A368108 - OEIS

A368108

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).

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, 33, 32, 35, 34, 36, 17, 38, 39, 19, 40, 42, 44, 45, 46, 48, 23, 49, 50, 51, 52, 54, 55, 56, 57, 58, 60, 29, 62, 63, 31, 64, 65, 66, 68, 69, 70, 72, 75, 74, 76, 37, 77

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OFFSET

1,2

COMMENTS

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

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

LINKS

Michael De Vlieger, Table of n, a(n) for n = 1..10000

Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^14, showing primes in red.

FORMULA

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

EXAMPLE

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

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

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.

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

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

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.

MATHEMATICA

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

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

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

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

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

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

s = Union[s, #];

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

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

Array[a, nn] (* Michael De Vlieger, Dec 12 2023 *)

CROSSREFS