A379883 a(1) = 1. Let j = a(n-1) and r = A046144(j). Then for n > 1, if j is novel and r > 0, a(n) = r. If j is novel and r = 0 then a(n) = 1. If j has occurred k (>1) times already then a(n) = k*j.
1, 1, 2, 1, 3, 1, 4, 1, 5, 2, 4, 8, 1, 6, 1, 7, 2, 6, 12, 1, 8, 16, 1, 9, 2, 8, 24, 1, 10, 2, 10, 20, 1, 11, 4, 12, 24, 48, 1, 12, 36, 1, 13, 4, 16, 32, 1, 14, 2, 12, 48, 96, 1, 15, 1, 16, 48, 144, 1, 17, 8, 32, 64, 1, 18, 2, 14, 28, 1, 19, 6, 18, 36, 72, 1, 20, 40, 1, 21, 1, 22, 4, 20, 60, 1, 23, 10, 30, 1, 24, 72, 144, 288, 1, 25, 8, 40, 80, 1, 26, 4
Offset: 1
Examples
a(2) = 1 since a(1)=1 and and 1 has one primitive root. Since 1 has been seen twice, a(3) = 2 and then a(4) = 1 since 2 is a novel term with one primitive root. a(9) = 5, a novel term with two primitive roots so a(10) = 2, which has appeared once before (a(3)=2), so a(11) = 4, the second occurrence of 4 so a(12) = 8, a novel term with no primitive roots, meaning that a(13) = 1. The count of 1's is now 6, so a(14) = 6, meaning 5 prior terms with one primitive root and one with none.
Links
- Michael De Vlieger, Table of n, a(n) for n = 1..10000
- Michael De Vlieger, Log log scatterplot of a(n), n = 1..2^20.
Programs
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Mathematica
nn = 120; c[_] := 0; j = 1; f[x_] := f[x] = Which[ x == 1, 1, IntegerQ[PrimitiveRoot[x]], Nest[EulerPhi, x, 2], True, 0]; {j}~Join~Reap[Monitor[Do[ If[c[j] == 0, Set[k, # + Boole[# == 0]] &[f[j]]; c[j]++, k = ++c[j]*j ]; j = Sow[k], {n, 2, nn}], n] ][[-1, 1]] (* Michael De Vlieger, Jan 09 2025 *)
Extensions
a(78)=1 inserted by David Radcliffe, Aug 03 2025
Comments