$$\begin{gathered} gay(i)=\left\{\begin{array}{ccc}0& if& i=k\times x\times x,x>k,k>1\\ i\times i& else& \end{array}\right\} \\ 求\sum _{i=1}^n\sum _{j=1}^{i}gay(j) \\ =\sum _{i=1}^n(n-i+1)gay(i) \\ =\sum _{i=1}^n(n-i+1){\mu }^2(i)i^2 \\ 因为{\mu }^2(n)=\sum _{i^2\mid n}\mu (i)，容斥定理显然得到 \\ 有原式=\sum _{i=1}^n(n-i+1)i^2\sum _{j^2\mid i}\mu (j) \\ =(n+1)\sum _{i=1}^n\sum _{j^2\mid i}\mu (j)-\sum _{i=1}^n{i}^3\sum _{j^2\mid i}\mu (j) \\ =(n+1)\sum _{j=1}^{\sqrt{n}}\mu (j)\sum _{j^2\mid i}{i}^2-\sum _{j=1}^{\sqrt{n}}\mu (j)\sum _{j^2\mid i}{i}^3 \\ =(n+1)\sum _{j=1}^{\sqrt{n}}\mu (j)\sum _{t=1}^{\frac{n}{j^2}}{t}^2{j}^4-\sum _{j=1}^{\sqrt{n}}\mu (j)\sum _{t=1}^{\frac{n}{j^2}}{t}^3{j}^6 \\ =\sum _{i=1}^{\sqrt{n}}\mu (i)\sum _{j=1}^{\frac{n}{i^2}}(i^4{j}^2(n+1)-i^6{j}^3) \end{gathered}$$