$6 \cdot 22$ For a given positive integer $k$, define $f_{1}(k)$ as the square of the sum of the digits of $k$, and let
$$
f_{n+1}(k)=f_{1}\left(f_{n}(k)\right),
$$

Find: $f_{1991}\left(2^{1990}\right)$.

[Solution] First, note that for large $k$ values, $f_{1}(k)$ is much smaller than $k$. Since $f_{1}$ is not monotonic, we express this fact in the following form:
If $A \leqslant B$, then the number of digits of $A$ $\leqslant$ the number of digits of $B$ $\leqslant 1+\lg B$,
$$
f_{1}(A)<(9 \times(1+\lg B))^{2}<\left(4 \log _{2} 16 B\right)^{2} .
$$

Using this inequality twice, we get an estimate for $f_{2}\left(2^{1990}\right)$:
$$
\begin{array}{l}
f_{1}\left(2^{1990}\right)<4^{2} \times(1994)^{2}<2^{26}, \\
f_{2}\left(2^{1990}\right)<(4 \times 30)^{2}=14400 .
\end{array}
$$

Thus, the sum of the digits of $f_{2}\left(2^{1990}\right)$ is at most $36(=4 \times 9)$, so
$$
\begin{array}{l}
f_{3}\left(2^{1990}\right) \leqslant 36^{2}=1296, \\
f_{4}\left(2^{1990}\right)<(9+9+9)^{2}=729, \\
f_{5}\left(2^{1990}\right)<(6+9+9)^{2}=576 .
\end{array}
$$

On the other hand, since
$$
f_{1}(k) \equiv k^{2}(\bmod 9),
$$

we have
$$
\begin{array}{l}
f_{1}\left(2^{1990}\right) \equiv\left(2^{1990}\right)^{2} \equiv(-2)^{2} \equiv 4(\bmod 9) \\
f_{2}\left(2^{1990}\right) \equiv 4^{2} \equiv-2(\bmod 9) \\
f_{3}\left(2^{1990}\right) \equiv(-2)^{2} \equiv 4(\bmod 9) \\
\cdots \cdots
\end{array}
$$

Therefore, when $n$ is odd,
$$
f_{n}\left(2^{1990}\right) \equiv 4(\bmod 9) .
$$

When $n$ is even,
$$
f_{n}\left(2^{1990}\right) \equiv-2(\bmod 9) .
$$

If $x^{2} \equiv 4(\bmod 9)$ and $x \leqslant 24$, then
$$
x^{2} \in H=\{4,49,121,256,400\},
$$

By simple calculation (for each $h \in H$),
$$
\begin{array}{l}
f_{3}(h)=f_{5}(h)=f_{7}(h)=\cdots=169, \\
f_{4}(h)=f_{6}(h)=f_{8}(h)=\cdots=256 .
\end{array}
$$

Since $f_{5}\left(2^{1990}\right) \in H$, for $n \geqslant 8$,
$$
f_{n}\left(2^{1990}\right)=\left\{\begin{array}{l}
169, \text { if } n \text { is even; } \\
256, \text { if } n \text { is odd. }
\end{array}\right.
$$

Therefore,
$$
f_{1991}\left(2^{1990}\right)=256 .
$$