3. Let $a_{1}=11^{11}, a_{2}=12^{12}, a_{3}=13^{13}$, and
$$
a_{n}=\left|a_{n-1}-a_{n-2}\right|+\left|a_{n-2}-a_{n-3}\right|, n \geqslant 4 \text {. }
$$

Find $a_{4^{4}}$.
For $n \geqslant 2$, define $s_{n}=\left|a_{n}-a_{n-1}\right|$. Then for $n \geqslant 5, a_{n}=s_{n-1}+s_{n-2}, a_{n-1}=s_{n-2}+s_{n-3}$. Thus, $s_{n}=$ $\left|s_{n-1}-s_{n-3}\right|$. Since $s_{n} \geqslant 0$, if $\max \left\{s_{n}, s_{n+1}, s_{n+2}\right\} \leqslant T$, then for all $m \geqslant n$, we have $s_{m} \leqslant T$. In particular, the sequence $\left\{s_{n}\right\}$ is bounded. We will prove the following proposition:

If for some $i, \max \left|s_{i}, s_{i+1}, s_{i+2}\right|=T \geqslant 2$, then $\max \left\{s_{1+6}, s_{1+7}, s_{1+8}\right\} \leqslant T-1$.

Proof by contradiction. If the above conclusion does not hold, then for $j=i, i+1, \cdots, i+6$, we have $\max \left\{s_{j}, s_{j+1}, s_{j+2}\right\}=T \geqslant 2$. For $s_{t}, s_{t+1}$, or $s_{t+2}=T$, take $j=i, i+1$, or $i+2$ accordingly, so the sequence $s_{j}, s_{j+1}, s_{j+2}, \cdots$ has the form $T, x, y, T-y, \cdots$, where $0 \leqslant x, y \leqslant T, \max \{x, y, T-y\}=T$. Therefore, either $x=T$, or $y=T$, or $y=0$.
(1) If $x=T$, then the sequence has the form $T, T, y, T-y, y, \cdots$. Thus $\max \mid y, T-y, y\}=T$. So $y=T$ or $y=0$.
(2) If $y=T$, then the sequence has the form $T, x, T, 0, x, T-x, \cdots$. Since $\max \mid 0, x, T-x=T$, so $x=0$ or $x=T$.
(3) If $y=0$, then the sequence has the form $T, x, 0, T, T-x, T-x, x, \cdots$. Since $\max \{T-x, T-x, x\}=T$, so $x=0$ or $x=T$.

In each of the above cases, $x$ and $y$ are either 0 or $T$. In particular, $T$ must divide every term in $s_{j}, s_{j+1}$, and $s_{j+2}$. Since $\max \left\{s_{2}, s_{3}, s_{4}\right\}=s_{3}=T$, for $n \geqslant 4$, $T$ divides $s_{n}$. However, when $s_{4}=11^{11}6(N-1)+4$, $a_{n}=s_{n-1}+s_{n-2}=0,1$ or 2. In particular, $a_{M}=0,1$ or 2. Since $M$ is a multiple of 7, $a_{M} \equiv a_{7} \equiv 1(\bmod 2)$, hence $a_{M}=1$.