Now that we have seen Dirichlet's hyperbola method, we try to use it find sub-linear algorithms for different functions.
Jordan's totient function, denoted as J_k(n), generalizes Euler's totient function. Let's use \mathbf{J}_k(n) to denote Jordan summatory function. That is
\mathbf{J}_k(n)=\displaystyle\sum\limits_{m=1}^nJ_k(m)
We know that,1*J_k(n)=n^k, where '*' denotes Dirichlet convolution.
Choosing f(n)=J_k(n) and g(n)=1, we have h(n)=f*g=n^k. The corresponding summatory functions are
F(n)=\mathbf{J}_k(n), G(n)=n and H(n)=S_k(n).
Using the results obtained in Dirichlet's hyperbola method, we get
\mathbf{J}_k(n)=S_k(n)-\displaystyle\sum\limits_{m=2}^n\mathbf{J}_k\left(\left\lfloor\frac{n}{m}\right\rfloor\right)
\mathbf{J}_k(n)=S_k(n)-\displaystyle\sum\limits_{m=1}^{n/(u+1)} \left\lfloor\frac{n}{m}\right\rfloor J_k(n) - \sum_{m=2}^u \mathbf{J}_k\left(\left\lfloor\frac{n}{m}\right\rfloor\right) +u^\text{ }\mathbf{J}_k\left(\left\lfloor\frac{n}{u+1}\right\rfloor\right)
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Jordan's totient function, denoted as J_k(n), generalizes Euler's totient function. Let's use \mathbf{J}_k(n) to denote Jordan summatory function. That is
\mathbf{J}_k(n)=\displaystyle\sum\limits_{m=1}^nJ_k(m)
We know that,1*J_k(n)=n^k, where '*' denotes Dirichlet convolution.
Choosing f(n)=J_k(n) and g(n)=1, we have h(n)=f*g=n^k. The corresponding summatory functions are
F(n)=\mathbf{J}_k(n), G(n)=n and H(n)=S_k(n).
Using the results obtained in Dirichlet's hyperbola method, we get
\mathbf{J}_k(n)=S_k(n)-\displaystyle\sum\limits_{m=2}^n\mathbf{J}_k\left(\left\lfloor\frac{n}{m}\right\rfloor\right)
\mathbf{J}_k(n)=S_k(n)-\displaystyle\sum\limits_{m=1}^{n/(u+1)} \left\lfloor\frac{n}{m}\right\rfloor J_k(n) - \sum_{m=2}^u \mathbf{J}_k\left(\left\lfloor\frac{n}{m}\right\rfloor\right) +u^\text{ }\mathbf{J}_k\left(\left\lfloor\frac{n}{u+1}\right\rfloor\right)
Yours Aye
Me
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