Liouville Summatory function

The Liouville Summatory function starts with the following well known identity

$\lambda*1=\epsilon_2$

where $\lambda$ is the Liouville's function and $\epsilon_2$ is the characteristic function of squares.

Like in the case of Merten's function, this property simplifies $\hat{F}(n)$ to a simple value.

Let $f(n)=\lambda(n)$. Then,

$F(n)=\displaystyle\sum\limits_{i=1}^n\lambda(i)=L(n)$ and $\hat{F}(n)=\displaystyle\sum\limits_{i=1}^n\sum_{d|i}\lambda(d)=\lfloor\sqrt n\rfloor$

Using these values in A special case of Dirichlet's Hyperbola method, we have

$\lfloor\sqrt n\rfloor=\displaystyle\sum\limits_{i=1}^{n/(u+1)} \left\lfloor\frac{n}{i}\right\rfloor \lambda(i) + \sum_{d=1}^u L\left(\left\lfloor\frac{n}{d}\right\rfloor\right) -u^{\text{ }}L\left(\left\lfloor\frac{n}{u+1}\right\rfloor\right)$, $u=\lfloor \sqrt{n}\rfloor$

Solving for the first term of the right summation,

$L(n)=\lfloor\sqrt n\rfloor-\displaystyle\sum\limits_{i=1}^{n/(u+1)} \left\lfloor\frac{n}{i}\right\rfloor \lambda(i) - \sum_{d=2}^u L\left(\left\lfloor\frac{n}{d}\right\rfloor\right) +u^{\text{ }}L\left(\left\lfloor\frac{n}{u+1}\right\rfloor\right)$, $u=\lfloor \sqrt{n}\rfloor$

We can use the intermediate result we obtained in Dirichlet's hyperbola method to write

$\lfloor\sqrt n\rfloor=\displaystyle\sum\limits_{k=1}^nL\left(\left\lfloor\frac{n}{k}\right\rfloor\right)$

$L(n)=\lfloor\sqrt n\rfloor-\displaystyle\sum\limits_{k=2}^nL\left(\left\lfloor\frac{n}{k}\right\rfloor\right)$


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