6. 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 $f_{2}\left(2^{1990}\right)$:
$$\begin{array}{l}
f_{1}\left(2^{1990}\right)<4^{2} \times(1994)^{2}<2^{26} \\
f_{2}\left(2^{1990}\right)<(4 \times 30)^{2}=14400
\end{array}$$

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

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

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

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

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

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