[Solution] From $a_{1}=a_{1+0}=\frac{1}{2}\left(a_{2}+a_{0}\right)-1$, we have $a_{2}=13$.
For $n \geqslant 1$, by the assumption,
$$a_{2 n}=a_{n+1+n-1}=\frac{1}{2}\left(a_{2 n+2}+a_{2 n-2}\right)-4 \text {, }$$

Thus, let $b_{n}=a_{2 n}$, we get
$$b_{0}=11, b_{1}=13 \text {, and } \quad b_{n+1}=2 b_{n}-b_{n-1}+8, n \geqslant 1 .$$

From this, we have $\sum_{k=1}^{n} b_{k+1}=2 \sum_{k=1}^{n} b_{k}-\sum_{k=1}^{n} b_{k-1}+8 n$,
which means $b_{n+1}=b_{n}+b_{1}-b_{0}+8 n=b_{n}+8 n+2$.
Therefore, we get $b_{n+1}=b_{1}+2 n+8 \sum_{k=1}^{n} k=4 n^{2}+6 n+13$.
Given $b_{0}=11, b_{1}=13$, we have
$$b_{n}=4(n-1)^{2}+6(n-1)+13=4 n^{2}-2 n+11, n=0,1,2, \cdots$$

Thus, $a_{2 n}=4 n^{2}-2 n+11, n=0,1,2, \cdots$.
For any $n \geqslant 0$, using the assumption, we get
$$\begin{aligned} a_{n} & =a_{n+0}=\frac{1}{2}\left(a_{2 n}+a_{0}\right)-n^{2} \\ & =\frac{1}{2}\left(4 n^{2}-2 n+11+11\right)-n^{2} \\ & =n^{2}-n+11 \end{aligned}$$

Therefore, $a_{45}=45 \cdot 44+11=1991$.