Parent & Child

A possible proof-by-contradiction opportunity

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

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

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

Connecting the left child to either the good parent or the right child remains open research!