A364800 The number of iterations that n requires to reach 1 under the map x -> A356874(x).
0, 1, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
Offset: 1
Examples
For n = 3 the trajectory is 3 -> 2 -> 1. The number of iterations is 2, thus a(3) = 2.
Links
- Amiram Eldar, Table of n, a(n) for n = 1..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, -1]]]]; (* A356874 *) a[n_] := Length@ NestWhileList[f, n, # > 1 &] - 1; Array[a, 100]
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PARI
f(n) = {my(b = binary(n), nb = #b); sum(i = 1, nb, b[i] * fibonacci(nb - i + 1));} \\ A356874 a(n) = if(n == 1, 0, a(f(n)) + 1);
Formula
a(n) = a(A356874(n)) + 1, for n >= 2.
Comments