Functional equation: $f(x)=\frac x{\prod_{i|x}^{x-1} f(i)}$

Question

Find an algebraic function $f:\Bbb N\to\Bbb N$ such that

$$f(x)=\frac x{\prod_{i|x}^{x-1} f(i)}$$

and

$$f(1)=1$$

for all $x\in\Bbb N$

I allready know two things:

I'm looking for a specific function, not just an algorithm.

$$f(x) = x\left(\prod_{i | x, i < x} f(i)\right)^{-1} = x f(x) \left(\prod_{i | x} f(i)\right)^{-1}$$ — reuns, Mar 16 '16 at 16:15
you get that if $p$ is prime : $f(p) = p$ hence $f(p^2) = p, f(p^k) = p, f(p q) = \ldots$ ? — reuns, Mar 16 '16 at 16:24
The stuff you "already know" entirely defines the function. @OskarVikhamar-Sandberg — Thomas Andrews, Mar 16 '16 at 16:41

Thomas Andrews · Answer 1 · 2016-03-16T16:39:33.687

You have $$n = \prod_{d\mid n} f(d)$$ so you can use the product version of Möbius inversion to get:

$$f(n)=\prod_{d\mid n} d^{\mu(n/d)}$$

Not sure if that helps, nor if there isn't some other, simpler closed form.

For $n=p^k$ with $p$ prime, we see this is $f(p^k)=\frac{p^k}{p^{k-1}}=p$.

For $n=p^aq^b$, this is $$\frac{(p^aq^b)(p^{a-1}q^{b-1})}{(p^{a-1}q^b)(p^aq^{b-1})}=1$$

I'm guessing it is $f(p^k)=p$ and $f(n)=1$ for $n$ not a prime power.

It certainly seems to work...

1 Answers1