G3.4 If $F_{n}$ is an integral valued function defined recursively by $F_{n}(k)=F_{1}\left(F_{n-1}(k)\right)$ for $n \geq 2$ where $F_{1}(k)$ is the sum of squares of the digits of $k$, find the value of $F_{2012}(7)$.
$$
\begin{array}{l}
F_{1}(7)=7^{2}=49 \\
F_{2}(7)=F_{1}\left(F_{1}(7)\right)=F_{1}(49)=4^{2}+9^{2}=97 \\
F_{3}(7)=9^{2}+7^{2}=130 \\
F_{4}(7)=1^{2}+3^{2}+0^{2}=10 \\
F_{5}(7)=1 \\
F_{2012}(7)=1
\end{array}
$$

Remark: the original question If $f$ is an integer valued function ... The recursive function is defined for $F_{n}$, not $f$.