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.

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    $\begingroup$ (off topic) I think $i=\sqrt {-1}$ is wrong notation. We can write only $i^2=-1$ $\endgroup$ Aug 13, 2018 at 20:42
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    $\begingroup$ 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, 2018 at 21:53
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    $\begingroup$ 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. $\endgroup$ Aug 14, 2018 at 3:05
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    $\begingroup$ 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, 2018 at 7:59
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    $\begingroup$ @cortek You're right because Brooks follows $-11$ with $-33 + i$, not $-33 - i$. $\endgroup$ Aug 15, 2018 at 4:09

3 Answers 3


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.


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 $3z + i$.

  • $\begingroup$ I don't understand the last two sentences. Did you prove that some Collatz sequences for $3z+i$ do not terminate? $\endgroup$ Aug 20, 2018 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... $\endgroup$ Aug 20, 2018 at 20:54
  • $\begingroup$ For what it's worth, I'm convinced, Mr. Soupe. $\endgroup$
    – Mr. Brooks
    Aug 22, 2018 at 20:19

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.


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