293 Find
$$
c_{n}=\frac{1}{2}\left[(2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}\right](n \in \mathbf{N})
$$

the highest power of the prime factor 7 in it.
Solve for the modulo $7,\left\{c_{n}\right\}$ as a modular periodic sequence, with a period of $8, c_{0} \sim c_{7}$ modulo 7 remainders are
$$
1,2,0,5,6,5,0,2 \text {. }
$$

Thus, $7 \mid c_{n} \Leftrightarrow n \equiv 2(\bmod 4)$.
Let $b_{n}=\frac{1}{2 \sqrt{3}}\left[(2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}\right]$
be the dual sequence of $c_{n}$. Then
$$
c_{n}+\sqrt{3} b_{n}=(2+\sqrt{3})^{n} .
$$

Hence $\left(c_{n}+\sqrt{3} b_{n}\right)\left(c_{m}+\sqrt{3} b_{m}\right)$
$$
=(2+\sqrt{3})^{m \cdot n}=c_{m \cdot n}+\sqrt{3} b_{m+n} \text {, }
$$

gives $\left\{\begin{array}{l}c_{m+n}=c_{m} c_{n}+3 b_{m} b_{n} \\ b_{m+n}=b_{n} c_{m}+b_{m} c_{n} .\end{array}\right.$
By reasoning, for $\left\{b_{n}\right\}$ modulo 7, it is also a modular periodic sequence, with a period of 8.
Thus, $7\left|b_{n} \Leftrightarrow 4\right| n$.
From equation (3), we get $b_{2 n}=2 b_{n} c_{n}$. Therefore,
$$
b_{2 \cdots \cdot 2_{4}}=2^{a+2} c_{2 a+1} c_{2 a_{1}} \cdots c_{d_{1}} c_{2 t} c_{1} b_{1}
$$

where $\uparrow, t$ are odd numbers, $\alpha \in \mathbf{N}$.
Also, $4 t, 8 t, \cdots, 2^{u+1} t \neq 2(\bmod 4)$, so,
$7 \times c_{2 n+1}, c_{2 n}, \cdots, c_{41}, 7 \times c_{1}, 7 \times b_{r}$.
Consider $c_{4 k+2}$ modulo 49.
From equation (1), we calculate
$$
\begin{array}{l}
c_{2} \equiv 7(\bmod 49), c_{6} \equiv 28(\bmod 49), \\
c_{10} \equiv 35(\bmod 49), c_{14} \equiv 0(\bmod 49), \\
c_{18} \equiv 14(\bmod 49), c_{22}=21(\bmod 49), \\
c_{26} \equiv 42(\bmod 49) .
\end{array}
$$

Also, $c_{28}=1(\bmod 49), b_{28}=0(\bmod 49)$,
$$
c_{n+28}=c_{n} c_{28}+3 b_{n} b_{2 \mathrm{x}} \equiv c_{n}(\bmod 49),
$$

thus, 49 divides $c_{n} \Leftrightarrow n \equiv 14(\bmod 28)$.
For $n=7 n_{1}, n_{1} \equiv 2(\bmod 4)$, we have
$$
c_{n}+\sqrt{3} b_{n}=\left(c_{n_{1}}+\sqrt{3} b_{n_{1}}\right)^{7} \text {. }
$$

Then $c_{n}=\left(c_{n_{1}}\right)^{7}+21 \times 3\left(c_{n_{1}}\right)^{5}\left(b_{n_{1}}\right)^{2}+$
$$
35 \times 9\left(c_{n_{1}}\right)^{3}\left(b_{n_{1}}\right)^{4}+7 \times 27 c_{n_{1}}\left(b_{n_{1}}\right)^{6} \text {. }
$$

Let $7^{s_{n}} \| c_{n}(n \in \mathrm{N})$. Then in equation (4), the exponents of 7 in the first term
are all less than $3 S_{n_{1}}+1>S_{n_{1}}+1$, while $7 \times 27 c_{n_{1}}\left(b_{n_{1}}\right)^{6}$ has an exponent of 7 as $S_{n_{1}}+1$, i.e., $S_{7_{n 1}}=S_{n_{1}}+1$.
Thus, $S_{7^{n_{n}}}=S_{n_{1}}+\alpha\left(n_{1} \equiv 2(\bmod 4)\right)$.
For $7 \times n_{1}, S_{n_{1}}=1$. Therefore, $S_{7_{n_{1}}}=\alpha+1$, i.e., the exponent of 7 in $c_{n}$ is
$$
\left\{\begin{array}{ll}
0, & n \neq 2(\bmod 4) ; \\
\alpha+1, & n \equiv 2(\bmod 4), 7^{a} \| n .
\end{array}\right.
$$