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

1,3

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.

Links

Rémy Sigrist, Table of n, a(n) for n = 1..10000

Index entries for sequences related to 3x+1 (or Collatz) problem

Formula

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

Example

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

Prog

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

Crossrefs

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

Sequence in context: A339333 A339334 A360259 * A033940 A286367 A196047

Adjacent sequences: A340870 A340871 A340872 * A340874 A340875 A340876

Keyword

nonn,base

Author

Rémy Sigrist, Jan 31 2021

Status

approved