$$
求\sum_{i=1}^{n}\gcd(\left\lfloor\sqrt[3]{i}\right\rfloor,i)\\令m=\sqrt[3]{n}\\
原式=\sum_{i=1}^{m-1}\sum_{j=i^3}^{(i+1)^3-1}\gcd(i,j)+\sum_{i=m^3}^{n}\gcd(m,i)\\
\begin{aligned}
其中\sum_{i=1}^{n}\gcd(i,a)&=\sum_{k|a}k\sum_{i=1}^{n}[\gcd(i,a)=k]\\
&=\sum_{k|a}k\sum_{i=1}^{n/k}[\gcd(i,a/k)=1]\\
&=\sum_{k|a}k\sum_{i=1}^{n/k}\sum_{d|\gcd(i,a/k)}\mu(d)\\
&=\sum_{k|a}k\sum_{d|(a/k)}\mu(d)\sum_{i=1}^{n/k}[d|i]\\
&=\sum_{k|a}k\sum_{d|(a/k)}\mu(d)(n/kd)\\
令l=a/k:\\
上式&=\sum_{l|a}\frac{a}{l}\sum_{d|l}\mu(d)\frac{nl}{ad}\\
&=\sum_{l|a}\frac{an}{la}\sum_{d|l}\mu(d)\frac{l}{d}\\
&=\sum_{l|a}\frac{n}{l}(\mu*id)l\\
&=\sum{l|a}\frac{n}{l}\varphi(l)\\
&=n\sum_{d|a}\frac{\varphi(d)}{d}
\end{aligned}\\
$$
$$
\begin{aligned}
原式&=\sum_{i=1}^{m-1}\left(\gcd(i^3,i)+\sum_{j=1}^{(i+1)^3-1}\gcd(i,j)-\sum_{j=1}^{i^3}\gcd(i,j)\right)+\left(\gcd(m,m^3)+\sum_{i=1}^{n}\gcd(m,i)-\sum_{i=1}^{m^3}\gcd(i,m)\right)\\
&=\sum_{i=1}^{m-1}\left(i+((i+1)^3-1)\sum_{d|i}\frac{\varphi(d)}{d}-i^3\sum_{d|i}\frac{\varphi(d)}{d}\right)+\left(m+n\sum_{d|m}\frac{\varphi(d)}{d}-m^3\sum_{d|m}\frac{\varphi(d)}{d}\right)\\
&=\sum_{i=1}^{m-1}\left(i+(i^3+3i^2+3i+1-1-i^3)\sum_{d|i}\frac{\varphi(d)}{d}\right)+m+\sum_{d|m}\frac{(n-m^3)\varphi(d)}{d}\\
&=\sum_{i=1}^{m-1}\left(i+3(i^2+i)\sum_{d|i}\frac{\varphi(d)}{d}\right)+m+S(可以\sqrt m即n^{\frac{1}{6}}算出)
\end{aligned}\\
\begin{aligned}
左边&=\sum_{i=1}^{m-1}i+3\sum_{i=1}^{m-1}(i^2+i)\sum_{d|i}\frac{\varphi(d)}{d}\\
&=\frac{m(m-1)}{2}+3\sum_{d=1}^{m-1}\frac{\varphi(d)}{d}\sum_{i=1}^{m-1}(i^2+i)[d|i]\\
&=\frac{m(m-1)}{2}+3\sum_{d=1}^{m-1}\frac{\varphi(d)}{d}\sum_{i=1}^{(m-1)/d}(i^2d^2+id)\\
&=\frac{m(m-1)}{2}+3\sum_{d=1}^{m-1}\varphi(d)\sum_{i=1}^{(m-1)/d}(i^2d+i)\\
&=\frac{m(m-1)}{2}+3\sum_{d=1}^{m-1}\varphi(d)\frac{(m-1)(\frac{m-1}{d}+1)(\frac{2(m-1)}{d}+1)+3((m-1)/d)((m-1)/d+1)}{6}\\
令t&=\frac{m-1}{d}\\
左&=\frac{m(m-1)}{2}+\sum_{i=1}^{m-1}\varphi(d)\left[(m-1)\frac{(t+1)(2t+1)}{2}+\frac{3t(t+1)}{2}\right]\\
\end{aligned}\\
$$
$$
设g(t)=(m-1)\frac{(t+1)(2t+1)}{2}+\frac{3t(t+1)}{2}\\
有原式=\frac{m(m-1)}{2}+\sum_{i=1}^{m-1}\varphi(d)g(\frac{m-1}{d})+m+S=\sum_{i=1}^{m-1}\varphi(d)g(\frac{m-1}{d})+\frac{m(m+1)}{2}+S
$$