Let $f \ge 1$ be a multiplicative arithmetic function and $F_f(n) = F(n)= \sum_{d|n}f(d)$. Define the entropy of $n$ with respect to $f$ to be

$H_f(n) = -\sum_{d|n} \frac{f(d)}{F(n)}\log(\frac{f(d)}{F(n)}) = \log(F(n)) - \frac{1}{F(n)}\sum_{d|n} f(d)\log(f(d))$

If $E_f(n) = \sum_{d|n} f(d)\log(f(d))$ then using Möbuis inversion one can show that from $E_f = E_g$ it follows that $f=g$. Similarily from $F_f = F_g$ it follows that $f=g$. ( As a byside: The function $E_f$ satisfies a Leibniz rule: $E_f(nm) = F_f(m)E_f(n)+E_f(m)F_f(n)$ for $m,n$ with $\gcd(m,n) = 1$.)

Now my question is: Does it from $H_f = H_g$ necessarily follow that $f=g$, or is there a counterexample?

  • $\begingroup$ $H_{f,g}(n) = \log(\sum_{d | n} g(d)) -\frac{\sum_{d | n} g(d) \log f(d)}{\sum_{d | n} g(d)}$ is injective in $f$. But $H_{f,f}$ is more complicated. Do you have a real problem behind those or is it like "anyone who can find something fun about it would be welcome" ? $\endgroup$
    – reuns
    Jul 28, 2017 at 14:26

1 Answer 1


This isn't an answer, but some thoughts that might be useful to someone later.

We can't proceed by Mobius inversion, but it's the same principle: you can determine $f(n)$ at a particular value, then exclude those terms later when solving for more complicated terms.

By definition, $f(1) = 1$. For a prime $p$, we have

$$H_f(p) = -\left(\frac{f(p)}{1+f(p)}\log\left(\frac{f(p)}{1+f(p)}\right) + \frac{1}{1+f(p)}\log\left(\frac{1}{1+f(p)}\right)\right)$$ which has derivative $$H_f'(p) = -\frac{\log(f(p))}{(1+f)^2}$$ which is never zero except singularly $f(p)=1$, so $H_f(p)$ is monotonic in $f(p)$. Thus we can recover $f(p)$ from $H_f(p)$.

But for higher powers as $p^k$, it goes $$H_f(p^k) = $$ $$-\Big(\frac{f(p^k)}{F(p^{k-1})\!+\!f(p^k)}\log\left(\frac{f(p^k)}{F(p^{k-1})\!+\!f(p^k)}\right)\!+$$ $$\sum_{d|p^{k-1}} \frac{f(d)}{F(p^{k-1})\!+\!f(p^k)}\log\left(\frac{f(d)}{F(p^{k-1})+f(p^k)}\right)\Big)$$ which has derivative $$\frac{dH_f(p^k)}{df(p^k)} = \frac{E_f(p^{k-1}) - F(p^{k-1})\log(f(p^k))}{\left(f(p^k) + F(p^{k-1})\right)^2}$$ ... which can be zero, so the entropy is no longer necessarily monotonic. But I suppose if the above for $H_f(p^k)$ can be shown to ever have only one integer solution for $f(p^k)$, that should suffice.

  • $\begingroup$ What do you mean by the last sentence? $\endgroup$
    – user6671
    Jul 28, 2017 at 11:42
  • $\begingroup$ I mean that (given $E_f(p^{k-1})$, $F(p^{k-1})$ and $H_f{(p^k)}$) there can be multiple real solutions for $f(p^k)$, but if you can show there's only ever one integer solution, then you're set. Also, I didn't say this, but if you can ever find a case where there are multiple solutions, then you can extend this to a complete function. $\endgroup$ Jul 28, 2017 at 12:07
  • $\begingroup$ Ok, Thanks for your answer and your comment! $\endgroup$
    – user6671
    Jul 28, 2017 at 12:08

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