Sum of Möbius functions

• Sum of [Möbius function ${\sum }_{d|n}\mu (d)=\{\begin{array}{lc}1& (n=1)\\ 0& (n>1)\end{array}$

• ${\sum }_{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})=0$ - binomial theorem - $(-1+1{)}^n={\sum }_{k=0}^n(-1{)}^k{1}^{n-k}\left(\genfrac{}{}{0ex}{}{n}{k}\right)={\sum }_{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})$ - $(-1+1{)}^n=0^n=0$

${\sum }_{d|n}\mu (n/d)g(d)={\sum }_{d|n}\mu (n/d){\sum }_{l|d}f(l)={\sum }_{d|n}{\sum }_{l|d}\mu (n/d)f(l)$

${\sum }_{d|n}{\sum }_{l|d}\mu (n/d)f(l)={\sum }_{l|n}f(l){\sum }_{(n/d)|(n/l)}\mu (n/d)$

From eq1 ${\sum }_{(n/d)|(n/l)}\mu (n/d)=\{\begin{array}{lc}1& (n/l=1)\\ 0& (n/l>1)\end{array}$

• Mobius inversion formula and three applications - Math and Other Days from Möbius transformation 20201105

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