Let $(A, +, *)$ be the commutative ring of arithmetic functions with Dirichlet convolution as the multiplicative operation *.
The element $$\sigma(n)=\prod_i \frac{p_i^{k_i+1}-1}{p_i-1}, \text { where } n=\prod_i p_i^{k_i}$$ has a multiplicative inverse because $\sigma(1)=1$. Using the recursive definition (Apostol, Introduction to Analytic Number Theory, Theorem 2.8 ) of the multiplicative inverse of an arithmetic function I found the following definition for $\sigma^{-1} $ :
$$\sigma^{-1}(n)=\prod_i f(p_i^{k_i}) \text { where }
f(p_i^{k_i}) = \begin{cases} -p_i-1 &\mbox{if } k_i = 1 \\
p_i & \mbox{if } k_i = 2 \\
0 & \mbox{if } k_i > 2. \end{cases}$$
Question: is there a shorter (computable) definition of $\sigma^{-1}$ than the one I gave above, preferably without the if cases and multiplicative notation as the one Apostol gave for the totient function. $$\varphi^{-1}=\prod_i (1-p_i).$$
Best Answer
Your question still doesn't make much sense. The inverse is unique, and you want a multiplicative definition so you are really asking to describe $\sigma^{-1}$ at each prime power. So I don't see how this has anything to do with number theory: you're just asking if there is a simpler way to write the function $f(p^k)$ even though you already know that $f(p^k)$ is exactly equal to $$\begin{cases} -p-1, &\mbox{if } k = 1; \\ p, & \mbox{if } k = 2; \\ 0, & \mbox{if } k > 2. \end{cases}$$
Within these constraints there is little you can expect to do besides trivial rearrangements such as $f(p^k) = \mu(p^k) - p \cdot\mu(p^{k-1})$ (only for $k \ge 1$). But this is morally equivalent to hiding if statements into indicator functions. I would say that your sense that it "does not look right" is misguided: the function is what it is.