-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).


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)?

  • $\begingroup$ For $7$, would the function return $13$ (last number) or $4$ (number of steps)? $\endgroup$ Commented Sep 12, 2019 at 0:58
  • $\begingroup$ Im looking for something that would return 4 in the case of inputting 7. $\endgroup$
    – JohnDNoone
    Commented Sep 12, 2019 at 1:04
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    $\begingroup$ 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 $n \ge 4$. $\endgroup$ Commented Sep 12, 2019 at 2:36
  • 1
    $\begingroup$ @JohnDNoone I think what you're looking for is similar to this Closed-Form Solution or Closed-form expression $\endgroup$
    – user366820
    Commented Sep 12, 2019 at 22:59
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    $\begingroup$ 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. $\endgroup$ Commented Sep 14, 2019 at 10:56

1 Answer 1


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).


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