6
$\begingroup$

Same question as in title:

What is sum of natural numbers that are coprime to $n$ and are $ \lt n$ ?

I know how to count number of them using Euler's function, but how to calculate sum?

$\endgroup$
5
$\begingroup$

Assume $n>2$. Then, if $n/2$ is an integer, then $n/2$ is certainly not a totative. Now it's easy to see that if $k$ is a totative, then $n-k$ is also a totative. So we can split $\phi(n)$ totatives into $\phi(n)/2$ pairs $\{k,n-k\}$, each containing two distinct elements (because $n/2$ isn't a totative) which sum to $n$. So sum of all totatives is $n\cdot\phi(n)/2=\frac{n\phi(n)}{2}$

$\endgroup$
5
$\begingroup$

Let's call this function $f$. Then $$f(n) = \sum_{i = 1}^{n - 1} \delta_{1}^{\gcd(i, n)} i = \frac{n \phi(n)}{2},$$ where $\delta$ is the Kronecker delta function and $\phi$ is Euler's totient function. Clearly if $n$ is prime, then $f(n) = T_{n - 1}$, where $T_n$ is the $n$th triangular number.

Work a few examples. I'll do two for you: $f(6) = 1 + 5 = \frac{6 \times 2}{2} = 2$ and $f(8) = 1 + 3 + 5 + 7 = \frac{8 \times 4}{2} = 16$.

Now, I didn't figure this out on my own. The answer comes from here: Sloane's OEIS A023896.

As for why I like the Kronecker delta function, that's because I'm a demon.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.