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I was asked to evaluate the sum $\sum_{k = 1}^n \frac{n}{\gcd(n,k)}$ in terms of the prime factorization of $n$. I know that there are $\phi(n)$ integers $k < n$ such that $\gcd(n,k) = 1$ so I tried writting $\sum_{k = 1}^n \frac{n}{\gcd(n,k)} = \phi(n)n + 1 + S$ where $S$ is the contribution from the terms where $k < n$ and $\gcd(n,k) > 1$.

I've also tried evaluating $\sum_{k = 1}^n \frac{\text{lcm}(n,k)}{k}$ but to no avail.

I would preferably like a hint as opposed to a full solution. Any help is appreciated.

lhf
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Jeevan Devaranjan
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1 Answers1

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If $G$ is the cyclic group of order $n$ and $G=\langle g\rangle $, then $ord(g^k)=\dfrac{n}{\gcd(n,k)}$.

Therefore, $$ \sum_{k = 1}^n \frac{n}{\gcd(n,k)} = \sum_{k = 1}^n ord(g^k) = \sum_{d\mid n} \sum_{x \in G, ord(x)=d} d = \sum_{d\mid n} d \phi(d) $$

For the last sum, see this question.

lhf
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