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%I #6 Oct 23 2022 23:37:17
%S 1,2,3,4,6,8,9,10,12,5,14,15,7,18,21,16,24,20,27,22,30,11,25,33,35,36,
%T 28,39,26,42,13,32,52,34,65,17,40,51,38,45,19,48,57,44,54,55,46,50,23,
%U 56,69,49,60,63,58,66,29,62,87,31,72,93,64,75,68,70,85,74
%N a(1)=1,a(2)=2,a(3)=3. Thereafter, if there are prime divisors p of a(n-2) which do not divide a(n-1), a(n) is the least novel multiple of any such p. Otherwise a(n) is the least novel multiple of the squarefree kernel of a(n-2).
%C In other words, if a(n-2) has k prime divisors p_j; 1 <= j <= k which do not divide a(n-1), where 1 <= k <= omega(a(n-2)), and if m_j*p_j is the least multiple of p_j which is not already a term, then a(n) = Min{m_j*p_j; 1 <= j <= k}. Otherwise, every prime divisor of a(n-2) also divides a(n-1), in which case a(n) is the least multiple of the squarefree kernel of a(n-2) which is not already a term.
%C Unlike in A064413, a prime p occurrence here is not directly flanked by multiples of p, but by numbers x, y sharing divisors other than p. The terms preceding and following x and y are divisible by p. Typically we observe m*p, x, p, y, r*p, where gcd(x, y) > 1, for multiples m,r of p which do not always follow the (2,3) pattern observed in A064413 and elsewhere. The case of p = 13 is remarkable for being preceded and followed by two multiples of itself (the first five multiples of 13 occur within the span of eight consecutive terms).
%C Conjecture 1: A permutation of the positive integers with primes in natural order.
%C Conjecture 2: The primes are the slowest numbers to appear (see also A352187).
%H Michael De Vlieger, <a href="/A357992/a357992.png">Scatterplot of a(n)</a>, n = 1..2^20.
%H Michael De Vlieger, <a href="/A357992/a357992_1.png">Log-log scatterplot of a(n)</a>, n = 1..2^12 labeling records in red, local minima in blue, highlighting primes in green and other prime powers in gold.
%e With a(2)=2, and a(3)=3, a(4) must be 4, the least unused multiple of 2.
%e Likewise, with a(3),a(4) = 3,4 a(5) must be the 6, the least unused multiple of 3.
%e Since every divisor of 4 also divides 6 a(6) = 8, the least unused multiple of 2, (squarefree kernel of 4).
%e Since a(8),a(9) = 10,12 and 5 is the only prime dividing 10 but not 12, it follows that a(10) = 5.
%t nn = 68; c[_] = False; q[_] = 1; Array[Set[{a[#], c[#]}, {#, True}] &, 2]; q[2] = 2; Do[m = FactorInteger[a[n - 1]][[All, 1]]; f = Select[m, CoprimeQ[#, a[n - 2]] &]; If[Length[f] == 0, While[Set[k, # * q[#]]; c[k], q[#]++] &[Times @@ m], Set[{k, q[#1]}, {#2, #2/#1}] & @@ First@ MaximalBy[Map[{#, Set[g, q[#]]; While[c[g #], g++]; # g} &, f], Last] ]; Set[{a[n], c[k]}, {k, True}], {n, 3, nn}]; Array[a, nn] (* _Michael De Vlieger_, Oct 23 2022 *)
%Y Cf. A001221, A064413, A352187, A357963.
%K nonn
%O 1,2
%A _David James Sycamore_, Oct 23 2022
%E More terms from _Michael De Vlieger_, Oct 23 2022
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