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A363444
a(n) = n for n <= 3; for n > 3, a(n) is the smallest positive number that has not yet appeared that includes as factors the distinct prime factors of a(n-2) and a(n-1) that are not shared between a(n-2) and a(n-1).
1
1, 2, 3, 6, 4, 9, 12, 8, 15, 30, 10, 18, 45, 20, 24, 60, 5, 36, 90, 25, 42, 210, 35, 48, 420, 70, 21, 120, 140, 63, 150, 280, 84, 75, 350, 126, 105, 40, 168, 315, 50, 252, 525, 80, 294, 630, 55, 462, 840, 110, 231, 1050, 220, 693, 1260, 330, 77, 1470, 660, 154, 735, 990, 308, 945, 1320, 616, 1155
OFFSET
1,2
COMMENTS
The last prime to appear in the first 10000 terms is a(17) = 5, and it is unknown if more appear. The largest terms increase rapidly in size, e.g., a(8924) = 2233642178577810, although subsequent terms can be significantly smaller. It is unknown is all numbers eventually appear.
LINKS
Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20.
Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^10, showing primes in red, composite prime powers in gold, squarefree composites in green, and other numbers in blue.
Michael De Vlieger, Plot p(i)^e(i) | a(n) at (x,y) = (n,i) for n = 1..2048, 12X vertical exaggeration, with a color function representing e(i), where black indicates e(i) = 1, red indicates e(i) = 2, yellow-green = 3, green = 4, and blue = 5. The bar at bottom indicates primes in red, composite prime powers in gold, squarefree composites in green, and numbers neither squarefree nor prime powers in blue.
EXAMPLE
a(4) = 6 as a(2) = 2 and a(3) = 3 contain the distinct prime factors 2 and 3 respectively, both of which only appear in one term. Therefore a(4) is the smallest unused number that contains both 2 and 3 as factors, which is 6.
a(6) = 9 as a(4) = 6 = 2*3 and a(5) = 4 = 2*2, so 3 is the only prime factor that is not shared between these terms. Therefore a(6) is the smallest unused number that contains 3 as a factor, which is 9.
MATHEMATICA
nn = 120; c[_] := False; m[_] := 1; Array[Set[{a[#], c[#]}, {#, True}] &, 3];
i = {a[2]}; j = {a[3]}; Do[q = Times @@ SymmetricDifference[i, j]; While[c[Set[k, q m[q]]], m[q]++]; Set[{a[n], c[k], i, j}, {k, True, j, FactorInteger[k][[All, 1]]}], {n, 4, nn}]; Array[a, nn] (* Michael De Vlieger, Jun 05 2023 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Scott R. Shannon, Jun 02 2023
STATUS
approved