How to compute the mean average exponent of the naturals? What is the limit for large numbers? With a friend I was trying to get an understanding for why the expected gap between primes is logarithmic. With that motivation I tried to express the average exponent of numbers. 
By average exponent of a number I mean the following rational: If $p,q,r$ are some primes and a number $n$ has factorization $n=p^aq^br^c$, then in this case the average exponent would be $(a+b+c)/3$. For example the number $3087$ equals $3^2\,7^3$, a number where all exponents are bigger than one, and the average exponent is $(2+3)/2=2.5$. Similarly, the average exponent of $156=2^2\,3^113^1$ is $(2+1+1)/3=1.\dot 3$. 
We computed the average exponent for the first $10000$ numbers and, not too surprisingly, the value jumps quite a bit and the function isn't nice to look at. So next we plot a function of $n$ with value being the mean from the numbers $2$ up to $n$  (of the average exponents). 
The plot is below, it seems to stabilize to $\approx 4/3$ in this interval. How to interpret this?
What is the behavior and more concretely the limit of the above construction? How to compute it?

 A: Define $\Omega(n)$ by the number of all prime divisors of $n$ counted with multiplicity, and 
$\omega(n)$ by the number of distinct prime divisors of $n$. Then the average that you are interested is $\Omega(n)/\omega(n)$. 
This function has lower limit $1$ since for $n=p$ prime, we have $\omega(p)=1$, $\Omega(p)=1$. 
On the other hand, this has upper limit $\infty$ since for $n=2^k$, we have $\omega(n)=1$ but $\Omega(n)= k$. We can make $k$ arbitrarily large. 
Thus, this function is very erratic as you noticed.
However, interesting behavior of this average is that we have the following:
$$
\sum_{n\leq x}\frac{\Omega(n)}{\omega(n)} = x+O\left(\frac{x}{\log\log x}\right).$$
So, the average of the average stabilizes to $1$. 
Proof of the mean value asymptotic
$$\sum_{n\leq x} \frac{\Omega(n)}{\omega(n)}=\sum_{n\leq x} \frac{\omega(n)+\Omega(n)-\omega(n)}{\omega(n)}
$$
$$
=x+O(1)+\sum_{k\geq 1} \sum_{\substack{{n\leq x}\\{\Omega(n)-\omega(n)=k}}} \frac{k}{\omega(n)}$$
We see from above that knowing density of numbers with $\Omega(n)-\omega(n)=k$ will be helpful. This is done by Renyi:
Reference: R´enyi, A. (1955). On the density of certain sequences of integers, Acad. Serbe Sci. Publ.
Inst. Math. 8, 157–162
The result is:
Let $b_k$ be the density of $n\leq x$ with $\Omega(n)-\omega(n)=k$. Then we have
$$
\sum_{k=0}^{\infty} b_k z^k = \prod_{p \textrm{ prime }} \left(1-\frac 1 p \right)\left( 1+\frac{1}{p-z}\right)$$
Note that this power series has radius of convergence $2$. 
Furthermore, a quantitative version of the density is 
$$
\sum_{\substack{{n\leq x}\\{\Omega(n)-\omega(n)=k}}} 1 = b_k x + O( 0.75^k x^{0.5}(\log x)^{4/3})
$$
Going back to the sum over $k$, we split the inner sum into two parts: 
$$
\sum_{k\geq 1} \left(\sum_{\substack{{n\leq x}\\{\Omega(n)-\omega(n)=k}\\{\Omega(n)<0.5\log\log x}}}\frac{k}{\omega(n)}+\sum_{\substack{{n\leq x}\\{\Omega(n)-\omega(n)=k}\\{\Omega(n)\geq 0.5\log\log x}}}\frac{k}{\omega(n)}\right)$$
The first inner sum can be treated by:
$$
\sum_{\substack{{n\leq x}\\{\Omega(n)-\omega(n)=k}\\{\Omega(n)<0.5\log\log x}}}\frac{k}{\omega(n)}\leq \sum_{\substack{{n\leq x}\\{\Omega(n)-\omega(n)=k}\\{\Omega(n)<0.5\log\log x}}}k
$$
$$
 \leq k(\log x)^{-0.5\log 0.5} \sum_{\substack{{n\leq x}\\{\Omega(n)-\omega(n)=k}}} 0.5^{\Omega(n)-\omega(n)+\omega(n)}$$
$$
 = k  \ 0.5^k (\log x)^{-0.5\log 0.5}\sum_{n\leq x} 0.5^{\omega(n)}
$$
$$\leq k  \ 0.5^k (\log x)^{-0.5\log 0.5} C x(\log x)^{-0.5}
$$ 
$$\leq k \ 0.5^k x (\log x)^{-c}$$
for absolute  constants $c>0, C>0$.
The second inner sum is treated by: (Case 1: $k\leq 0.25\log\log x$ )
$$
\sum_{\substack{{n\leq x}\\{\Omega(n)-\omega(n)=k}\\{\Omega(n)\geq 0.5\log\log x}}}\frac{k}{\omega(n)}\leq k \sum_{\substack{{n\leq x}\\{\Omega(n)-\omega(n)=k}}}\frac{1}{0.5\log\log x -k}
$$
$$\leq  4 k b_k \frac{x}{\log\log x} + 4 k \frac{0.75^k x^{0.5} (\log x)^{4/3}}{\log\log x}
$$
and for Case 2: $k>0.25\log\log x$
$$
\sum_{\substack{{n\leq x}\\{\Omega(n)-\omega(n)=k}\\{\Omega(n)\geq 0.5\log\log x}}}\frac{k}{\omega(n)}\leq k \sum_{\substack{{n\leq x}\\{\Omega(n)-\omega(n)=k}}}1
$$
$$
\leq k b_k x + k \ 0.75^k x^{0.5} (\log x)^{4/3}
$$
By the way, the $b_k$ also has an expression:
$$
b_k = \sum_{\substack{{f\in \mathcal{F}}\\{\Omega(f)-\omega(f)=k}}}\frac 1 f \prod_{p|f} \left(1+\frac 1 p \right)^{-1}
$$
where $\mathcal{F}$ is the set of all powerfull numbers ($p|f$ implies $p^2|f$).
Also, it satisfies 
$$
\sum_{\substack{{f\geq x}\\{f\in \mathcal{F}}\\{\Omega(f)-\omega(f)=k}}}\frac 1 f \prod_{p|f} \left(1+\frac 1 p \right)^{-1} \ll x^{-0.5} 0.75^k (\log x)^{4/3}
$$
We use the above for $2^{0.25\log\log x}$ because that is the lower bound of powerfull numbers when $k\geq 0.25\log\log x$. Then we obtain
$$
kb_k\ll k (\log x)^{-c} \ 0.75^k (\log\log\log x)^{4/3}$$
for some positive $c$. 
Thus, if we sum everything over $k\geq 1$, then we have
$$
\sum_{k\geq 1} \sum_{\substack{{n\leq x}\\{\Omega(n)-\omega(n)=k}}} \frac{k}{\omega(n)}
\leq B \frac{x}{(\log x)^b}\sum_{k\geq 1} k 0.75^k + B \frac{x}{\log\log x} \sum_{k\geq 1} k b_k
$$
Since the power series has radius of convergence $2$, the sum $\sum_{k\geq 1} kb_k$ should converge.  Hence, we are done. 
