Using Schramm's result (2008): $$\varphi(n)=\sum_{k=1}^n{\gcd(k,n)\cos\left(\frac{-2\pi k}{n}\right)} $$ Let the partial sum of this series be represented by:

$$\varphi(m,n)=\sum_{k=1}^m{\gcd(k,n)\cos\left(\frac{-2\pi k}{n}\right)}\qquad1\leq m\leq n$$

I came across this relationship:

$$\varphi(n)=\varphi(n,n)=\frac{n}{2}+n\left(\frac{n}{2}-\lfloor{\frac{n}{2}}\rfloor\right)+2\varphi\left( \lfloor\frac{n-1}{2}\rfloor,n\right)$$

Has anyone seen a recurrence like this before, or something similar? I wonder if this can be solved for $\varphi(n)$ without involving a gcd...

Edit: Oops this is not a recurrence, still, maybe there are more relations that can be combined with this?

By numerical evaluation of quite a few values of the totient and cumulative sums, I found that $\varphi(n)$ can be computed from partial summing of the series plus a function of $n$, but there seems to be 2 different patterns, depending on whether $n$ is an even or odd integer.

$$\varphi(n)=\varphi(n,n)=\frac{n}{2}+2\varphi\left(\frac{n}{2}-1, n\right)= \frac{n}{2}+2\sum_{k=1}^{\frac{n}{2}-1}\gcd(n,k) \cos\left(\frac{2\pi k}{n}\right)\qquad n \text{ even }$$ $$\varphi(n)=\varphi(n,n)=n+2\varphi\left(\frac{n-1}{2}, n\right)= n+2\sum_{k=1}^{\frac{n-1}{2}}\gcd(n,k) \cos\left(\frac{2\pi k}{n}\right)\qquad n \text{ odd }$$

I then combined these two relations into the one above using the floor function to cover both conditions.