%I #24 Feb 09 2021 11:00:58

%S 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,

%T 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,

%U 15,15,15,7,15,7,7,6,22,22,22,14,14,14,14,14,22,14

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

%C This sequence is well defined.

%C Sketch of proof:

%C - we focus on odd numbers n > 1,

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

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

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

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

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

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

%C than that of n,

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

%H Rémy Sigrist, <a href="/A340873/b340873.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>

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

%e For n = 10:

%e - the trajectory of 10 is 10 -> 5 -> 14 -> 7 -> 8 -> 4 -> 2 -> 1,

%e - so a(10) = 7.

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

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

%K nonn,base

%O 1,3

%A _Rémy Sigrist_, Jan 31 2021