A364801 The number of iterations that n requires to reach a fixed point under the map x -> A022290(x).

0, 0, 0, 0, 1, 2, 3, 4, 3, 4, 5, 4, 4, 5, 6, 5, 4, 5, 6, 5, 5, 5, 6, 7, 6, 7, 6, 5, 5, 6, 7, 6, 6, 7, 6, 5, 5, 6, 7, 6, 7, 6, 6, 6, 6, 7, 8, 7, 6, 7, 8, 7, 7, 8, 7, 6, 7, 6, 6, 7, 7, 8, 7, 7, 6, 7, 8, 7, 7, 8, 7, 6, 7, 6, 6, 7, 7, 8, 7, 7, 7, 8, 7, 7, 7, 8, 7

Offset: 0

Author

Amiram Eldar, Aug 08 2023

Keywords

• nonn
• base
• easy

Comments

a(n) is well-defined since A022290(n) = n for n <= 3 (the fixed points), and A022290(n) < n for n >= 4.

Examples

			For n = 4 the trajectory is 4 -> 3. The number of iterations is 1, thus a(4) = 1.
For n = 6 the trajectory is 6 -> 5 -> 4 -> 3. The number of iterations is 3, thus a(6) = 3.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A022290.

Similar sequences: A003434, A364800.

Programs

Mathematica

f[n_] := f[n] = Module[{d = IntegerDigits[n, 2], nd}, nd = Length[d]; Total[d * Fibonacci[Range[nd + 1, 2, -1]]]]; (* A022290 *)
a[n_] := -2 + Length@ FixedPointList[f, n]; Array[a, 100, 0]

PARI

f(n) = {my(b = binary(n), nb = #b); sum(i = 1, nb, b[i] * fibonacci(nb - i + 2)); } \\ A022290
a(n) = if(n < 4, 0, a(f(n)) + 1);

Python

def A364801(n):
    if n<4: return 0
    a, b, s = 1, 2, 0
    for i in bin(n)[-1:1:-1]:
        if int(i):
            s += a
        a, b = b, a+b
    return A364801(s)+1 # Chai Wah Wu, Aug 10 2023

Formula

a(n) = a(A022290(n)) + 1, for n >= 4.