Let f be the arithmetic function defined by $f(n)$ = $3^{w(n)}$, where $w(n)$ is the number of distinct prime factors of n. Let $f^{-1}$ be the inverse of f with respect to the convolution product. Find $f^{-1}(n)$ for n = 1, 2, 3, 4, 5, 6.
Recursion formula:
$\dfrac{-1}{f(1)}\sum_{d|n} f(d)f^{-1}(\frac{n}{d})$
I think the recursion formulas you require are $~f^{-1}(1)=\frac{1}{f(1)}$ and for $n >1$, $\displaystyle f^{-1}(n)=\frac{-1}{f(1)}\sum_{\substack {d|n \\d<n}}f\left(\frac{n}{d}\right)f^{-1}(d)$. In the sum for $n=6$, for example, $~d=1,2,3$ and you would have already calculated the inverse values for $n=1,2,3$.
You could also use
$\displaystyle f^{-1}(n)=\frac{-1}{f(1)}\sum_{\substack {d|n \\d>1}}f^{-1}\left(\frac{n}{d}\right)f(d)$.