China Mathematical Competition (Extra Test)

For arbitrary $a,k\in {\mathbb{N}}^{+}$, if $k^2<a<(k+1{)}^2$, we set $$a=k^2+m,m=1,2,\dots ,2k,$$ $$\sqrt{a}=k+\theta ,0<\theta <1,$$ $$\left[\frac{1}{\sqrt{a}}\right]=\left[\frac{1}{\sqrt{a-k}}\right]=\left[\frac{\sqrt{a+k}}{a-k^2}\right]=\left[\frac{2k+\theta }{m}\right].$$ $$\text{For}0<\frac{2k+\theta }{m}-\frac{2k}{\theta }<1,$$ if there exists an integer $t$ between $\frac{2k}{m}$ and $\frac{2k+\theta }{m}$, then $$\frac{2k}{m}<t<\frac{2k+\theta }{m}.$$ On one hand $2k<mt$, thus $2k+1\le mt$. On the other hand, $mt<2k+\theta <2k+1$, a contradiction. $$\begin{gathered} \text{Thus}\left[\frac{2k+\theta }{m}\right]=\left[\frac{2k}{m}\right], \\ \sum _{k<a<k+1}\left[\frac{1}{\{a\}}\right]=\sum _{m=1}^{2k}\left[\frac{2k}{m}\right], \\ \sum _{a=1}^{n+1}f(a)=\sum _{k=1}^n\sum _{i=1}^{2k}\left[\frac{2k}{i}\right].(1) \end{gathered}$$

Thus $$\begin{array}{rl}\sum _{a=1}^{(n+1{)}^2}f(a)& =\sum _{k=1}^n\sum _{j=1}^{2k}T(j)\\ & =n[T(1)+T(2)]+(n-1)[T(3)+T(4)]\\ & +\cdots +[T(2n-1)+T(2n)]\end{array}(2)$$ $$\text{From (2), }\sum _{k=1}^{16^2}f(k)=\sum _{k=1}^{15}(16-k)[T(2k-1)+T(2k)].(3)$$

k	1	2	3	4	5	6	7	8	9	10	11	12	13	14	15
$a_k$	3	5	6	6	7	8	6	9	8	8	8	10	7	10	10

$$\text{Therefore, }\sum _{k=1}^{256}f(k)=\sum _{k=1}^{15}(16-k)a_k=783.(4)$$ Note that $f(256)=f(16^2)=0$ by definition. When $k\in \{241,242,\dots ,255\}$, denote $k=15^2+r(16\le r\le 30)$, then $$\sqrt{k}-15=\sqrt{15^2+r}-15=\frac{r}{\sqrt{15^2+r}+15},$$ $$\frac{r}{31}<\frac{r}{\sqrt{15^2+r+15}}<\frac{r}{30},1\le \frac{30}{r}<\frac{1}{\{\sqrt{15^2+r}\}}<\frac{31}{r}<2.$$ Thus $$\left[\frac{1}{\sqrt{k}}\right]=1,k\in \{241,242,\dots ,255\}.(5)$$ Therefore, $$\sum _{k=1}^{200}f(k)=783-\sum _{k=201}^{256}f(k)=783-15=768.$$