Xrfgtjtk

Let $a,binmathbb{N}$ such that $(a,b)=1$. If $2nmid a$, then one can show that
$$rho_{a/b}(n)=sum_{k=1}^n (-1)^{lfloor ak/brfloor}$$
is periodic. Is there any explicit way to give the maximum of $|rho_{a/b}(n)|$ in terms of $a$ and $b$? As a first attempt we can show that
$$rho_{a/b}(n+b)=sum_{k=1}^{n+b}(-1)^{lfloor ak/brfloor}=rho_{a/b}(b)+sum_{k=1}^n(-1)^{lfloor a(k+b)/brfloor}=rho_{a/b}(b)+(-1)^arho_{a/b}(n).$$
Since we assume that $2nmid a$ we have that $rho_{a/b}(n+2b)=rho_{a/b}(n)$ meaning the period is $le 2b$. Since the function starts at $0$ and $rho_{a/b}(2b)=0$, combined with the fact that it moves in discrete steps, gives
$$max|rho_{a/b}(n)|le b.$$
However, I've noticed that the maximum doesn't seem to depend too reliably on just $b$. If you fix $b$ you can achieve a whole spectrum of bounds by adjusting $a$. Hence, I've wondering if we can improve this bound, or even better explicitly find the max.

Edit
Since $2b$ is a valid period, if $T_0$ is a fundamental period we know that $T_0mid 2b$. Moreover, we know that any period must be even due to $rho_{a/b}(T_0)=0$ so that there can be an equal number of negative terms as even terms. Thus $T_0=2d$ for some $dmid b$. We know that for whatever $T_0$ we find that
$$max |rho_{a/b}(n)|le T_0/2$$
however, the function rarely achieves this maximum of $T_0/2$.

Edit 2 Let $M(a,b)=max |rho_{a/b}(n)|$. As noted we have that $M(a_1,b)=M(a_2,b)$ if $a_1equiv a_2mod 2b$. Hence for fixed $b$, it suffices to analyze $ain [-b,b]$. Notice that we have
$$leftlfloorfrac{-ak}{b}rightrfloor=begin{cases}
-leftlfloorfrac{ak}{b}rightrfloor - 1 & bnmid k \
-leftlfloorfrac{ak}{b}rightrfloor & bmid k
end{cases}.$$
Hence,
$$rho_{a/b}(n)+rho_{-a/b}(n)=2sum_{substack{k=1 \ bmid k}}^n(-1)^{leftlfloorfrac{ak}{b}rightrfloor}=2sum_{k=1}^{lfloor n/brfloor}(-1)^k=begin{cases}
-2 & 2nmid lfloor n/brfloor \
0 & 2mid lfloor n/brfloor
end{cases}.$$
Thus
$$|M(a,b)-M(-a,b)|le 2.$$
Thus to properly analyze bounds on $M(a,b)$ it suffices to analyze $M(a,b)$ for $ain [1,b]$.

A pretty solid estimation is as follows. By playing around with the fourier series of $rho_{a/b}$ we find that for all $xin mathbb{N}$,
$$rho_{a/b}(x)=frac{1}{2b}sum_{n=1}^{2b}rho_{a/b}(n)+frac{1}{2b}cdotfrac{1}{pi}sum_{n=1}^{2b}psi^{(0)}left(frac{n}{2b}right)left[a_ncosleft(frac{npi x}{b}right)-b_nsinleft(frac{npi x}{b}right)right]$$
where
$$begin{aligned}
a_n &= sum_{k=1}^{2b}(-1)^{lfloor ak/brfloor}sinleft(frac{npi k}{b}right) \
b_n &= sum_{k=1}^{2b} (-1)^{lfloor ak/brfloor}cosleft(frac{npi k}{b}right).
end{aligned}$$
and $psi^{(0)}$ is the digamma function. Thus,
$$|rho_{a/b}|_{infty}le frac{1}{2b}left(left|sum_{n=1}^{2b}rho_{a/b}(n)right|+frac{1}{pi}sum_{n=1}^{2b}left|psi^{(0)}left(frac{n}{2b}right)right|left[|a_n|+|b_n|right]right)$$
simply using the triangle inequality and $sin$ and $cos$'s upper bounds.