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
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
Programs
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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.
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