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Let $n,m \in \mathbb{N}$ $$n=\prod_{i=1}^{r}p_{i}^{a_i}$$ where $p_i$ are prime factors and $f$ , $g$ and $h$ are the functions $$f(n,m)=\sum_{j=1}^{n}j^m$$ And $$g(n)=\sum_{i=1}^{r}a_i.p_i$$ If we put $m=1,n=21$ then $$g(f(21,1))=g(231)=21.$$

21 is only number satisfy $g(f(n,1))=n$.

proof for 21

Now let

$$h(m) = \sum_{g(f(n,m))=n}1$$

So $h(1)=1$.

Question

If $m$ have finite $n$ satisfied $g(f(n,m))=n$ then what is formula for $h(m)$?

Can we prove there are infinitely many $m$ satisfying above statement?

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  • $\begingroup$ What makes you suspect there is any kind of closed form for $h(m)$? $\endgroup$ – Servaes Jul 30 at 9:32
  • $\begingroup$ @servaes Actually I also don't think there is any close form for $h(m)$ but maybe more interesting is, if $h(m)$ have finite value then how many such $m$ exist in which pattern $\endgroup$ – Pruthviraj Jul 30 at 10:08
  • $\begingroup$ If $n$ and $n+1$ are square free $g({n(n+1)\over 2})$ is simply a sum of primes if $m$ is 1. $\endgroup$ – Roddy MacPhee Aug 5 at 15:09

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