A368133 a(1,2,3) = 1,2,3; let j = a(n-1), M(n) = Product_{i = 1..n-2} { p a distinct prime: p | a(i), gcd(p, j) = 1 }. For n > 3, a(n) is the least novel multiple of M(n) if M(n) > 1; otherwise a(n) is the least novel multiple of A053669(j), the smallest prime which does not divide j.
1, 2, 3, 4, 6, 5, 12, 10, 9, 20, 15, 8, 30, 7, 60, 14, 45, 28, 75, 42, 25, 84, 35, 18, 70, 21, 40, 63, 50, 105, 16, 210, 11, 420, 22, 315, 44, 525, 66, 140, 33, 280, 99, 350, 132, 175, 198, 245, 264, 385, 24, 770, 27, 1540, 36, 1155, 26, 2310, 13, 4620, 39, 3080
Offset: 1
Keywords
Examples
a(1, 2, 3) = 1, 2, 3. M(4) = 2 because 2 | a(2) but does not divide a(3); 2 is the only a(m), m < 3, with this property, so a(4) = 4, the least novel multiple of 2. Now we have a(1,2,3,4) = 1,2,3,4. M(5) = 3 because 3 | a(3) but does not divide a(4); 3 is the only a(m), m < 4, with this property, so a(5) = 2*3 = 6, the least novel multiple of 3. We now have a(1..5) = 1, 2, 3, 4, 6. M(6) = 1, the empty product, because there is no prime which divides some a(m), m < 5, which does not also divide a(n-1) = 6. This situation invokes the second condition of the definition, so a(6) = 1*5, the least novel multiple of A053669(6) = 5, the smallest prime which does not divide 6. Consequently a(7) = 2*6 = 12 because no prime dividing a(1..5) also divides 5. The same situation arises again at a(13) = 30 = 2*3*5; every prime divisor of a(m), m < 13, is 2, 3, or 5, which again invokes the second condition, M(14) = 1, the empty product, so a(14) = 1*7, since A053669(30) = 7. Consequently a(15) = 2*7 = 14. a(91307) = 61 (whereas A362855(91307) = 53; point of divergence from A362855).
Links
- Michael De Vlieger, Log log scatterplot of a(n) and b(n), n = 1..2^20, where a(n) is shown in red, b(n) = A362855(n) is shown in blue. Black indicates where a(n) = b(n).
Programs
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Mathematica
nn = 10^5; c[] := False; m[] := 1; Array[Set[{a[#], c[#], m[#]}, {#, True, 2}] &, 3]; j = 3; s = {2}; r = Max[s]; c[3] = False; q[x_] := Block[{qq = 2}, While[Divisible[x, qq], qq = NextPrime[qq]]; qq]; Do[(If[# == 1, Set[k, NextPrime[r]], While[Or[c[#], # == j] &[# m[#]], m[#]++]; Set[k, # m[#]]] &[Times @@ Complement[s, #]]; 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, Jan 05 2024 *)
Formula
Extensions
More terms from Michael De Vlieger, Jan 05 2024
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