Interesting property related to the sums of the remainders of integers

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:

1. Is $z(x)$ always positive? If so, then how do we prove this?
2. 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?

Solution 1:

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.

Solution 2:

Claim: For every positive integer $b$ we have $r(b)\equiv r(b-1)\pmod{2}$ if and only if either

1. $b\equiv3\pmod{4}$, or
2. $b=k^2$ for some integer $k$, or
3. $b=2k^2$ for some integer $k$.

The main ingredient in proving this claim is the following lemma:

Lemma: For every positive integer $b$ we have $$r(b)-r(b-1)=c(b)-\sigma(b),$$ where $\sigma(b)$ denotes the sum of all positive divisors of $b$, and $$c(b):=\begin{cases} 2b-1&\text{ if }\ b\equiv0\pmod{2},\\ \tfrac{3b-1}{2}&\text{ if }\ b\equiv1\pmod{2}\ \text{ and }\ b\neq3,\\ 4&\text{ if }\ b=3. \end{cases}$$

This reduces the question to a question on the parity of $\sigma(b)$.

Proof. For every pair of positive integers $b$ and $k$ there exist unique nonnegative integers $q(b,k)$ and $r(b,k)$ such that $r(b,k)<k$ and $$b=q(b,k)\cdot k+r(b,k).$$ This is simply dividing $b$ by $k$ with remainder $r(b,k)$. With this, your function $r$ can be expressed as $$r(b)=\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}r(b,k).$$ To find a more manageable form for $r(b)$, note that $q(b,k)=\lfloor\frac bk\rfloor$, so that \begin{eqnarray*} r(b)&=&\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}r(b,k) =\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}\Big(b-q(b,k)k\Big)=\big\lfloor\tfrac{b-1}{2}\big\rfloor b -\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}\big\lfloor\tfrac{b}{k}\big\rfloor k. \end{eqnarray*} Then the difference of two consecutive terms can be simplified. If $b$ is even: \begin{eqnarray*} r(b)-r(b-1)&=&\left(\big\lfloor\tfrac{b-1}{2}\big\rfloor b -\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}\big\lfloor\tfrac{b}{k}\big\rfloor k\right) -\left(\big\lfloor\tfrac{b-2}{2}\big\rfloor(b-1) -\sum_{k=1}^{\lfloor\frac{b-2}{2}\rfloor}\big\lfloor\tfrac{b-1}{k}\big\rfloor k\right)\\ &=&\left(\frac{b-2}{2}b -\sum_{k=1}^{\frac{b-2}{2}}\big\lfloor\tfrac{b}{k}\big\rfloor k\right) -\left(\left(\frac{b-2}{2}\right)(b-1) -\sum_{k=1}^{\frac{b-2}{2}}\big\lfloor\tfrac{b-1}{k}\big\rfloor k\right)\\ &=&\frac{b-2}{2}-\sum_{k=1}^{\frac{b-2}{2}}\left(\big\lfloor\tfrac{b}{k}\big\rfloor-\big\lfloor\tfrac{b-1}{k}\big\rfloor\right)k,\\ \end{eqnarray*} and similarly if $b$ is odd: \begin{eqnarray*} r(b)-r(b-1)&=&\left(\big\lfloor\tfrac{b-1}{2}\big\rfloor b -\sum_{k=1}^{\lfloor\frac{b-1}{2}\rfloor}\big\lfloor\tfrac{b}{k}\big\rfloor k\right) -\left(\big\lfloor\tfrac{b-2}{2}\big\rfloor(b-1) -\sum_{k=1}^{\lfloor\frac{b-2}{2}\rfloor}\big\lfloor\tfrac{b-1}{k}\big\rfloor k\right)\\ &=&\left(\frac{b-1}{2}b -\sum_{k=1}^{\frac{b-1}{2}}\big\lfloor\tfrac{b}{k}\big\rfloor k\right) -\left(\frac{b-3}{2}(b-1) -\sum_{k=1}^{\frac{b-3}{2}}\big\lfloor\tfrac{b-1}{k}\big\rfloor k\right)\\ &=&\frac{3}{2}(b-1)-\big\lfloor\tfrac{2b}{b-1}\big\rfloor\frac{b-1}{2}-\sum_{k=1}^{\frac{b-3}{2}}\left(\big\lfloor\tfrac{b}{k}\big\rfloor-\big\lfloor\tfrac{b-1}{k}\big\rfloor\right)k, \end{eqnarray*} where the extra term $-\big\lfloor\tfrac{2b}{b-1}\big\rfloor\frac{b-1}{2}$ appears because the summation in $r(b)$ has one more term than the summation in $r(b-1)$, with $k=\tfrac{b-1}{2}$. For odd $b>3$ this further simplifies to $$r(b)-r(b-1)=\frac{b-1}{2}-\sum_{k=1}^{\frac{b-3}{2}}\left(\big\lfloor\tfrac{b}{k}\big\rfloor-\big\lfloor\tfrac{b-1}{k}\big\rfloor\right)k.$$ Now these inner summations have more familiar closed forms; to see this, note that $$\big\lfloor\tfrac{b}{k}\big\rfloor-\big\lfloor\tfrac{b-1}{k}\big\rfloor=\begin{cases} 0&\text{ if }\ k\nmid b\\ 1&\text{ if }\ k\mid b\\ \end{cases}.$$ So effectively these summations sum precisely the divisors of $b$, up to $\lfloor\tfrac{b-2}{2}\rfloor$. A routine check shows that the only divisors not counted are $b$, and if $b$ is even also $\tfrac{b}{2}$, and if $b=3$ also $1$. Then we can further simplify the differences for even $b$ as \begin{eqnarray*} r(b)-r(b-1)&=&\frac{b}{2}-1-\Big(\sigma(b)-b-\tfrac{b}{2}\Big)\\ &=&2b-1-\sigma(b), \end{eqnarray*} and for odd $b>3$ as \begin{eqnarray*} r(b)-r(b-1)&=&\frac{b-1}{2}-\Big(\sigma(b)-b\Big)\\ &=&\frac{3b-1}{2}-\sigma(b), \end{eqnarray*} and the latter is easily verified to also hold for $b=3$.$\quad\square$

It is a well known fact (or nice exercise) to prove that for a positive integer $m$ with prime factorization $m=\prod_{i=1}^np_i^{a_i}$, where $p_1,\ldots,p_n$ are distinct prime numbers and $a_1,\ldots,a_n$ are positive integers, we have $$\sigma(m)=\prod_{i=1}^n\sum_{j=0}^{a_i}p_i^j.$$ In particular this shows that $\sigma(m)$ is odd if and only if for every odd prime $p_i$ dividing $m$ we have $a_i\equiv0\pmod{2}$, or equivalently either $m=k^2$ or $m=2k^2$ for some integer $k$. In particular we see that if $b\equiv3\pmod{4}$ then $\sigma(b)$ is even and hence $r(b)\equiv r(b-1)\pmod{2}$.

Here are some small values for $r(b)$: $$\begin{array}{r|ccccccccc} b&0&1&2&3&4&5&6&7\\ \hline r(b)&0&0&0&0&0&1&0&2\\ &&&&&&&&\\ b&8&9&10&11&12&13&14&15\\ \hline r(b)&2&2&3&7&2&7&10&8\\ &&&&&&&&\\ b&16&17&18&19&20&21&22&23\\ \hline r(b)&8&15&11&19&16&15&22&32\\ \end{array}$$