A383777 a(n) is the number of steps that n requires to reach 0 under the map: x -> 2*x + 1 if x is even; 0 if x = 1; x - lpf(x) otherwise where lpf(x) is the least prime factor of x. a(n) = -1 if 0 is never reached.

0, 1, 2, 1, 4, 1, 2, 1, 2, 3, 4, 1, 4, 1, 2, 5, 4, 1, 2, 1, 2, 3, 10, 1, 10, 3, 2, 11, 4, 1, 2, 1, 10, 3, 12, 3, 2, 1, 6, 3, 4, 1, 8, 1, 2, 9, 4, 1, 2, 9, 2, 3, 6, 1, 2, 3, 2, 3, 4, 1, 8, 1, 4, 9, 10, 9, 10, 1, 2, 11, 4, 1, 4, 1, 2, 5, 10, 5, 2, 1, 6, 3, 6, 1

Offset: 0

Author

Ya-Ping Lu, May 17 2025

Keywords

• nonn

Comments

Conjecture: a(n) != -1.

Examples

			a(10) = 4 because it takes 4 steps for 10 to reach 1 by iterating the map: 10 -> 2*10+1=21 -> 21-3=18 -> 2*18+1=37 -> 37-37=0.
		

Links

• Paolo Xausa, Table of n, a(n) for n = 0..10000

Crossrefs

Cf. A005408, A006577, A046666.

Programs

Mathematica

A383777[n_] := Length[NestWhileList[If[OddQ[#], # - FactorInteger[#][[1,1]], 2*# + 1] &, n, # >0 &]] - 1;
Array[A383777, 100, 0] (* Paolo Xausa, May 22 2025 *)

Python

from sympy import primefactors; mp = lambda x: (0 if x ==1 else x - min(primefactors(x)) if x%2 else 2*x+1)
def A383777(n, c = 0):
    while n != 0: n = mp(n); c += 1
    return c