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a(1,2,3) = 1,2,3. For n > 3, a(n) is the smallest of the least novel multiples of all primes which divide an earlier term but do not divide a(n-1). If the prime divisors of all prior terms also divide a(n-1), a(n) is the least novel multiple of the smallest prime which does not divide a(n-1).
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%I #14 Dec 25 2023 10:03:02

%S 1,2,3,4,6,5,8,9,10,12,15,14,18,7,16,20,21,22,24,11,25,26,27,13,28,30,

%T 33,32,35,34,36,17,38,39,19,40,42,44,45,46,48,23,49,50,51,52,54,55,56,

%U 57,58,60,29,62,63,31,64,65,66,68,69,70,72,75,74,76,37,77

%N a(1,2,3) = 1,2,3. For n > 3, a(n) is the smallest of the least novel multiples of all primes which divide an earlier term but do not divide a(n-1). If the prime divisors of all prior terms also divide a(n-1), a(n) is the least novel multiple of the smallest prime which does not divide a(n-1).

%C Conjectured to be a permutation of the positive integers with primes in order.

%C Same as A351495 for the first 13 terms; diverges thereafter.

%H Michael De Vlieger, <a href="/A368108/b368108.txt">Table of n, a(n) for n = 1..10000</a>

%H Michael De Vlieger, <a href="/A368108/a368108.png">Log log scatterplot of a(n)</a>, n = 1..2^14, showing primes in red.

%F If a(m) = 2*p where p is a prime > 5 which is not already a term, then a(m+2) = p.

%e a(4) = 4, least novel multiple of 2, the smallest prime which does not divide 3.

%e a(5) = 6, least novel multiple of 3, the smallest prime which does not divide 4.

%e There is only one occasion where the second condition of the definition applies, namely a(5) = 6, where 2 and 3 have already occurred; therefore a(6) = 5, the smallest prime which does not divide 6.

%e a(7) = 8 since 2 and 3 do not divide 5, and their least novel multiples are 8, and 9 respectively.

%e Since a(7) = 8, a(8) is the least novel multiple of 3 (9) or 5 (10), so a(8) = 9.

%e a(13) = 18 and 5, 7 are the primes which divide prior terms but don't divide 18. The least novel multiple of 5 is 20, and the least novel multiple of 7 is 7, therefore a(14) = 7.

%t nn = 120; c[_] := False; m[_] := 1;

%t Array[Set[{a[#], c[#], m[#]}, {#, True, 2}] &, 3];

%t j = 3; s = {2}; r = Max[s]; c[3] = False;

%t Do[(If[Length[#] == 0, Set[k, NextPrime[r]],

%t Set[k, Min[#]]] &@

%t DeleteCases[Map[(While[c[# m[#]], m[#]++]; # m[#]) &, s], j];

%t s = Union[s, #];

%t If[Last[#] > r, r = Last[#]]) &@ FactorInteger[j][[All, 1]];

%t Set[{a[n], c[j], j}, {k, True, k}], {n, 4, nn}];

%t Array[a, nn] (* _Michael De Vlieger_, Dec 12 2023 *)

%Y Cf. A351495.

%K nonn

%O 1,2

%A _David James Sycamore_, Dec 12 2023

%E More terms from _Michael De Vlieger_, Dec 12 2023