# What is the value of the sum $\sum_{n=3}^N(\mu(n))^22^{\nu(n)}=\sum_{p=3}^N2+\sum_{pq=15}^N4+\sum_{pqr=105}^N8+\dots$

Consider the sum $$\sum_{n=3}^N(\mu(n))^22^{\nu(n)}=\sum_{\rho_1=3}^N2+\sum_{\rho_1\rho_2=15}^N4+\sum_{\rho_1\rho_2\rho_3=105}^N8+\dots$$ where for each $i$ we have $\rho_i$ in the set of primes and $\ell=\prod_{i=1}^m\rho_i$ ranges over the set of squarefree values having $m=\nu(\ell)$ factors.

The simple upper bound I have come up with so far is

$$\sum_{n=3}^N2^k=2^k(N-2)$$

where $k$ is the largest integer such that $p_k\#\le N$. I believe that this simple upper bound is many orders of magnitude larger than the value of the sum for large $N$, but I don't know how to improve it.

Is there anything known about the value of this sum or a better upper bound on it?

• This may be helpful: oeis.org/A069201 – Matthew Conroy Feb 1 '18 at 0:23
• @MatthewConroy: $O(n\log n)$ sounds about right, thank you for the find! – abiessu Feb 1 '18 at 1:05

The comment by Matthew Conroy (https://math.stackexchange.com/users/2937/matthew-conroy) supplied the answer to my question, and also reminded me of the usual notation for the quantity which I labeled $\nu(n)$ in my question. This usual notation is $\omega(n)$ (http://oeis.org/wiki/Omega(n),_number_of_prime_factors_of_n_(with_multiplicity)) and represents the number of distinct prime factors of a number $n$.

The sequence listed at http://oeis.org/A069201 lists the asymptotic formula as $Cn\log n+O(n)$, where $C$ is the constant given by

$$C=\prod_p\left(1-\frac 1p\right)^2\left(1+\frac 2p\right).$$