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Liy 10 :
Liy 10 :
<math>\tau\left(n\right)=-\sum_{i=1}^{n}{\left[\frac{n}{i}-\left[\frac{n}{i}\right]\right]}</math>
<math>\tau\left(n\right)=-\sum_{i=1}^{n}{\left[\frac{n}{i}-\left[\frac{n}{i}\right]\right]}</math>
<math>\tau\left(n\right)=\sum_{i=1}^{n}{-\left[\frac{n}{i}-\left[\frac{n}{i}\right]\right]}</math>
<math>\tau\left(n\right)=\sum_{i=1}^{n}{\left| \left[\frac{n}{i}-\left[\frac{n}{i}\right]\right]\right| }</math>
Vèsyon jou 24 mas 2013 à 13:57
Fonksyon
τ
{\displaystyle \tau }
ki kantite divizè positif yon antye pozitif ( cf : Approche Élémentaire de l'Étude des Fonctions Arithmétiques : A. Mercier et J. M. De Koninck )
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{\displaystyle \tau \left(n\right)=\sum _{d|n}1}
Ekpresyon fonksyon
τ
{\displaystyle \tau }
selon Lainé Jean Lhermite Junior
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{\displaystyle \tau \left(n\right)=\sum _{i=1}^{n}{\left[{\frac {\left[{\frac {n}{i}}\right]}{\left({\frac {n}{i}}\right)}}\right]}}
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{\displaystyle \tau \left(n\right)=-\sum _{i=1}^{n}{\left[{\frac {n}{i}}-\left[{\frac {n}{i}}\right]\right]}}
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{\displaystyle \tau \left(n\right)=\sum _{i=1}^{n}{\left|\left[{\frac {n}{i}}-\left[{\frac {n}{i}}\right]\right]\right|}}
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{\displaystyle \tau \left(n\right)=\left|\sum _{i=1}^{n}{\left[{\frac {n}{i}}-\left[{\frac {n}{i}}\right]\right]}\right|}