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Like the question title reads, in the Collatz conjecture, why are $\max(\textrm{collatz}(n))$ and $\textrm{var}(\textrm{collatz}(n))$ so closely related? See the Figure below for a log-log plot.

enter image description here

I refer with $\textrm{collatz}(n)$ to a sequence starting from $n$ (so $\max(\textrm{collatz}(n))$ is the maximum value of the sequence etc). EDIT: oops, I just noticed thet x- and y-labels are in wrong order. Maximum value is on the x-axis. The plot shows the data for sequences starting with $n = 1, ..., 100000$.

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    $\begingroup$ Also, it seems that $\textrm{max}(\textrm{log}_{2}(\textrm{collatz}(n)))<3\textrm{log}_{2}(n)$ for all $n$ using the same notation. So the left hand side says that base-two logarithm of maximum value of a sequence starting from $n$ is less than three times base-two logarithm of $n$. $\endgroup$ – mmh May 17 '17 at 21:18
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    $\begingroup$ It may seem that way to you, but if you could prove that then you would have a proof than no Collatz sequence continues indefinitely without looping (repeating a number already encountered), which would be a big step in proving the Collatz Conjecture. $\endgroup$ – Mark Fischler May 17 '17 at 21:33
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    $\begingroup$ Are the variance and maximum more closely related than we'd expect just from the fact that the variance of sampling from $\{1, 2, \dots, M\}$ increases as $M$ increases? $\endgroup$ – Misha Lavrov May 17 '17 at 21:49
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It's natural that the variance of a chaotic function of this nature will positively correlate with its maximum, particularly a function in which every known sample, shares the same minimum (1). The maximum therefore measures the spread of the sequence, which is exactly what the variance does.

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