---
description: A possible proof-by-contradiction opportunity
---

# Parent & Child

Recall that there are two cases that disprove the conjecture:

* A number is **looping**: there is a number $$n = F^j(n)$$ such that $$n\ne1$$ and $$j > 0$$.
* A number is **diverging**: there is a number $$n < F^j(n)$$ for $$j > 0$$.

Now, we can argue as follows:

* The root number in a PIPTree is always a power of two, which is trivially known to **converge** (i.e. reach 1).
* If we can prove that **whatever behavior a parent node has, the children will have the same**; then, we come to conclusion that all numbers must have the same behavior with powers of two, which are known to converge.

## Right Child

We know that for a parent with number $$n$$, the right child has $$n/2$$. It is trivial to see that these will have the same behavior, as the right child is just the result of a single iteration of Collatz function $$F(n)$$.

## Left Child

This is a tricky case. For a parent with number $$n$$, the left child has $$n/2+r$$ where $$r$$ is the root number. We know from the previous section the terminating numbers:

* **Good parent**

$$
g_L' = 3^{m+1} + \frac{3n_m + 1}{2^{p_r + 1 - p_m}}
$$

$$
g_R' = n_m
$$

* **Bad parent**

$$
b_L' = n_m + 3^m2^{p_r+1-p_m}
$$

$$
b_R' = \frac{3n_m+1}{2^{p_r+1-p_m}}
$$

{% hint style="warning" %}
Connecting the left child to either the good parent or the right child remains **open research**!
{% endhint %}
