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

Author

Rok Cestnik, Jul 13 2023

Keywords

• nonn

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

Cf. A364198 (indices of record highs).

Cf. A339929, A340224.

Cf. A281130, A330772, A340134, A002516, A003056, A004001, A005185.

Programs

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 *)

Python

a = [0]; [a.append(a[abs(n-a[n]**2)]+1) for n in range(100)]

Formula

a(n) ~ (3n)^(1/3) (conjectured).