11
$\begingroup$

Given a Gaussian integer $z = a + bi$, where $a, b \in \mathbb{Z}$, $i = \sqrt{-1}$, iterate the function $$f(z) = \frac{z}{1 + i}$$ if $z$ has even Gaussian norm (that is, both $a$ and $b$ are odd, or they're both even), otherwise $f(z) = 3z + i$.

I conjecture that iterating this function eventually leads, if not to $1$, to one of the other Gaussian units ($-1, i, -i$).

For example, starting with $z = 14$, we get $$7 - 7i, -7i, -22i, -11 - 11i, -11, -33 + i, -16 + 17i, \ldots$$

(this is wrong, see edit below)

I have tried a few different values of $z$ with small norm, some purely real, with pencil and paper, haven't gotten far. Also I have tried it in Mathematica, but either I've made some mistakes in my programming that crash the program, or there really are a lot of values that escape to some infinity.

Surely someone else has studied this variant? If so, have they been able to determine anything (like finding a periodic orbit that doesn't include any units)?

EDIT: I made a mistake. Mathematica did save my notebook file at some point before it crashed, and I could have gotten the correct sequence from there instead of having to recalculate it anew. It should go like this: $$7 - 7i, -7i, -20i, 10 + 10i, 10, 5 - 5i, -5i, -14i, -7 - 7i, -7, -20, \ldots$$

Thanks to Mr. Cortek for pointing this out.

$\endgroup$
  • 4
    $\begingroup$ (off topic) I think $i=\sqrt {-1}$ is wrong notation. We can write only $i^2=-1$ $\endgroup$ – Elvin Aug 13 '18 at 20:42
  • 4
    $\begingroup$ Few quick tests in Maple seem to suggest that it does not behave nicely, all examples I tried just blowed up to large numbers quite quickly. Some numbers hit unit before that, but some dont, for example $3$ does not seem to... $\endgroup$ – Sil Aug 13 '18 at 21:53
  • 2
    $\begingroup$ Hmm... 14, some number, $-7i$, $-22i$, some other number, $-11$, ... that sounds like a familiar $3n + 1$ sequence with some of the numbers multiplied by Gaussian units and some other numbers inserted in between. $\endgroup$ – Robert Soupe Aug 14 '18 at 3:05
  • 5
    $\begingroup$ For $z=2+3i$, we get a loop without any units. Also for $z=14$, I think $3(-7i)+i=-20i$, not $-22i$. In that case, we do go to $i,-i,-1$, but then it goes wild, doesn't seem to return to any units or even any loop. It looks like the value of $\max a,b$ goes to infinity. And not only $z=14$ but also for lots of other $z$ as well. $\endgroup$ – cortek Aug 14 '18 at 7:59
  • 1
    $\begingroup$ @cortek You're right because Brooks follows $-11$ with $-33 + i$, not $-33 - i$. $\endgroup$ – Robert Soupe Aug 15 '18 at 4:09
5
$\begingroup$

The hope here is apparently to find that all Gaussian integers eventually lead to a number of the form $(1 + i)^n u$, where $n$ is a positive real integer and $u$ is a Gaussian unit.

Looking at a list of powers of $1 + i$, it seems to me like they take one of these forms: $\pm 2^n$, $\pm 2^n i$, $\pm 2^n \pm 2^n i$.

Then it's definitely possible for, say, $-3i$ to be followed by $-8i$ and form there it's a spiral drop to 1.

But when the real and imaginary parts are both nonzero, we run into the problem that $3z + i$ affects the real and imaginary parts unevenly.

For example, $5 + 5i$ multiplied by 3 is $15 + 15i$, but then you're only adding $i$ and so $15 + 16i$ falls short of $16 + 16i$ by 1.

Since $\mathbb Z[i]$ is a unique factorization domain and 3 is prime, a number with real and imaginary parts both nonzero is divisible by 3 only if each part is itself also divisible by 3.

Hence $3z$ fails to lead us to a number of the form $2^n + (2^n - 1)i$. A similar problem occurs with $3z + 1$ instead of $3z + i$.

$\endgroup$
  • $\begingroup$ I don't understand the last two sentences. Did you prove that some Collatz sequences for $3z+i$ do not terminate? $\endgroup$ – punctured dusk Aug 20 '18 at 20:06
  • $\begingroup$ @barto Maybe I did prove it, but something's holding me back from making that assertion. Looks like either way I'll have to edit the last paragraph... $\endgroup$ – Robert Soupe Aug 20 '18 at 20:54
  • $\begingroup$ For what it's worth, I'm convinced, Mr. Soupe. $\endgroup$ – Mr. Brooks Aug 22 '18 at 20:19
10
$\begingroup$

The Collatz conjecture works heuristically because the map is contracting on average (to prove this it is easier to use $x \mapsto (3x+1)/2$ when $x$ is odd)

For your procedure this is no longer the case because dividing by $1+i$ divides the modulus by $\sqrt 2$.

If $z$ is "odd" then you multiply it by approximately $3/ \sqrt 2$ in two steps, and if it is "even" you divide it by $\sqrt 2$ in one step, so on average, you multiply a number by $3/2$ in $3$ steps, which means a multiplication by $(3/2)^{1/3}$ per step. This is greater than $1$ so heuristically, most of the time the sequence should diverge to infinity.

For example if you start at $1+2i$, after 200 steps you are at $-673097903812-335968886130i$, whose norm is approximately $7.523 \times 10^{11}$ and it doesn't look like it wants to get back to small numbers.

For comparison with the heuristic, $\sqrt 5 \times (3/2)^{200/3} \approx 1.227 \times 10^ {12}$, which is decently close.

$\endgroup$
4
+250
$\begingroup$

Not a full answer, just some remarks and considerations.

In the norm language we can see that division of $z$ with $1+i$ produces some complex number $z^*$ that has the norm smaller by a factor of $\sqrt{2}$, and $3z+1$ approximately triples the norm.

Now since ${(\sqrt{2}})^3 \approx 2.8<3$ we see that for every $z \in \mathbb Z$ his associated Collatzian sequence must be generated in such a way that, on average, on every appliance of a function $f(z)=3z+1$ there are more than three appliances of a function $f(z)=\frac {z}{1+i}$ if we want that all subsequences of a Collatzian sequence associated to $z$ do not escape to infinity.

But this seems very unlikely to ever happen because there seems to be no reason that division $\frac {z}{1+i}$ will favour numbers of the form $O+Oi$ and $E+Ei$ quite more often than numbers $E+Oi$ and $O+Ei$, where $E$ stands for even and $O$ for odd integers.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.