$\forall g, h, i, n \in \Bbb N, \ g, i > 1, g \neq 10^n, 3h, 3h - 1, (3h)^i \pm 1$. Square $g$, sum its digits, square result, etc...you'll reach $13$ $$\LARGE \text{The $13$ Conjecture}$$

Let $i$ be a positive integer and $g$ a positive integer which is not of the forms $3h, \ 3h-1, \ (3h)^m \pm 1$ for some natural $h, \ m$ greater than $1$, and $g \neq 10^n$ for some natural $n$. $δ(n)$ is the sum of digits of $n$ in base $10$. Consider the two sequences created by applying sum of digits and rising to the $2i$-th power alternatively to $g$.

Conjectures:

*

*If $i=1$, the two sequences reach the periodic point with $13$.


*If $i > 1$ then the two sequences reach a fixed point of the function $δ(n^{2i})$

An example to illustrate:
Let $g = 13, \ i = 1$, the first sequence is: $$13^2 = 169 \to 1 + 6 + 9 = 16 \to 16^2 = 256 \to 2 + 5 + 6 = \boxed{13}$$
The second sequence is: $$1 + 3 = 4 \to 4^2 = 16 \to 1 + 6 = 7 \to 7^2 = 49 \to 4 + 9 = \boxed{13}$$
Now let $g = 13, \ i = 2$, the first sequence is: $$13^4 = 28561 \to 2 + 8 + 5 + 6 + 1 = 22 \to 22^4 = 234256 \to 2 + 3 + 4 + 5 + 6 = 20$$ $$\to 20^4 = 16000 \to 1 + 6 + 0 + 0 + 0 = 7 \to 7^4 = 2401 \to 2 + 4 + 0 + 1 \to 7$$
Indeed, $7$ is a fixed point.
the second sequence is: $$1 + 3 = 4 \to 4^4 = 256 \to 2 + 5 + 6 = 13 \to \ldots \to 7$$
 A: As with many (but not all!) dynamical systems on the integers, this one can be completely analyzed with a bit of computer brute force.
Let $f(x)=\delta(x^2)$. Then for sufficiently large $x$, we have $f(x)\leq x$. Proof: since squaring a number at most doubles the number of digits, then denoting $n(x)$ the number of digits of $x$, we have $f(x)\leq 9\cdot2n(x)$. But we also have $x\geq 10^{n(x)-1}$. We end up with $f(x)\leq x$ as long as $x$ has at least $3$ digits.
Let $A$ denote the set of integers with at most $2$ digits, so $A=\{1, ..., 99\}$. The previous paragraph shows that any orbit of $f$ passes infinitely often through $A$, therefore it must visit some element of $A$ twice, and thus it will be eventually periodic. To understand orbits of $f$, we therefore only need to understand the orbits of elements of $A$, which can be calculated on a computer. We find that $f$ has two fixed points, $1$ and $9$, and one cycle, $\{13, 16\}$.
Since $\delta(n)\equiv n\mod 9$, we have $f(n)\equiv n^2\mod 9$. By looking at what happens modulo $9$ under repeated squaring, we obtain, for any $n$, not just $n\in A$:

  
*
  
*If $n\equiv 0, 3, 6\mod 9$, then the orbit of $n$ eventually hits the fixed point $9$.
  
*If $n\equiv 1, 8\mod 9$, then the orbit of $n$ eventually hits the fixed point $1$.
  
*If $n\equiv 2, 4, 7\mod 9$, then the orbit of $n$ eventually enters the cycle $\{13, 16\}$.
  

I imagine a similar approach would solve the case $f(x)=\delta(x^{2i})$ as well.
A: $$\LARGE \text{The $13$ Conjecture}$$

For every natural number $g > 1$, if you square it, then take the sum of its digits, square the result, take the sum of digits of that, and so on, you will ultimately reach $13$.
Since the digit sum of $g$ is divisible by $3$ if and only if $3$ divides $g$, my conjecture will thus lead to a different kind of loop, never reaching $13$ since $13 \neq 3h$ for all positive $h$.
Thus $3$ cannot divide $g$.
Since $1^2 = 1$, my conjecture will thus lead to another different loop if the digit sum of $g$ is equal to $1$. We established that $g > 1$, but $g \neq 10, 100, 1000, \ldots$
Thus $g \neq 10^n$, and in fact for some positive $i$, $g \neq (3h)^i \pm 1$. Of course if $h = 1$ and $i = 1$ then this is unsuitable, so $h, i > 1$. But $g \neq 3$ so we have no restrictions on $h$ apart from the fact that it is positive (with other added restrictions on $g$).

We can now re-write the conjecture by adding the necessary restrictions on $g$.
For all natural numbers $h, n$ and all $g, i > 1$:
$$
g \neq
\begin{cases}
 10^n, &\ 6(2n + 1) \pm 1 &\text{ for $n$}\\
 3h, &\ (3h)^i \pm 1&\text{ for $h$}
\end{cases}
$$
Allowing these restrictions on $g$, square it, then take the sum of its digits, then square the result, take the sum of digits of that, and so on. Ultimately, you will reach $13$.

SECOND EDIT
