A358921 a(1) = 1; a(n) is the smallest positive number not among the terms a(n-c .. n-1) where c = the number of times a(n-1) has occurred.
1, 2, 1, 3, 1, 2, 3, 1, 4, 1, 5, 1, 2, 3, 4, 1, 6, 1, 7, 1, 5, 2, 3, 4, 1, 8, 1, 9, 1, 6, 2, 3, 4, 1, 5, 2, 6, 1, 7, 2, 3, 4, 5, 1, 8, 2, 6, 3, 7, 1, 9, 2, 4, 5, 3, 6, 1, 10, 1, 11, 1, 12, 1, 13, 1, 14, 1, 8, 2, 3, 4, 5, 6, 1, 7, 2, 9, 1, 15, 1, 16, 1, 17, 1
Offset: 1
Examples
For a(6), a(5) = 1 has occurred 3 times, so the smallest positive integer not in {a(5), a(4), a(3)} = {1, 3, 1} is 2, thus a(6) = 2. Next, for a(7), a(6) = 2 has occurred 2 times, so the smallest positive integer not in {a(6), a(5)} = {2, 1} is 3, thus a(7) = 3. Then, for a(8), a(7) = 3 has occurred 2 times, so the smallest positive integer not in {a(7), a(6)} = {3, 2} is 1, thus a(8) = 1. Now, for a(9), a(8) = 1 has occurred 4 times, so the smallest positive integer not in {a(8), a(7), a(6), a(5)} = {1, 3, 2, 1} is 4, thus a(9) = 4. The first terms, alongside the number of times they have occurred o(n), are: n a(n) o(n) - ---- ---- 1 1 1 2 2 1 3 1 2 4 3 1 5 1 3 6 2 2 7 3 2 8 1 4 9 4 1 10 1 5
Links
- Samuel Harkness, Table of n, a(n) for n = 1..10000
- Samuel Harkness, Scatterplot of the first 2000000 terms
Programs
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Mathematica
V = {1} While[Length[V] < 84, b = 1; While[MemberQ[Take[V, -Count[V, Last[V]]], b], b++ ]; AppendTo[V, b]]; Print[V]
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PARI
{ a = o = vector(84); v = 1; for (n=1, #a, print1 (a[n]=v", "); v=setminus([1..n+1], Set(a[n-o[a[n]]+++1..n]))[1]) } \\ Rémy Sigrist, Jan 09 2023
Comments