2022 01:15:02
%S 3,6,9,12,15,18,10,2,14,4,22,8,24,16,20,28,26,30,34,32,36,40,5,50,60,
%T 38,42,44,46,48,54,52,58,56,62,64,66,68,70,72,74,78,39,90,57,21,51,69,
%U 84,87,33,27,96,81,63,7,126,45,168,99,93,75,95,19,114,105,77,198,153,165,11,132,55,595
%N a(1) = 3; for n > 2, a(n) is the smallest distinct positive integer such that gcd(a(n), a(n-k)) > 1 for each divisor k of a(n) such that k < n.
%C The majority of terms are concentrated along a line whose slope is approximately 1.3. Occasionally though there are terms which correspond to the smallest unused number up to that point, and these tend to lead to a subsequent very large term. For example a(499) = 628, a(500) = 682, a(501) = 31, a(502) = 14322. Other large terms appear seemingly at random, for example a(15449) = 19880, a(15450) = 19099, a(15451) = 74962230.
%C It is likely all numbers > 1 eventually appear. The smallest number not seen after 20000 terms is 89.
%C Note that if the sequence starts with 2 then the terms are just all the increasing even numbers.
%H Michael De Vlieger, <a href="/A347404/b347404.txt">Table of n, a(n) for n = 1..10000</a>
%H Michael De Vlieger, <a href="/A347404/a347404.png">Annotated log-log scatterplot of a(n)</a>, n = 1..15450, showing records in red, local minima in blue, primes in green, highlighting fixed points in amber.
%e a(2) = 6. As a(1) = 3 the next term must be a multiple of 3, and the smallest unused such number is 6. Note that as a(2-2), a(2-3) and a(2-6) are not defined these are ignored.
%e a(7) = 10. As a(6) = 18 the next term must have 2 and/or 3 as divisors. If it has 2 as a divisor is must also have 3 and/or 5 as a divisor as a(8-2) = a(6) = 15. The smallest unused number satisfying these is 10. Note that as 5 is a divisor of 10 it must be that a(7-5) = a(2) = 6 has 2 or 5 as a divisor, which is true.
%e a(8) = 2. As a(7) = 10 the next term must have 2 and/or 5 as a divisor. As a(6) = 18 also has 2 as a divisor a(8) = 2 is the next smallest unused term.
%t nn = 74; c[_] = 0; a[1] = 3; c[3] = 1; u = 2; While[c[u] > 0, u++]; Do[k = u; While[Nand[c[k] == 0, AllTrue[TakeWhile[Divisors[k], # < i &], ! CoprimeQ[a[i - #], k] &]], k++]; Set[{a[i], c[k]}, {k, i}], {i, Length[s] + 1, nn}]; Array[a, nn] (* _Michael De Vlieger_, Apr 12 2022 *)
%Y Cf. A347179, A347619, A347406, A027750, A098550, A336957.
%K nonn
%O 1,1
%A _Scott R. Shannon_, Aug 30 2021
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