5. Given the function $f:\{1,2, \cdots, 10\} \rightarrow\{1,2,3,4,5\}$, and for all $k=1,2, \cdots, 9$, we have $|f(k+1)-f(k)| \geqslant 3$. Then the number of functions $f$ that satisfy this condition is $\qquad$.
Obviously, $f(k) \neq 3$. Let $A=\{1,2, \cdots, n\}, B=\{1,2,3,4,5\}$, and the number of functions $f: A \rightarrow B$ be $c_{n}$. When $n=2$, $f($ ordered by $f(1), f(2)$ $): 14,15,25,41,51,52$;
When $n=3$, $f: 141,151,152,251,252,414,415,514,515,525$.
Thus, when $f(n)=2,4$, $f(n+1)$ can only be 5,1; when $f(n)=1,5$, $f(n+1)$ can be 4,5; 1,2. Therefore, let the number of $c_{n}$ that satisfies $f(n)=2,4$ be $a_{n}$, and the number that satisfies $f(n)=1,5$ be $b_{n}$. Then we have $a_{n+1}=b_{n}, b_{n+1}=a_{n}+b_{n}$, and $c_{n}=a_{n}+b_{n}$.
Rearranging the above equations, we get $c_{n}=b_{n+1}=b_{n-1}+b_{n}, b_{2}=4, b_{3}=6$, so $b_{4}=10, b_{5}=16, b_{6}=26$, $b_{7}=42, b_{8}=68, b_{9}=110, b_{10}=178, c_{10}=b_{11}=288$.