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A153330 Differences in adjacent elements of the sequence quantifying the steps needed for n to converge to 1 in the Collatz Conjecture.

%I A153330 #15 May 04 2024 14:58:07
%S A153330 1,6,-5,3,3,8,-13,16,-13,8,-5,0,8,0,-13,8,8,0,-13,0,8,0,-5,13,-13,101,
%T A153330 -93,0,0,88,-101,21,-13,0,8,0,0,13,-26,101,-101,21,-13,0,0,88,-93,13,
%U A153330 0,0,-13,0,101,0,-93,13,-13,13,-13,0,88,0,-101,21,0,0,-13,0,0,88,-80,93
%N A153330 Differences in adjacent elements of the sequence quantifying the steps needed for n to converge to 1 in the Collatz Conjecture.
%C A153330 Collatz Conjecture: Starting with any positive integer n and continually halving it when even and tripling and adding 1 to it when odd, n will always converge to 1. A006577 is the number of iterations required to turn n into 1.
%C A153330 The sequence may be of interest because showing that all of its elements are finite is tantamount to proving the Collatz Conjecture. However there is no obvious reason to believe that demonstrating the property for this sequence would be any simpler than showing it for A006577!
%C A153330 Conjecture 1: More than half of the terms are 0. Conjecture 2: 1, 6 and 16 appear only once and 3 appears twice in the sequence, i.e., a(1) = 1, a(2) = 6, a(4) = a(5) = 3, and a(8) = 16. Conjecture 3: Except 1, 3 and 6, all terms can be written as 5x + 8y, where x and y are integers. For example, 62 = 5*6 + 8*4 and -101 = 5*(-9) + 8*(-7). Conjecture 4: The ratio of the number of terms with the value of m to that of -m approaches 1 as n tends to infinity, where m != 1, 3, 6, or 16. - _Ya-Ping Lu_, May 04 2024
%H A153330 Ian Kent, <a href="/A153330/b153330.txt">Table of n, a(n) for n = 1..10000</a>
%F A153330 a(n) = A006577(n+1) - A006577(n) for n>0.
%t A153330 Differences[Table[Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>1&]], {n,80}]] (* _Harvey P. Dale_, Oct 10 2011 *)
%o A153330 (Python)
%o A153330 def A006577(n):
%o A153330     ct = 0
%o A153330     while n != 1: n = 3*n + 1 if n%2 == 1 else n//2; ct += 1
%o A153330     return ct
%o A153330 b = 0
%o A153330 for n in range(1, 73): b_next = A006577(n+1); a = b_next - b; print(a, end = ", "); b = b_next # _Ya-Ping Lu_, May 04 2024
%K A153330 easy,sign
%O A153330 1,2
%A A153330 _Ian Kent_, Dec 23 2008