Density of outputs of divisor function We define the function:
$$\sigma_k{(n)}=\sum_{d \mid n} d^k$$
as the sum of divisors function of the $k$th power, where $k \in \mathbb{N_0}$, for $n \in \mathbb{N}$. Now, we define:
$$S_k=\{a \space | \space \exists \space m \in \mathbb{N} \space ; \sigma_k{(m)}=a\}$$
It is clear that $S_0$ is the same as $\mathbb{N}$ since for any $n \in \mathbb{N}$, we have $\sigma_0{(2^{n-1})}=n$. However, it is clear than this is not true for $S_k$ where $k&gt;0$.


  What is the density of $S_k$ in $\mathbb{N}$ for $k \in \mathbb{N}$?


Edit: As pointed out by @ThomasAndrews in the comments below, $k&gt;1$ has zero density by bounding. Now, the problem is solved if anybody is able to find the density of the sum of divisors function $\sigma(x)$ ($1$st power).
The point is that $\sigma(n)$, as $n$ ranges, is usually very even (i.e. many powers of $2$ divide it).

Fix $M \ge 1$ large. Since $\sigma(n) \ge n$ for each $n$, $$\#\{1 \le m \le M : m = \sigma(n) \text{ for some } n\} \le \#\{\sigma(n) : 1 \le n \le M\} \cap [M].$$ Up to a set of density 0, every positive integer $n$ has at least $\frac{1}{2}\log\log n$ prime divisors. Therefore, we expect there to be at least $\frac{1}{4}\log\log n$ prime divisors of $n$ with highest exponent dividing $n$ being odd. The point is that for such prime divisors, since $\sigma$ is multiplicative, they contribute a power of $2$ to $\sigma(n)$, so we expect $\frac{1}{2}\log\log n$ powers of $2$ dividing $\sigma(n)$, so $$\#\{\sigma(n) : 1 \le n \le M\} \cap [M] \le \#\{m \le M : 2^{\frac{1}{4}\log\log M} \mid m\}$$ has density $0$. I don't think it should be hard to make this more rigorous, but I'm watching the finals.
Below shows that $\sigma_k(n) &gt; n^k$ for $n&gt;1$ and $k&gt;0$. $\sigma_1(n) = \sum_{d|n}d = \sigma(n)$ is the very well known divisor function. Basic properties of $\sigma_k(n)$ and fundamental theorem of arithmetic will lead us to the result. 

$\textbf{Property 1:}$ If $p$ is a prime number and $q \in \mathbb{N}$, then

$\sigma_k(p^q) = 1+p^k+p^{2k}+...+p^{qk}$

Proof of above property is straight forward as any integer of the form $p^i$ for $i \in \mathbb{N}$ and $i \leq q $ will divide $p^q$ with a quotient of $p^{q-i}$

$\textbf{Property 2:}$ If $gcd(a,b)=1$ then $\sigma_k(ab) = \sigma_k(a)\sigma_k(b)$

Proof: Let $d_a,d_b$ be such that $d_a|a$ and $d_b|b$. As $gcd(a,b) =1$ we also have $gcd(d_a,d_b) =1$ and thus there's a bijection between the sets $(d_a,d_b)\rightarrow d_ad_b$. Denote $d_{ab} = d_ad_b$ and we also have $d_{ab}|ab$.

Now $$\sigma_k(a)\sigma_k(b) = (\sum_{d_a | a} d_a^k)(\sum_{d_b | b} d_b^k) = \sum_{d_a | a, \ d' | b} (d_ad_b)^k = \sum_{d_{ab} | ab} d_{ab}^k=\sigma_k(ab)$$

Look at a different approach for $k=1$ here: The sigma function (sum of divisors ) multiplicative proof

Let me know in the comments if the bijection is not obvious, I write some more on that if needed.

Now using the fundamental theorem of arithmetic, for any $n\in \mathbb{N}$, there exists $p_1,p_2,...,p_q,a_1,a_2,...,a_q$ where $p_1,p_2,...,p_q$ are prime numbers and $a_1,a_2,...,a_q$ are positive integers such that, 

$$n = p_1^{a_1}p_2^{a_2}...p_q^{a_q}$$

Then, $\sigma_k(n) = \sigma_k(p_1^{a_1}p_2^{a_2}...p_q^{a_q}) = \sigma_k(p_1^{a_1})\sigma_k(p_2^{a_2})...\sigma_k(p_q^{a_q})$

$=(1+p_1^k+p_1^{2k}+...+p_1^{a_1k})(1+p_2^k+p_2^{2k}+...+p_2^{a_2k})...(1+p_q^k+p_q^{2k}+...+p_q^{qk}) &gt; p_1^{a_1k}p_2^{a_2k}...p_q^{a_qk}$

$= (p_1^{a_1}p_2^{a_2}...p_q^{a_q})^k = n^k$

Note that the inequality holds true only if $k \neq 0$ and $n \neq 1$.

Thus the Natural Density (https://en.wikipedia.org/wiki/Natural_density) of $\sigma_k(n)$ is $0$ for $k&gt;0$