For the first 20 positive integers, the recursive digit sum (mod 9) of their squares follow a pattern of repeating 1, 4, 9, 7, 7, 9, 4, 1, 9. I wonder whether this pattern applies to all integer numbers. If it does, what will be the reason behind this phenomenon? Any suggestions?
Some examples below:
1, 1, 1
2, 4, 4
3, 9, 9
4, 16, 7
5, 25, 7
...
9, 81, 9
10, 100, 1
11, 121, 4
...
19, 361, 1
To summarize the discussion in the comments:
Let's denote the iterated digit sum by $S(n)$. It is well known, and easy to demonstrate, that $n\equiv S(n)\pmod 9$. Indeed, $S(n)$ is essentially equal to the remainder you get on dividing $n$ by $9$, with the proviso that if the remainder is $0$, $S(n)=9$.
Now, consider the sequence $\mathscr S=\{S(n^2)\}$. Since $(n+9)^2\equiv n^2\pmod 9$ we see that $\mathscr S$ is periodic with period $9$. A priori there might be a period smaller than $9$ but in this case there is not.
The same argument would apply to any power, not just squares. For instance the sequence of the digit sum of cubes is $\{1,8,9,1,8,9,\cdots\}$ which has period $3$ (note that $3$ divides $9$ so $9$ is still a period, just not the least one). Or you could take fifth powers to get the periodic sequence $$\{\overline {1,5,9,7,2,9,4,8,9}\}$$
and so on.