\[\sum\limits_{p\in prime}\sum\limits_{i=1}^{n}\sum\limits_{j=1}^{n}\gcd(i,j)=p\\ \]

\[\sum\limits_{p\in prime}\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\sum\limits_{j=1}^{\lfloor\frac{n}{p}\rfloor}\gcd(i,j)=1\\ \]

\[\sum\limits_{p\in prime}\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\left(\sum\limits_{j=1}^i2\times [\gcd(i,j)=1]\right)-1\\ \]

\[\sum\limits_{p\in prime}\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\left(\sum\limits_{j=1}^i2\times [\gcd(i,j)=1]\right)-1\\ \]

\[\sum\limits_{p\in prime}\sum\limits_{i=1}^{\lfloor\frac{n}{p}\rfloor}\left(\sum\limits_{j=1}^i2\times \varphi(i)\right)-1\\ \]