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Let $a,b\in\mathbb{N}$ such that $(a,b)=1$. If $2\nmid a$, then one can show that $$\rho_{a/b}(n)=\sum_{k=1}^n (-1)^{\lfloor ak/b\rfloor}$$ 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 $$\rho_{a/b}(n+b)=\sum_{k=1}^{n+b}(-1)^{\lfloor ak/b\rfloor}=\rho_{a/b}(b)+\sum_{k=1}^n(-1)^{\lfloor a(k+b)/b\rfloor}=\rho_{a/b}(b)+(-1)^a\rho_{a/b}(n).$$ Since we assume that $2\nmid a$ 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 $T_0\mid 2b$. 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 $d\mid b$. 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 $a_1\equiv a_2\mod 2b$. Hence for fixed $b$, it suffices to analyze $a\in [-b,b]$. Notice that we have $$\left\lfloor\frac{-ak}{b}\right\rfloor=\begin{cases} -\left\lfloor\frac{ak}{b}\right\rfloor - 1 & b\nmid k \\ -\left\lfloor\frac{ak}{b}\right\rfloor & b\mid k \end{cases}.$$ Hence, $$\rho_{a/b}(n)+\rho_{-a/b}(n)=2\sum_{\substack{k=1 \\ b\mid k}}^n(-1)^{\left\lfloor\frac{ak}{b}\right\rfloor}=2\sum_{k=1}^{\lfloor n/b\rfloor}(-1)^k=\begin{cases} -2 & 2\nmid \lfloor n/b\rfloor \\ 0 & 2\mid \lfloor n/b\rfloor \end{cases}.$$ 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 $a\in [1,b]$.

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  • $\begingroup$ It’s interesting to note that $\rho_{(a+2b)/b}(x)=\rho_{a/b}(x)$ since $(-1)^{\lfloor (a+2b)k/b \rfloor}=(-1)^{\lfloor ak/b + 2k\rfloor}=(-1)^{\lfloor ak/b \rfloor}$, so we have another period of $2b$. Also, by comparing terms, we have $\rho_{a/b}(x)+\rho_{a/b+1}(x)=\rho_{2a/b}(\lfloor x/2 \rfloor)$, which may be of use as well. $\endgroup$ – Jacob Nov 19 '18 at 20:58
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A pretty solid estimation is as follows. By playing around with the fourier series of $\rho_{a/b}$ we find that for all $x\in \mathbb{N}$, $$\rho_{a/b}(x)=\frac{1}{2b}\sum_{n=1}^{2b}\rho_{a/b}(n)+\frac{1}{2b}\cdot\frac{1}{\pi}\sum_{n=1}^{2b}\psi^{(0)}\left(\frac{n}{2b}\right)\left[a_n\cos\left(\frac{n\pi x}{b}\right)-b_n\sin\left(\frac{n\pi x}{b}\right)\right]$$ where $$\begin{aligned} a_n &= \sum_{k=1}^{2b}(-1)^{\lfloor ak/b\rfloor}\sin\left(\frac{n\pi k}{b}\right) \\ b_n &= \sum_{k=1}^{2b} (-1)^{\lfloor ak/b\rfloor}\cos\left(\frac{n\pi k}{b}\right). \end{aligned}$$ 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.

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