A340873 a(n) is the number of iterations of A245471 needed to reach 1 starting from n.

0, 1, 3, 2, 6, 4, 4, 3, 11, 7, 7, 5, 9, 5, 5, 4, 12, 12, 12, 8, 8, 8, 8, 6, 10, 10, 10, 6, 14, 6, 6, 5, 21, 13, 13, 13, 13, 13, 13, 9, 17, 9, 9, 9, 17, 9, 9, 7, 19, 11, 11, 11, 11, 11, 11, 7, 15, 15, 15, 7, 15, 7, 7, 6, 22, 22, 22, 14, 14, 14, 14, 14, 22, 14

Offset: 1

Author

Rémy Sigrist, Jan 31 2021

Keywords

• nonn
• base

Comments

This sequence is well defined.

Sketch of proof:

- we focus on odd numbers n > 1,

- if the binary representation of n ends with k 0's and one 1:

in two steps we obtain a number with the same binary length as n

and ending with k-1 0's and one 1,

iterating again will eventually give a number ending with two or more 1's,

- if the binary representation of n ends with k 1's (k > 1):

in k+1 steps we obtain a number with a binary length strictly smaller

than that of n,

- so any odd number > 1 will eventually reach the number 1.

Examples

			For n = 10:
- the trajectory of 10 is 10 -> 5 -> 14 -> 7 -> 8 -> 4 -> 2 -> 1,
- so a(10) = 7.
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006577, A245471, A341194, A341218, A341220, A341231, A341235.

Programs

PARI

a(n) = for (k=0, oo, if (n==1, return (k), n%2, n=bitxor(n, 2*n+1), n=n/2))

Formula

a(2*n) = a(n) + 1.