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Question Number 196062 by qaz last updated on 17/Aug/23

$$\underset{i=1}{\sum ^n}\underset{j=1}{\sum ^n}[gcd(i,j)=1]=?,\left[D\right]=\{\begin{array}{l}1,Disture.\\ 0,Disfalse.\end{array}$$

Commented by mr W last updated on 17/Aug/23

$$i{don}^{\prime }tthinkwecangetitinaclosed$$$$form.$$$$n=10:\Rightarrow 63$$$$n=100:\Rightarrow 6087$$$$n=1000:\Rightarrow 808383$$

Answered by witcher3 last updated on 17/Aug/23

$${\mathrm{A}}_{\mathrm{n}}=\sum _{\mathrm{i},\mathrm{j}⩽\mathrm{n}}[\gcd (\mathrm{i},\mathrm{j})=1]$$$${\mathrm{A}}_{\mathrm{n}+1}=\sum _{\mathrm{i},\mathrm{j}⩽\mathrm{n}+1}[\gcd (\mathrm{i},\mathrm{j})=1]$$$$={\mathrm{A}}_{\mathrm{n}}+2\underset{\mathrm{m}=1}{\sum ^{\mathrm{n}+1}}\left[\gcd (\mathrm{n}+1,\mathrm{m})\right]$$$$={\mathrm{A}}_{\mathrm{n}}+2\phi (\mathrm{n}+1)..\phi \mathrm{euler}\mathrm{indecatrice}\mathrm{function}$$$$\phi \left(\mathrm{n}\right)=\mathrm{card}(\mathrm{i},\gcd (\mathrm{i},\mathrm{n})=1)$$$$\Rightarrow {\mathrm{A}}_{\mathrm{n}+1}-{\mathrm{A}}_{\mathrm{n}}=2\phi (\mathrm{n}+1)$$$$\Rightarrow \underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}{\mathrm{A}}_{\mathrm{k}+1}-{\mathrm{A}}_{\mathrm{k}}=2\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}\phi (\mathrm{k}+1),\mathrm{n}⩾2$$$$\Rightarrow {\mathrm{A}}_{\mathrm{n}}-{\mathrm{A}}_1=2\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}\phi \left(\mathrm{k}\right)$$$${\mathrm{A}}_{\mathrm{n}}=1+2\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}-1}}\phi \left(\mathrm{k}\right)$$

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