omega function

[afif mansib ch]

let \[\omega (n)\]denote the number of distinct prime divisor of n>1 with\[\omega (1)=0\]
for instance \[\omega (360=2^3.3^2.5)=3\]
a.show that \[2^{\omega (n)}\]
is a multiplicative function.
b.\[\pi (n^2)=\sum2^{\omega (d)}\]
d means the divisors of n.

[Phlembac Adib Hasan]

Well, no a. is very easy.Because if $a$ and $b$ are co-prime then $\omega (a)+\omega(b)=\omega (ab)$.So the result is obvious as $2^{\omega (a)}2^{\omega(b)}=2^{\omega (a)+\omega(b)}=2^{\omega (ab)}$.I'll try b now.

[afif mansib ch]

well i just solved both.
a.\[\omega (mn)=(\prod_{i=1}^{x}pi\times \prod_{r=1}^{y}qr)=x+y\]
so\[2^{mn}=2^{m}.2^{n}\]
b.from a \[\sum_{d/n}2^{\omega(d)}\]
is multiplicative function.so it's enough to establish n for a prime power.
\[\pi ((p^k)^2)=\pi (p^{2k})=2k+1\]
now \[2^{\omega (pi)}=2\]
for k+1 this sum is exactly 2k+1
i can't find a sol for b using pi function.can it be done?

[*Mahi*]

What did you mean by $\pi (n)$?
http://mathworld.wolfram.com/PrimeCountingFunction.html

[afif mansib ch]

the total number of divisors.
the real sign i didn't find in eq editor.

[Phlembac Adib Hasan]

Hei, then you should told before. In number theory $\pi (n)$ means the number of primes that are less then or equal to $n$.
My solution:

Let $n=\Pi_{m \ge k\ge 1} p_k ^{q_k}$
Then, \[\pi (n)=\Pi_{m \ge k\ge 1} (q_k+1) \]
\[ \pi (n^2)=\Pi_{m \ge k\ge 1} (2q_k+1) \]
\[=\sum_{w=1}^{n}2^w.\sum_{p_1,p_2,,,,p_w\epsilon 1,2,3,,,n}\prod_{z=1}^{w} q_{p_z}\]
Here the set $p_1,p_2,,,,,,p_n$ is any possible permutation of the set $1,2,3,,,,n$.
Now let's find what $\sum_{d|n}2^ {\omega (d)}$'s accurate value.${\omega (d)}=2$ happens if $d=p_i^x.p_j^y$
So for such divisors $\sum_{d|n}2^ {\omega (d)} =2^2.\sum_{p_1,p_2 \epsilon 1,2,3,,,n}\prod_{z=1}^{2} q_{p_z}$
From this we can say that, \[\sum_{d|n}2^ {\omega (d)}=\sum_{w=1}^{n}2^w.\sum_{p_1,p_2,,,,p_w\epsilon 1,2,3,,,n}\prod_{z=1}^{w} q_{p_z} \]
So \[ \pi(n^2)=\sum_{d|n}2^ {\omega (d)}\]

[afif mansib ch]

ok sorry.i posted a similiar problem b4 bt did't have any problem then.