Is it true this formula for $g_f(x)$ is independent of the value of $f\in\mathbb{Z}_{>0}$

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This question is related to analytic formulas for $\sigma_0(x)$ and it's first-order derivative $\sigma_0'(x)$ which are defined and illustrated in this answer I posted to a question on Math Overflow. In particular, $g_f(N)$ defined in formula (1) below represents the evaluation of $\sigma_0'(x)$ at $x=1$. I suspect the independence of $\sigma_0'(x)$ with respect to $f$ extends to all integer values of $x$, but hopefully an answer to this question for the specific case $x=1$ will lead to a proof (or disproof) of more general independence.


Consider formula (1) below for $g_f(N)$ where $f$ is assumed to be a positive integer.


$$g_f(N)=-\frac{(2\pi)}{f}\sum_{n=1}^N\frac{1}{n^2}\sum\limits_{k=1}^{f\ n} k \sin\left(\frac{2 \pi k}{n}\right)\tag{1}$$


I've noticed formula (1) for $g_f(N)$ above seems to evaluate exactly the same independent of the value of $f$, and $g_1(N)$ is approximated by the linear function $x$ which is illustrated in Figure (1) below.


Illustration of g_1(N)

Figure (1): Illustration of $g_1(N)$


Question: Is it true that $g_m(N)=g_1(N)$ for all positive integer values of $m$?