A364197 a(n+1) = a(|n-a(n)^2|) + 1, a(0) = 0.
0, 1, 1, 2, 2, 1, 3, 3, 2, 3, 1, 4, 2, 3, 3, 2, 5, 4, 2, 4, 3, 5, 3, 4, 4, 3, 6, 2, 5, 3, 4, 4, 3, 5, 3, 4, 5, 5, 3, 4, 5, 3, 4, 7, 4, 6, 4, 5, 4, 4, 6, 4, 5, 3, 5, 4, 5, 5, 4, 5, 4, 5, 6, 7, 4, 5, 6, 5, 5, 8, 2, 7, 4, 6, 6, 4, 6, 6, 4, 7, 5, 5, 6, 5, 5, 6, 5, 6, 5, 8, 4, 7, 5, 6, 6, 5, 3, 7, 5, 7
Offset: 0
Keywords
Examples
a(1) = a(|0-a(0)^2|)+1 = a(|0-0|)+1 = a(0)+1 = 1. a(2) = a(|1-a(1)^2|)+1 = a(|1-1|)+1 = a(0)+1 = 1. a(3) = a(|2-a(2)^2|)+1 = a(|2-1|)+1 = a(1)+1 = 2. a(4) = a(|3-a(3)^2|)+1 = a(|3-4|)+1 = a(1)+1 = 2. a(5) = a(|4-a(4)^2|)+1 = a(|4-4|)+1 = a(0)+1 = 1.
Links
- Michael De Vlieger, Table of n, a(n) for n = 0..10000
- Rok Cestnik, Term-referencing tree for 1000 terms.
Crossrefs
Programs
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Mathematica
a[0] = 0; a[n_] := a[n] = a[Abs[n - 1 - a[n - 1]^2]] + 1; Array[a, 100, 0] (* Amiram Eldar, Jul 13 2023 *)
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Python
a = [0]; [a.append(a[abs(n-a[n]**2)]+1) for n in range(100)]
Formula
a(n) ~ (3n)^(1/3) (conjectured).
Comments