# Asymptotics of 3n+1 stopping time

It is a well-known open problem whether the following process terminates for every positive integer:

Experiments suggest that it does, possibly with unintended side effects.

Since for any odd integer ${n}$ the number ${3n+1}$ is even, it is convenient to replace the step ${n:=3n+1}$ with ${n:=(3n+1)/2}$, as show below:

As a function of ${n}$, the stopping time has a nice patterned graph:

An odd integer ${n}$ is of the form ${4k+1}$ or ${4k+3}$. In the first case, ${(3n+1)/2 = 6k+2}$ is even, while in the second case ${(3n+1)/2 = 6k+5}$ is odd. So, if ${n}$ is picked randomly from all odd integers of certain size, the probability of ${(3n+1)/2}$ being even is ${1/2}$. Similarly, for an even ${n}$, the number ${n/2}$ can be even or odd with probability ${1/2}$.

This leads to a stochastic model of the process:

The graph of stopping time in the stochastic model is, of course random. It looks nothing like the nice pattern of the deterministic process.

However, smoothing out both graphs by moving window of width ${200}$ or so, we see the similarity:

The stochastic process is much easier to analyze. Focusing on the logarithm ${\log x}$, we see that it changes either by ${\log(1/2)}$ or by approximately ${\log (3/2)}$. The expected value of the change is ${\displaystyle \frac12\log (3/4)}$. This suggests that we can expect the logarithm to drop down to ${\log 1=0}$ in about ${\displaystyle \frac{2}{\log (4/3)}\log x}$ steps. (Rigorous derivation requires more tools from probability theory, but is still routine.)

The curve ${\displaystyle \frac{2}{\log (4/3)}\log x}$ fits the experimental data nicely. (The red curve, being randomly generated, is different from the one on the previous graph.)

For an in-depth investigation, see Lower bounds for the total stopping time of 3X+1 iterates by Applegate and Lagarias.

For the computations, I used Scilab. The function hail(n,m) calculates the stopping times up to given value of n, and takes moving average with window size m (which can be set to 1 for no averaging).

 function hail(n,m) steps=zeros(1:n); steps(1)=0 for i=2:n k=i; s=0; while k>=i if modulo(k,2)==0 then k=k/2; s=s+1; else k=(3*k+1)/2; s=s+1; end end steps(i)=s+steps(k); end total = cumsum(steps) for i=1:n-m average(i)=(total(i+m)-total(i))/m; end plot(average,'+'); endfunction 

As soon as the result of computations drops below the starting value, the number of remaining steps is fetched from the array that is already computed. This speeds up the process a bit.

The second function follows the stochastic model, for which the aforementioned optimization is not available. This is actually an interesting point: it is conceivable that the stochastic model would be more accurate if it also used the pre-computed stopping time once ${x}$ drops below the starting value. This would change the distribution of stopping times, resulting in wider fluctuations after averaging.

 function randomhail(n,m) rsteps=zeros(1:n); rsteps(1)=0 for i=2:n k=i; s=0; while k>1 if grand(1,1,"bin",1,1/2)==0 then k=k/2; s=s+1; else k=(3*k+1)/2; s=s+1; end end rsteps(i)=s; end rtotal = cumsum(rsteps) for i=1:n-m raverage(i)=(rtotal(i+m)-rtotal(i))/m; end plot(raverage,'r+'); endfunction 

## 3 thoughts on “Asymptotics of 3n+1 stopping time”

1. Upps, there was a silly error in the “stopping-time” graph. Please remove my entry… Gottfried Helms

1. Corrected. However, the initially “impressive pattern” is no more existent in the “Stopping-time”-picture