Answer.

(big?) Hint.

Obviosly, that $3|a_{n}\to 3|a_{n}'$, therefore, if $3|a_{1}$, then $3|a_{n}\ \forall n$ (exactly $a_{n}=a_{1}2^{n-1}(mod \ 9)).$ If $11|a_{n}$, then the sum of digites in even positions equal to sum of digits in odd positions by mod 11, therefore $11|a_{n}'$ too. It give $11|a_{n}\to 11|a_{n+1}$. If $10^{k-1}\le a_{n}<10^{k}$ had odd number of digits (k is odd ), then $a_{n+r}, r\le 5$ had k+1 digits. It is easy to chek, that $11|a_{n+r+1}$, therefore $11|a_{7}$ for any $a_{1}$.

Rust wrote: If $ 10^{k - 1}\le a_{n} < 10^{k}$ had odd number of digits (k is odd ), then $ a_{n + r}, r\le 5$ had k+1 digits. It is easy to chek, that $ 11|a_{n + r + 1}$, therefore $ 11|a_{7}$ for any $ a_{1}$. I didn't seem to understand this step Can s.o explain it please?

If $ a$ had even number of digits $ a=a_0+10*a_1+...+a_{m}10^{m}$ (m is odd), then $ a=a_0-a_1+a_2-...-a_m\mod 11$, $ a'=a_m-a_{m-1}+...+a_1-a_0\mod 11 \equiv -a\mod 11$. Therefore $ 11(a+a')$ or if $ a_n$ had odd number of digits, then $ 11|a_{n+1}$.

excuse me, how do you know $ a_6$ has an odd number of digits, I think it's not necessary.

Can anyone help me to prove $ 11 / a_7$ I don't see how.

Answer: No. We will prove first that one of $a_1,a_2,\ldots,a_6$ has an even number of digits. Suppose the contrary. Let $a_1$ has $k$ digits, where $k$ is odd. If $a_n$ has $k$ digits, then $a_{n+1}$ has at most $k+1$ digits, which has different parity with $k$. The numbers of digits in $a_1,a_2,\ldots,a_6$ are even, hence each of $a_1,a_2,\ldots,a_6$ has exactly $k$ digits. Let $p$ and $q$ be the first and the last digits of $a_1$. So $a_2$ starts and ends with $p+q$. Similarly, we get that $a_3,a_4,a_5,a_6$ start and end with $2p+2q,4p+4q,8p+8q,16p+16q$ respectively. But $16p+16q>10$, which implies that $a_6$ has more than $k$ digits, a contradiction. Therefore our claim holds, let $a_i$ ($i\le6$) be the integer with an even number of digits. The digits in odd positions of $a_i$ become digits in even positions of $a_i'$, and vice versa, therefore the sums of digits in even positions and odd positions of $a_i+a_i'$ are equal. Thus $11|a_{i+1}$. Note that if $11|a_k$ then $11|a_k'$ and hence $11|a_{k+1}$, which further implies $11|a_n$ for all $n\ge k$. Since $11|a_{i+1}$ where $i\le6$, then $11|a_7$. If $a_7$ is prime, then $a_7=11$. If $a_6$ has only one digit, then $a_6=a_6'$ and $11=a_7=2a_6$, which is impossible. Hence $a_6$ has two digits, $a_6=10$, then $a_5=5$, which leaves no possible value for $a_4$, a contradiction.