Multiples of 11 in a Fibonacci-like sequence formed by concatenation instead of addition Let $A_1=0$ and $A_2=1$ and suppose that the number $A_n$ is obtained from the decimal expansions of $A_{n-1}$ and $A_{n-2}$. For example $A_3=A_2A_1=10$; $A_4=A_3A_2=101$; $A_5=A_4A_3=10110$.
Determine the integers $n$ that satisfy the condition $A_n \equiv 0 \pmod{11}$.

Progress
If I want a number congruent with 0 then I want multiples of 11. The problem is how to find the $n$'s that satisfy the terms. I could use the fact that a number is divisible by 11 when the sum of even order of digits $S_p$ less the sum of  odd order digits $S_i$ are a number divisible by 11. But how?
 A: The key is to recognize that each $A_n$ has $F_n$ digits, so that
$$A_n=10^{F_{n-2}}A_{n-1}+A_{n-2}$$
for $n\ge 3$. Thus,
$$A_n\equiv 10^{F_{n-2}}A_{n-1}+A_{n-2}\equiv(-1)^{F_{n-2}}A_{n-1}+A_{n-2}\pmod{11}\;.$$
It’s easy to show by induction that $F_n$ is even if and only if $n$ is a multiple of $3$, so
$$A_n\equiv\begin{cases}
A_{n-2}+A_{n-1}\pmod{11},&\text{if }n\equiv2\pmod3\\
A_{n-2}-A_{n-1}\pmod{11},&\text{otherwise}\;.
\end{cases}$$
Let $B_1=A_1=0$ $B_2=A_2=1$, and 
$$B_n=\begin{cases}
B_{n-2}+B_{n-1},&\text{if }n\equiv2\pmod3\\
B_{n-2}-B_{n-1},&\text{otherwise}\;;
\end{cases}$$
clearly $B_n\equiv A_n\pmod{11}$ for $n\in\Bbb Z^+$. Now calculate a few values of $B_n$
$$\begin{array}{rcc}
n:&1&2&3&4&5&6&7&8&9&10&11&12&13&14\\
B_n:&0&1&-1&2&1&1&0&1&-1&2&1&1&0&1
\end{array}$$
At this point it’s pretty obvious that $\langle B_n:n\in\Bbb Z^+\rangle$ is periodic with period $6$, and it’s not at all difficult to prove this by induction. It follows that $A_n$ is divisible by $11$ precisely when $n\equiv 1\pmod 6$.
By the way, it’s also not hard to show by induction on $n$ that $B_n$ is the sum of the digits of $A_n$ corresponding to even powers of $10$ minus the sum of the digits of $A_n$ corresponding to odd powers of $10$.
