Something interesting that I found about some numbers - and would like to see if it's known Well I am quite sure it's known (I mean number theory exists thousands of years), warning beforehand, it may look like numerology, but I try not to go to mysticism.
So I was in a bus, and from boredom I started just adding numbers in the next way:
$$1+1=2$$
$$2+2=4$$
$$4+4=8$$
$$....$$
etc. up to $32,768$ (it was quite boring, I can tell... :-)), I didn't have calculator.
And notice that if I keep adding the digits until I get a number from 1 to 10, I get that for example for $8+8=16$, $1+6=7$, now seven steps after this, at $512+512=1024$, which $1+2+4=7$, and again after $7$ steps $32768+32768=65536$, and adding $6+5+5+3+6=10+12+3=25$, $2+5=7$. So this led me to the conjecture that this repetition may occur endlessly.
Now, of course I can program some code that will check for large numbers, but I am tired, long day. So, if this indeed the case (which could be disproved, but even then I would wonder when this repetition stops) then why?
As I said, I am tired, it may make no sense, and I might have done mistakes in my calculations, and it may be trivial.
Either way, if you have some answer, I would like to hear it.
 A: "Adding the digits until [you] get a number from $1$ to $10$" is the same as finding the remainder when dividing by $9$ (except that you would get $9$ instead of $0$ by adding digits, and there is no reason to stop with $10$: you can add the digits to get $1$).
It turns out, by something called Euler's Theorem, that if $a$ is not divisible by $3$, then $a^6$ always leaves a remainder of $1$ when divided by $9$. In particular, $2^6$ has remainder $1$ when divided by $7$.
Another property of remainders is that if $a$ leaves a remainder of $r$ when divided by $9$, and $b$ leaves a remainder of $s$, then $ab$ leaves the same remainder as $rs$. 
So, that means that if $2^n$ leaves a remainder of $r$, then $2^n\times 2^6 = 2^{n+6}$ will leave the same remainder as $r\times 1 = r$; that is, the same remainder as $2^n$. So adding the digits of $2^n$ until you get a single number between $1$ and $9$ will give you the same answer as doing it for $2^{n+6}$. This is what you observed: $8=2^3$, and $512=2^9 = 2^{3+6}$. You will get the same answer ($7$) with $2^{15}$, $2^{21}$, $2^{27}$, etc.
I note that you were slightly off in describing $512$ as being "seven steps after" $8$: it's really only six steps later: 
$$8\stackrel{1}{\mapsto} 16 \stackrel{2}{\mapsto}32 \stackrel{3}{\mapsto}64\stackrel{4}{\mapsto}128\stackrel{5}{\mapsto}256\stackrel{6}{\mapsto}512.$$
A: There is indeed a period of $6$. The sums of digits cycles through the numbers $1,2,4,8,7,5$.
You are doubling the number each time, but then the sum of the digits of that number will double as well, since it is a linear function of the number. This then shows that if you start with $1$ you will go through the cycle
$$1 \to 2 \to 4 \to 8 \to 16=7 \to 14=5 \to 10=1 \to \ldots$$
A: A repetition of this sort was bound to happen, and it always happens even under more general circumstances.
First, as others have pointed out, the sum of the base-10 digits of a number $N$ is congruent to $N$ modulo $9$. The reason for this is that $10\equiv 1 \bmod 9$, and so
$$\begin{align*} N &= a_t \cdot 10^t + a_{t-1} \cdot 10^{t-1} +\cdots +a_2 \cdot 10^2 + a_1 \cdot 10 +a_0 \\
& \equiv a_t\cdot  1^t + a_{t-1}\cdot 1^{t-1} +\cdots +a_2\cdot  1^2 + a_1 \cdot 1 +a_0 \bmod 9\\
& \equiv a_t + a_{t-1} +\cdots +a_2 + a_1 +a_0 \bmod 9.
\end{align*}$$ 
Thus, if you add all the digits of $N$ and obtain $N_1$, then $N\equiv N_1\bmod 9$. If now we add all the digits of $N_1$ and obtain $N_2$, then $N\equiv N_1\equiv N_2 \bmod 9$. In this way we may create a sequence $N>N_1>N_2> \cdots$, and since all $N_i$ are natural numbers, we end up at some $1\leq N_t \leq 9$, such that $N_t\equiv N\bmod 9$, so $N_t$ is simply the remainder of division of $N$ by $9$.
Let $a>1$ be any natural number relatively prime to $3$, whose sum of base-10 digits is $b$, and let $s$ be the order of $b\bmod 9$, i.e., $s$ is the smallest positive number such that $b^s\equiv 1 \bmod 9$. Then:


*

*The sum of the base-10 digits of $a$ is $b\bmod 9$.

*The sum of the base-10 digits of $a^{1+sk}$ is also $b\bmod 9$, for all $k\geq 0$, because
$$a^{1+sk}\equiv b^{1+sk}\equiv b\cdot b^{sk}\equiv b\cdot (b^s)^k\equiv b \cdot 1\equiv b \bmod 9.$$

*For a fixed $t\geq 1$, the sum of the base-10 digits of $a^{t+sk}$ is $b^t\bmod 9$, for all $k\geq 0$, for similar reasons as above.

*For a fixed $t\geq 1$, the sum of the base-10 digits of 
$$a^{t+sk} + a^{t+sk}$$
is $2\cdot b^t \bmod 9$, for all $k\geq 0$. 

*And more generally, for fixed $r\geq 1$ (relatively prime to $3$) and $t\geq 1$, the sum of the base-10 digits of 
$$a^{t+sk} + \cdots + a^{t+sk} = r\cdot a^{t+sk},$$
where we have added $r$ copies of $a^{t+sk}$, is $r\cdot b^t \bmod 9$, for all $k\geq 0$. 


Your example is the case where $a=2$, $b=2$, $s=6$, $t=3$ and $r=2$. According to the formula above, the sum of the digits must be
$$r\cdot b^t \equiv 2\cdot 2^3\equiv 7 \bmod 9.$$
But any other choice works just as well. Pick $a=11$, $b\equiv a\equiv 2 \bmod 9$, $s=6$, $t=2$ and $r=5$. Then, the sum of the digits of the numbers 
$$11^{2+6k}+11^{2+6k}+11^{2+6k}+11^{2+6k}+11^{2+6k},$$
for all $k\geq 0$, is congruent to 
$$r\cdot b^t \equiv 5\cdot 2^2\equiv 2 \bmod 9.$$
For instance:


*

*$11^{2}+11^{2}+11^{2}+11^{2}+11^{2}=605,$ and $6+5=11$ and $1+1=2$.

*$11^{8}+11^{8}+11^{8}+11^{8}+11^{8}=1071794405,$ and 
$$1+0+7+1+7+9+4+4+5=38$$ and $3+8=11$, and $1+1=2$.

*Etc.
A: This is indeed the case.  
The value you get out at the end of all your summations is just the value of your number mod 9 (this is because any natural number is congruent to the sum of its digits mod 9).  For example, $1024=9\cdot113+7$ is congruent to 7 mod 9.  Also note that the numbers you're getting out by your procedure of successive doubling are just the powers of 2.  Finally, I think you're off by one in your counting of sevens -- the powers of two that are working for you are $2^4=16$, $2^{10}=1024$, $2^{16}=65536$, etc.
So your fact is true because it's true for $2^4$, and each subsequent term differs by a factor of $2^6$ which is congruent to 1 mod 9 (either by direct computation or by Euler's Theorem -- $\phi(9)=6$).
So in general, the sum of the sum of the .... sum of the digits of your values $2^{4+6k}$ is given by 
$$
2^{4+6k}\equiv 2^4\cdot (2^6)^k\equiv 7\cdot 1^k\equiv \boxed{7}\pmod{9}.
$$
