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I was playing around with collatz sequence stuff recently, and I made a plot of seed values vs the maximum value the seed's collatz sequence will reach. Highlighting primes in a logarithmic graph gives an interesting result

enter image description here

In this image, we evaluate the collatz sequences of all numbers ("seed values") from $1$ to $10^5$, and find the maximum number within each sequence. We plot each seed value on the $x$ axis with its corresponding maximum number in its sequence on the $y$ axis.

Then, highlight all points corresponding to a prime seed value red, and a composite blue.

As seen in the graph, the prime numbers seem to have a consistently larger "lower bound" for their maximum values than composites.

What causes this behavior?

My initial thought was that for composites highly divisible by 2 and have numbers in its sequence that are highly divisible by 2, the maximum value is itself, hence making up the very bottom of the graph. Meanwhile, a prime is always odd so the smallest possible number in its sequence is itself scaled up by $\frac32$.

Is this the correct explanation, or is there something else I am missing?

thanks!

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  • $\begingroup$ "a prime is always odd" - except $2$ of course , but I understand the point. I see no clear reason why prime numbers should reach higher maximums as odd composite numbers. You forgot to explain the meaning of the colours in the graph. $\endgroup$
    – Peter
    Jul 14 at 7:22
  • $\begingroup$ @Peter ah let me do that real quick $\endgroup$
    – Max0815
    Jul 14 at 7:36
  • $\begingroup$ I think your explanation fully proves the diagonal completely blue band at the bottom. It probably also explains why the area directly above it is predominantly red. It is a bit surprising that the red part gets so wide towards the right, but that may be an artifact. The dots overlap each other so it may look different if the dots are displayed in a different order. $\endgroup$
    – Jaap Scherphuis
    Jul 14 at 9:41
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    $\begingroup$ You should create a graph only with odd numbers it is obvious that for each $n$ all numbers $2^a\cdot n$ are plotted with a smaller lower bound $\endgroup$
    – user140242
    Jul 14 at 11:14
  • $\begingroup$ Note that almost every prime will share the record for the maximum with numerous odd composite numbers. $\endgroup$
    – Peter
    Jul 14 at 11:56

1 Answer 1

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Thanks to @user140242 and @Peter for the insight.

Plotting a the same aforementioned graph but with odd numbers only shows that my explanation was pretty much correct.

The lowest possible maximum for each odd number's collatz sequence is at the very least $\frac32$ times itself. This makes up the bottom boundary for the primes as well.

Meanwhile for even composites, the lowest possible maximum could be itself, or arise later in the sequence such that it's greater than itself but still less than $\frac32$ times itself &c, making up the various other lines seen below the bottom prime boundary (shown below).

enter image description here

Below are some graphs, and you can see that the bottom "max number" for each odd number seed's collatz sequence is the same for composites and primes.

Primes highlighted and plotted on top

prime on top

Primes just highlighted

prime

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    $\begingroup$ If you also highlight the numbers $n=\frac{2^{2\cdot x}-1}{3}+k\cdot 2^{2\cdot x+1}$ and $n=\frac{5\cdot 2^{2\cdot x+1}-1}{3}+k\cdot 2^{2\cdot x+2}$ for $x>0$ and $k\geq 0$ these correspond to the lower limit $\endgroup$
    – user140242
    Jul 15 at 15:57
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    $\begingroup$ Referring to the previous comment, the lower limit are the numbers $n \equiv 1 \pmod 4$ which have the maximum in the sequence equal to $3\cdot n+1$ $\endgroup$
    – user140242
    Jul 16 at 16:25

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