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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Ian Kent

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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.

Original entry on oeis.org

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, -93, 0, 0, 88, -101, 21, -13, 0, 8, 0, 0, 13, -26, 101, -101, 21, -13, 0, 0, 88, -93, 13, 0, 0, -13, 0, 101, 0, -93, 13, -13, 13, -13, 0, 88, 0, -101, 21, 0, 0, -13, 0, 0, 88, -80, 93
Offset: 1

Author

Ian Kent, Dec 23 2008

Keywords

Comments

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.
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!
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

Programs

  • Mathematica
    Differences[Table[Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,n,#>1&]], {n,80}]] (* Harvey P. Dale, Oct 10 2011 *)
  • Python
    def A006577(n):
        ct = 0
        while n != 1: n = 3*n + 1 if n%2 == 1 else n//2; ct += 1
        return ct
    b = 0
    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

Formula

a(n) = A006577(n+1) - A006577(n) for n>0.