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

%I A340873 #24 Feb 09 2021 11:00:58
%S A340873 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 A340873 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 A340873 15,15,15,7,15,7,7,6,22,22,22,14,14,14,14,14,22,14
%N A340873 a(n) is the number of iterations of A245471 needed to reach 1 starting from n.
%C A340873 This sequence is well defined.
%C A340873 Sketch of proof:
%C A340873 - we focus on odd numbers n > 1,
%C A340873 - if the binary representation of n ends with k 0's and one 1:
%C A340873        in two steps we obtain a number with the same binary length as n
%C A340873        and ending with k-1 0's and one 1,
%C A340873        iterating again will eventually give a number ending with two or more 1's,
%C A340873 - if the binary representation of n ends with k 1's (k > 1):
%C A340873        in k+1 steps we obtain a number with a binary length strictly smaller
%C A340873        than that of n,
%C A340873 - so any odd number > 1 will eventually reach the number 1.
%H A340873 Rémy Sigrist, <a href="/A340873/b340873.txt">Table of n, a(n) for n = 1..10000</a>
%H A340873 <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%F A340873 a(2*n) = a(n) + 1.
%e A340873 For n = 10:
%e A340873 - the trajectory of 10 is 10 -> 5 -> 14 -> 7 -> 8 -> 4 -> 2 -> 1,
%e A340873 - so a(10) = 7.
%o A340873 (PARI) a(n) = for (k=0, oo, if (n==1, return (k), n%2, n=bitxor(n, 2*n+1), n=n/2))
%Y A340873 Cf. A006577, A245471, A341194, A341218, A341220, A341231, A341235.
%K A340873 nonn,base
%O A340873 1,3
%A A340873 _Rémy Sigrist_, Jan 31 2021