# Sum of the multiplicative function $\lambda_2$

Do we know any precise evaluation of the sum: $$\sum_{m \leqslant X} \lambda_2(m)$$

where $\lambda_2 = \mu \star \mu$ ?

That is a first part to my question, which seems to me quite classical but I don't succeed in finding anything (even with the hyperbola principle, and I prefer ask more than sinking into technicalities possibly useless with Perron-like sumatory formulas...). The other part is : do we know a similar estimation for the sum not over integers, but over ideals on a number field ?

Thank you in advance,

D

## 1 Answer

We may use the fact that $\sum_{n = 1}^{\infty} \frac{\mu * \mu(n)}{n^s} = \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} \sum_{n = 1}^{\infty} \frac{\mu(n)}{n^s} = \frac{1}{\zeta(s)^2}.$ This allows one to use the Perron inversion formula and mimic the proof of the prime number theorem to show that $\sum_{n \leq x} \mu * \mu(n) = o(x),$ and indeed something stronger along the lines of $\sum_{n \leq x} \mu * \mu(n) = O(x \exp(-c\sqrt{\log x})).$ The Riemann hypothesis implies that $\sum_{n \leq x} \mu * \mu(n) = O_{\varepsilon}(x^{1/2 + \varepsilon})$ for every $\varepsilon > 0$, while one can also show that $\sum_{n \leq x} \mu * \mu(n) = \Omega_{\pm}(\sqrt{x})$ via the same methods as for $M(x) = \sum_{n \leq x} \mu(n)$. In particular, this function ought to be quite oscillatory and satisfy almost square-root cancellation, but have some occasional large fluctuations.

All this also works over number fields via the exact same methods.

To give a classical proof of the first bound, use the Dirichlet hyperbola method to write $\sum_{n \leq x} \mu * \mu(n) = 2 \sum_{n \leq \sqrt{x}} \mu(n) M\left(\frac{x}{n}\right) - M(\sqrt{x})^2.$ Now using the fact (equivalent to the prime number theorem) that $M(x) = O(x \exp(-c\sqrt{\log x}))$ gives you the desired result.