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?

• The efficient calculation is based on the formula for $d(n):=r(n)-r(n-1)$ , which is $$d(n)=2n-1-\sigma(n)$$ for even $n$ and $$d(n)=\frac{3n-1}{2}-\sigma(n)$$ for odd $n$ and the fact that $\sigma(n)$ is odd if and only if $n=k^2$ or $n=2k^2$ for some positive integer $k$. – Peter Mar 19 at 21:50
• $r(n)$ and $r(n-1)$ have the same parity if and only if $n$ is of the form $k^2,2k^2$ or $4k+3$ for some positive integer $k$ (in the case $4k+3$ , $k=0$ is also possible). – Peter Mar 19 at 21:54
• If my calculation is correct, we have $z(31988856^2)=80$ and the function does not really feel like it would be bounded from above. – Peter Košinár Mar 29 at 2:02
• @PeterKošinár thanks for the calculation. Was $z(n)$ ever negative in that range? – Mathphile Mar 29 at 4:48
• @Mathphile No, I have not found any case where $z(n)$ would be negative or zero. It has reached $1$ on multiple occasions, though. (again, assuming my computation was correct). – Peter Košinár Mar 29 at 16:43

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.

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

• It seems that $r(b)\equiv r(b+1)\pmod{2}\iff\sigma(b+1)\equiv1\pmod{2}$ is false. Take $b=6$. – mathlove Mar 23 at 15:24
• @mathlove I had spotted the same inconsistency. I have removed the error in my argument. I will try to complete this answer soon (likely tomorrow). – Servaes Mar 23 at 15:53