141 Write an $n$-digit number using only the digits 1 and 2, where no two adjacent digits are both 1, and let the number of such $n$-digit numbers be $f(n)$. Then $f(10)=$ $\qquad$ .
141 144. $f(1)=2, f(2)=3$. $f(n+1)$ numbers of $n+1$ digits can be divided into two categories: one ends with 2, and there are $f(n)$ such numbers; the other ends with 1, and the second last digit of this category must be 2, thus there are $f(n-1)$ such numbers. Therefore,
$$
f(n+1)=f(n)+f(n-1) . f(10)=55+89=144 .
$$