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Definitions:

-Collatz: odd: $\frac{3x+1}{2}$ | even: $\frac{x}{2}$

-Branch: Starting at an odd number and increasing until reaching an even number. Then decreasing until reaching another odd number (ex: 7,11,17,26,13 -> 4 steps).

Question:

As far as I'm aware, there is not an equation where one can plug in a specific starting number and get the stopping time of Collatz for that number, but is there a known equation that gives the stopping time for the first 'branch' of Collatz for a given number (or in the case of 5, the full stopping time)?

JohnDNoone
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  • For $7$, would the function return $13$ (last number) or $4$ (number of steps)? – Varun Vejalla Sep 12 '19 at 00:58
  • Im looking for something that would return 4 in the case of inputting 7. – JohnDNoone Sep 12 '19 at 01:04
  • one could use a computer and make a table (with $n$ and its stopping time), and look for patterns. – Mirko Sep 12 '19 at 01:16
  • Letting $f(n)$ be the required function, it seems that $f(2^n-1) =$ A093052$(n)$. – Varun Vejalla Sep 12 '19 at 01:35
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    Also $f(2^n-k) = c$, where $c$ is constant, $n \ge$ some value, and $k \ge 2$. For example $f(2^n-3) = 3$ for $n \ge 4$. – Varun Vejalla Sep 12 '19 at 02:36
  • Are you asking for a formula? I dont understand what you mean by an equation in this sense. –  Sep 12 '19 at 21:15
  • @NaturalNumberGuy I didn't know there was a technical difference between the word equation and formula. I've always used them interchangeably...

    The scenario is: I have the number 7 and I want to know how many steps in Collatz before it reaches 13, I'm looking for an equation/function that when I plug in 7, it will equal 4.

    I'm hoping this equation/function won't require me to know that 13 is the ending number in the sub-sequence. Ideally it would only require me to input 7 as the only variable to make it equal to 4.

     – JohnDNoone Sep 12 '19 at 21:32
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    @JohnDNoone I think what you're looking for is similar to this Closed-Form Solution or Closed-form expression –  Sep 12 '19 at 22:59
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    Just for reference: Ray Steiner called this form of trajectory (what you call "branch") in 1978 a "circuit" and investigated (and disproved) the possibility of cyclic "circuits" other than the trivial one $1 \to 2 \to 1 \to \cdots $ (This should also be mentioned in wikipedia and is in all of Lagarias' overviews) The cyclic form of this form of trajectory has also been called "1-cycle" by J. Simons/B.deWeger (who build on Steiner and extended btw.) The articles of Simons/deWeger are also online and easily accessible to ones who are interested. – Gottfried Helms Sep 14 '19 at 10:56

1 Answers1

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There are two steps:

It is well known and proven that any odd number can be written as $a\cdot2^k-1$ and reaches $a\cdot3^k-1$ (which is even) in exactly $k$ steps of the condensed Collatz function $\frac {3x+1}{2}$.

Than you need to divide by $2^n$ to get your final odd number (adding $n$ steps to the $k$ steps)

Now from here you can find the specific numbers that meet some criterias: You can target those specific numbers which divide by $2^n$ exactly to reach your target number (the one being odd), or you find the specific numbers for which the stopping point is reached (in this case the target number can still be even).

For the second case, there is a formula here ( Showing $3^iq-2^{i-p}\neq2^pq-1$, with $p:=\lceil i\,\log_23\rceil$, $q:=\left\{\frac{2^{i}\,3^{2^{p-i-2}}}{2^p3^i}\right\}$, $i>6$) But unless you prove the equality is not possible (this is the question of the post), you can't prove this is the stopping point (which needs strict inequality).

Collag3n
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