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Let $S(n)$ be the number of steps required for $n$ to reach $1$ in the 3n+1 problem (A006577). As showed in other posts, $S(n)$ is locally random, but a local average/estimate $s(n)$ can be calculated (this can also be seen 'empirically'), such that $$s(n)=b\ln(n)$$ Where $b=\frac{3}{\ln(4/3)}\approx10.43$. This can be derived by noting that $$s(n)=\frac{1}{2}\left(1+s\left(\frac{n}{2}\right)\right)+\frac{1}{2}\left(2+s\left(\frac{3n}{2}\right)\right)$$ However, the value of the constant $b$ does not change if $s(n)=b\ln(n)+c$, for any constant $c$, that is, when substituting the expression in the above equation, all $c$ terms get cancelled out. Therefore, the problem is to derive the 'correct' value for $c$ with another approach.

Due to the random behavior of $S(n)$, I figured that considering the sum $$f(N)=\sum_{k=1}^{N}S(k)$$ would provide a better behavior, as the fluctuations in $S(n)$ would be negligible for big $N$. In this sense, $$f(N)=\sum_{k=1}^{N}S(k)\sim\sum_{k=1}^{N}s(k)=b\ln(N!)+cN$$ By calculating the value of $f(N)$ at every power of two from $2^0$ to $2^{32}$ and doing a least squares regression, I get that $$f(N)\approx\frac{3}{\ln(4/3)}\ln(N!)-2.4544N$$ Has an $R^2$ of $0.9999999999611$. Just to demonstrate, $f(2^{32})=938,111,615,297$, while the approximation gives $\approx938,114,543,463$, an error of only 0.0003%.

Is there a way to derive a value for $c$ other than empirically?

Edit

It seems that a heuristic or probabilistic argument is not sufficient. Consider the following probabilistic procedure on a given number $n$: return $n/2$ with probability $50$% or return $\frac{3n+1}{2}$ with probability $50$%. Repeat with the result until it gets $\leq1$ while counting the number of steps (incrementing by 1 when $n/2$ and by 2 when $(3n+1)/2$). If you use this probabilistic version of the Collatz conjecture, the value of $b$ remains unchanged, as expected, but, by doing a Python simulation, the value of $c$ changes to $11.19\pm0.01$, which is extremely different to the value of $-2.4544$ found in the original problem, suggesting that something else is going on.

ordptt
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    What is the question ? You have apparently derived a function with great accuracy. What else do you want ? – Peter Oct 20 '22 at 09:29
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    @Peter it is at the end of the question. I want to derive the value of $c$ in a way other than empirically, in the same way it was done for the value of $b$. – ordptt Oct 20 '22 at 13:25
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    Even after averaging, the right constant $c$ (if there is one) retains a very sensitive dependence on the behavior at small values of $n$. For example, if you were to redefine $S(n)$ to be the number of steps to reach $2$ instead of $1$, all of its values except $S(1)$ would decrease by $1$, so the averages would decrease by nearly $1$. So I suspect that any method to compute $c$ with increasing precision will need to examine an increasing portion of the actual Collatz tree rather than making an approximation. – Anders Kaseorg Oct 25 '22 at 01:10
  • @AndersKaseorg - here is my observation, given as a comment to my answer in the previous version of the question: "yes, I've already seen this; but I've so far no real idea for the consturction of the constant c. When I change the set of points-to-regress-on I only observed that it comes out much variable and my guess is it would tend to zero for larger and larger n - but I've no idea of its true structure. I'm not going deeper in this, but am curious as to what you'll find..." – Gottfried Helms Oct 30 '22 at 07:50

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Maybe we can use the following series, which is just the generalization of the original one (it means: half of the numbers are odd, $\frac{1}{4}$ of the numbers are divisible by 2 and not by 4, $\frac{1}{8}$ of the numbers are divisible by 4 and not by 8, etc.)

$$ s(n) = \frac{1}{2}\left(2+s(\frac{3n}{2})\right) + \sum_{k = 0}^{\infty}\frac{1}{2^{2+k}}\left(1+k+s(\frac{n}{2^{1+k}})\right) $$

in order to approximate c, we slightly modify the above infinite series to a finite one:

$$ s(n) = \frac{1}{2}\left(2+s(\frac{3n}{2})\right) + \sum_{k = 0}^{N-1}\frac{1}{2^{2+k}}\left(1+k+s(\frac{n}{2^{1+k}})\right) + \frac{1}{2^{2+N}}\left(1+N+S(1)\right) $$

then, try plugging the equation $s(n) = b\ln n+c$. It should give you the value of c depending on N.

Now I'll explain the assumption of the modification. Assume $2^N\leq n< 2^{N+1} $, then, the infinite series must stop when $ k $ goes to $ N $ (there is no number divisible by $2^{N+1}$).

wsxmble
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  • Interesting idea. I will try to see if I can derive $c$ from it. But, shouldn't there be an $N$ in the $\frac{1}{2^{2+N}}(1+S(1))$ term? that is, $\frac{1}{2^{2+N}}(1+N+S(1))$ – ordptt Oct 29 '22 at 14:49
  • I've tested it, and it really doesn't seem to give correct results, apparently the value of $c$ goes to infinity as $n\rightarrow\infty$. I believe it has to do with the fact that your finite sum is essentially the infinite one for large $n$, and note that the $c$ terms are cancelled in it, that is, we get a $c\cdot0$. In the case of the finite sum, it is $c$ times a really small number, and when isolating $c$ it will, consequently, result in a really large number. – ordptt Oct 29 '22 at 15:17
  • I edited the question to include an observation I've made that also adds to my previous comment. – ordptt Oct 29 '22 at 17:32