TITLE: Interesting property related to the sums of the remainders of integers
QUESTION [10 upvotes]: Cross-posted on Math Overflow too

Let us define,
$$r(b)=\sum_{k=1}^{\lfloor \frac{b-1}{2} \rfloor} (b \bmod{k})$$
After playing around with the $r(b)$ function for sometime I noticed that $r(b)$ appreared to be more even than odd. So to see the difference between the number of even and odd terms of $r(b)$, I defined a function,
$$z(x)=\sum_{n=1}^x(-1)^{r(n)}$$
When user Peter ran a program for computing values of $z(x)$ in PARI, I observed that for $x\le 10^{10}$, $z(x)\gt 0$. This suggests that there are always more even terms of $r(n)$ than odd terms for any $x$.
This leads to my two questions:

Is $z(x)$ always positive? If so, then how do we prove this?
Is $|z(x)|$ bounded by some maximum value? If so, then what is this maximum value? Till now the maximum value of $|z(x)|$ found was $49$ for $x = 5424027859$. I find it odd that $|z(x)|$ goes to these large values and then returns back to small values as small as $1$. 

Edit:
With the help of user Vepir I was able to plot $z(x)$ and noticed that the function has a sinusoidal-fractal-like appearance and it seems to grow without bounds, although very very slowly.


Is there any reason for this sinusoidal and fractal nature?

REPLY [5 votes]: I can't prove/disprove your conjectures, but I proved a claim which might be useful to prove/disprove your conjectures.
This answer proves the following claim : 
Claim : 
$$\small\begin{align}z(8m)&=(-1)^{c(8m)}+S(m)
\\\\z(8m+1)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+S(m)
\\\\z(8m+2)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+(-1)^{c(8m+2)}+S(m)
\\\\z(8m+3)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}+S(m)
\\\\z(8m+4)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m)
\\\\z(8m+5)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+(-1)^{c(8m+5)}+S(m)
\\\\z(8m+6)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m)
\\\\z(8m+7)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}-(-1)^{c(8m+7)}+S(m)\end{align}$$
where $$c(n)=\#\{x:\text{ $1\le x\le n$, and $x$ is either a square or twice a square}\}$$
and $$S(m):=\sum_{k=0}^{m-1}\bigg((-1)^{c(8k)}-(-1)^{c(8k+1)}+2(-1)^{c(8k+2)}-(-1)^{c(8k+4)}-(-1)^{c(8k+7)}\bigg)$$
According to oeis.org/A071860, one has
$$c(n)=\lfloor\sqrt{n}\rfloor+\left\lfloor\sqrt{\frac n2}\right\rfloor$$
which might make the problem easier to deal with.
Also, it follows from the above claim that
$$z(8m+4)=z(8m+6)$$
for every non-negative integer $m$.

The claim follows from the following lemmas :
Lemma 1 : $$r(n)=n^2-\frac{1}{2}\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n}\sigma(k)$$ 
Lemma 2 : $$r(n)\stackrel{\text{mod $2$}}\equiv\begin{cases}\displaystyle\sum_{k=1}^{n}\sigma(k)&\text{if $n\equiv 0,2,3,5\pmod 8$}\\\\1+\displaystyle\sum_{k=1}^{n}\sigma(k)&\text{if $n\equiv 1,4,6,7\pmod 8$}\end{cases}$$
Lemma 3 : $$\text{$\sigma(n)$ is odd $\iff$ $n$ is either a square or twice a square}$$
Lemma 4 : $$\sum_{k=1}^{n}\sigma(k)\equiv c(n)\pmod 2$$where $$c(n)=\#\{x:\text{ $1\le x\le n$, and $x$ is either a square or twice a square}\}$$
Lemma 5 : If $n\equiv 3\pmod 4$, then $c(n)=c(n-1)$.
Lemma 6 : $$z(x)=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{c(8k)}-\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{c(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{c(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{c(8k+2)}-\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{c(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{c(8k+7)}$$
Lemma 7 : 
$$\small\begin{align}z(8m)&=(-1)^{c(8m)}+S(m)
\\\\z(8m+1)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+S(m)
\\\\z(8m+2)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+(-1)^{c(8m+2)}+S(m)
\\\\z(8m+3)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}+S(m)
\\\\z(8m+4)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m)
\\\\z(8m+5)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+(-1)^{c(8m+5)}+S(m)
\\\\z(8m+6)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m)
\\\\z(8m+7)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}-(-1)^{c(8m+7)}+S(m)\end{align}$$
where $$S(m):=\sum_{k=0}^{m-1}\bigg((-1)^{c(8k)}-(-1)^{c(8k+1)}+2(-1)^{c(8k+2)}-(-1)^{c(8k+4)}-(-1)^{c(8k+7)}\bigg)$$

Lemma 1 : $$r(n)=n^2-\frac{1}{2}\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n}\sigma(k)$$
Proof : For $n=1$, it is true. In the following, let us use $$r(n+1)-r(n)=\begin{cases}2n+1-\sigma(n+1)&\text{if $n$ is odd}\\\\\frac{3n+2}{2}-\sigma(n+1)&\text{if $n$ is even}\end{cases}$$
Suppose that it is true for $n=2m+1$. Then, we get
$$\begin{align}&r(n+1)
\\\\&=r(n)+2n+1-\sigma(n+1)
\\\\&=n^2-\frac{1}{2}\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor +1\bigg)
-\sum_{k=1}^{n}\sigma(n)+2n+1-\sigma(n+1)
\\\\&=(2m+1)^2-\frac{m(m+1)}{2}+2(2m+1)+1-\sum_{k=1}^{n+1}\sigma(k)
\\\\&=(2m+2)^2-\frac{m(m+1)}{2}-\sum_{k=1}^{n+1}\sigma(k)
\\\\&=(2m+2)^2-\frac{1}{2}\bigg(\left\lfloor\frac{2m-1}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{2m-1}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n+1}\sigma(k)
\\\\&=(n+1)^2-\frac{1}{2}\bigg(\left\lfloor\frac{n+1-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n+1-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n+1}\sigma(k)\end{align}$$
Suppose that it is true for $n=2m$. Then, we get
$$\begin{align}&r(n+1)
\\\\&=r(n)+\frac{3n+2}{2}-\sigma(n+1)
\\\\&=n^2-\frac{1}{2}\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n-3}{2}\right\rfloor +1\bigg) -\sum_{k=1}^{n}\sigma(k)+\frac{3n+2}{2}-\sigma(n+1)
\\\\&=(2m)^2-\frac{m(m-1)}{2}+3m+1-\sum_{k=1}^{n+1}\sigma(k)
\\\\&=(2m+1)^2-\frac{(m+1)m}{2}-\sum_{k=1}^{n+1}\sigma(k)
\\\\&=(2m+1)^2-\frac{1}{2}\bigg(\left\lfloor\frac{2m+1-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{2m+1-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n+1}\sigma(k)
\\\\&=(n+1)^2-\frac{1}{2}\bigg(\left\lfloor\frac{n+1-3}{2}\right\rfloor+2\bigg)\bigg(\left\lfloor\frac{n+1-3}{2}\right\rfloor +1\bigg)-\sum_{k=1}^{n+1}\sigma(k)\end{align}$$
So, it is also true for $n+1$.$\quad\square$

Lemma 2 : $$r(n)\stackrel{\text{mod $2$}}\equiv\begin{cases}\displaystyle\sum_{k=1}^{n}\sigma(k)&\text{if $n\equiv 0,2,3,5\pmod 8$}\\\\1+\displaystyle\sum_{k=1}^{n}\sigma(k)&\text{if $n\equiv 1,4,6,7\pmod 8$}\end{cases}$$
Proof : It follows from Lemma 1 that$$r(8m)=64m^2-2m(4m-1)-\sum_{k=1}^{8m}\sigma(k)\equiv \sum_{k=1}^{8m}\sigma(k)\pmod 2$$ 
$$r(8m+1)=(8m+1)^2-2m(4m+1)-\sum_{k=1}^{8m+1}\sigma(k)\equiv 1+\sum_{k=1}^{8m+1}\sigma(k)\pmod 2$$
$$r(8m+2)=(8m+2)^2-2m(4m+1)-\sum_{k=1}^{8m+2}\sigma(k)\equiv \sum_{k=1}^{8m+2}\sigma(k)\pmod 2$$ 
$$r(8m+3)=(8m+3)^2-(2m+1)(4m+1)-\sum_{k=1}^{8m+3}\sigma(k)\equiv \sum_{k=1}^{8m+3}\sigma(k)\pmod 2$$ 
$$r(8m+4)=(8m+4)^2-(2m+1)(4m+1)-\sum_{k=1}^{8m+4}\sigma(k)\equiv 1+\sum_{k=1}^{8m+4}\sigma(k)\pmod 2$$
$$r(8m+5)=(8m+5)^2-(2m+1)(4m+3)-\sum_{k=1}^{8m+5}\sigma(k)\equiv \sum_{k=1}^{8m+5}\sigma(k)\pmod 2$$ 
$$r(8m+6)=(8m+6)^2-(2m+1)(4m+3)-\sum_{k=1}^{8m+6}\sigma(k)\equiv 1+\sum_{k=1}^{8m+6}\sigma(k)\pmod 2$$ 
$$r(8m+7)=(8m+7)^2-(2m+2)(4m+3)-\sum_{k=1}^{8m+7}\sigma(k)\equiv 1+\sum_{k=1}^{8m+7}\sigma(k)\pmod2$$
So, the claim follows.$\quad\square$

Lemma 3 : $$\text{$\sigma(n)$ is odd $\iff$ $n$ is either a square or twice a square}$$
Proof : See here or here.

Lemma 4 : $$\sum_{k=1}^{n}\sigma(k)\equiv c(n)\pmod 2$$where $$c(n)=\#\{x:\text{ $1\le x\le n$, and $x$ is either a square or twice a square}\}$$
Proof : It follows from Lemma 3 that 
$$\sum_{k=1}^{n}\sigma(k)=\underbrace{\sum_{k\in A}\sigma(k)}_{\text{sum of odd numbers}}+\underbrace{\sum_{k\not\in A}\sigma(k)}_{\text{sum of even numbers $=$ even}}\equiv \sum_{k\in A}\sigma(k)=c(n)\pmod 2$$
where $A=\{n\ :\ \text{$n$ is either a square or twice a square}\}$.

Lemma 5 : If $n\equiv 3\pmod 4$, then $c(n)=c(n-1)$.
Proof : Since we have
$$\text{(a square)}\equiv 0,1\pmod 4\qquad\text{and}\qquad \text{(twice a square)}\equiv 0,2\pmod 4$$
we see that if $n\equiv 3\pmod 4$, then $n$ is neither a square nor twice a square.

Lemma 6 : $$z(x)=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{c(8k)}-\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{c(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{c(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{c(8k+2)}-\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{c(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{c(8k+7)}$$
Proof : It follows from Lemma $1,2,3,4,5$ that $$z(x)=\sum_{k=1}^{x}(-1)^{r(k)}=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{r(8k)}+\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{r(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{r(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{r(8k+3)}+\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{r(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{r(8k+5)}+\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{r(8k+6)}+\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{r(8k+7)}$$
$$=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{c(8k)}-\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{c(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{c(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{c(8k+3)}-\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{c(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{c(8k+6)}-\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{c(8k+7)}$$
$$=\sum_{k=0}^{\lfloor x/8\rfloor}(-1)^{c(8k)}-\sum_{k=0}^{\lfloor (x-1)/8\rfloor}(-1)^{c(8k+1)}+\sum_{k=0}^{\lfloor (x-2)/8\rfloor}(-1)^{c(8k+2)}+\sum_{k=0}^{\lfloor (x-3)/8\rfloor}(-1)^{c(8k+2)}-\sum_{k=0}^{\lfloor (x-4)/8\rfloor}(-1)^{c(8k+4)}+\sum_{k=0}^{\lfloor (x-5)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-6)/8\rfloor}(-1)^{c(8k+5)}-\sum_{k=0}^{\lfloor (x-7)/8\rfloor}(-1)^{c(8k+7)}$$

Lemma 7 : 
$$\small\begin{align}z(8m)&=(-1)^{c(8m)}+S(m)
\\\\z(8m+1)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+S(m)
\\\\z(8m+2)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+(-1)^{c(8m+2)}+S(m)
\\\\z(8m+3)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}+S(m)
\\\\z(8m+4)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m)
\\\\z(8m+5)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+(-1)^{c(8m+5)}+S(m)
\\\\z(8m+6)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}+S(m)
\\\\z(8m+7)&=(-1)^{c(8m)}-(-1)^{c(8m+1)}+2(-1)^{c(8m+2)}-(-1)^{c(8m+4)}-(-1)^{c(8m+7)}+S(m)\end{align}$$
where $$S(m):=\sum_{k=0}^{m-1}\bigg((-1)^{c(8k)}-(-1)^{c(8k+1)}+2(-1)^{c(8k+2)}-(-1)^{c(8k+4)}-(-1)^{c(8k+7)}\bigg)$$
Proof : This immediately follows from Lemma 6.