The given expression equals to $\left\lceil x-\frac12\right\rceil+\big[(2x+1)/4\text{ is integer}\big]$. This is the nearest integer to $x$, if $\{x\}\neq\frac12$; otherwise it is the nearest even integer to $x$.
If $n$ is an integer in the interval, $\alpha\lt n\lt\beta$$\iff$$\lfloor\alpha\rfloor\lt n\lt\lceil\beta\rceil$. The number of such $n$’s is $(\beta-\alpha-1)$, which could be negative if $\alpha,\beta$ are integers and $\alpha=\beta$.
$$\begin{aligned}\left\lceil\frac nm\right\rceil&=\left\lfloor\frac{n+m-1}m\right\rfloor\\\left\lfloor\frac nm\right\rfloor+\left\lceil\frac{n\,{\rm mod}\,m}m\right\rceil&=\left\lfloor\frac nm\right\rfloor+\left\lfloor\frac{(m-1)+n\,{\rm mod}\,m}m\right\rfloor.\end{aligned}$$
Both sides are equal to $\left\lfloor\frac nm\right\rfloor+[n\,{\rm mod}\,m>0]$.
Use the definition of $N(\alpha,n)$, by $(3.14)$. We also have a useful property that $\lceil x\rceil-\lfloor x\rfloor=1$, if $x$ is irrational. Then
$$\begin{aligned}N(\alpha,n)+N(\beta,n)&=\left\lceil\frac{n+1}\alpha\right\rceil-1+\left\lceil\frac{n+1}\beta\right\rceil-1\\&=\left\lfloor\frac{n+1}\alpha\right\rfloor+\left\lfloor\frac{n+1}\beta\right\rfloor\\&=\frac{n+1}\alpha-\left\{\frac{n+1}\alpha\right\}+\frac{n+1}\beta-\left\{\frac{n+1}\beta\right\}\\&=n+1-\left\{\frac{n+1}\alpha\right\}-\left\{\frac{n+1}\beta\right\}.\end{aligned}$$
Since $\alpha,\beta$ are irrational, the last two terms are both non-zero; and they must add up to one. Finally
$$N(\alpha,n)+N(\beta,n)=n.$$
$$\begin{aligned}(x\,{\rm mod}\,ny)\,{\rm mod}\,y&=x-ny\left\lfloor\frac x{ny}\right\rfloor-y\left\lfloor\frac{x-ny\left\lfloor\frac x{ny}\right\rfloor}y\right\rfloor\\&=x-y\Bigg\lfloor\frac xy-n\left\lfloor\frac x{ny}\right\rfloor+n\left\lfloor\frac x{ny}\right\rfloor\Bigg\rfloor\\&=x-y\left\lfloor\frac xy\right\rfloor=x\,{\rm mod}\,y.\end{aligned}$$
$$\lceil mx\rceil=\lceil x\rceil+\left\lceil x-\frac1m\right\rceil+\cdots+\left\lceil x-\frac{m-1}m\right\rceil.$$
According to the explanation on Page 93, and $\gcd(n,3)=1$ if integer $n$ is not an integer multiple of $3$, the sequence of numbers
$$0\,{\rm mod}\,3,\;n\,{\rm mod}\,3,\;2n\,{\rm mod}\,3$$
must hit $1,2,3$ exactly once, respectively. So we have
$$\begin{aligned}n\,{\rm mod}\,3=0&\implies a+b+c=0;\\n\,{\rm mod}\,3=1&\implies a+b\omega+c\omega^2=1;\\n\,{\rm mod}\,3=2&\implies a+b\omega^2+c\omega=2.\end{aligned}$$
This gives
$$\begin{aligned}a&=1,\\b&=\frac12\left(-1+\frac i{\sqrt3}\right)=\frac13(\omega-1),\\c&=\frac12\left(-1-\frac i{\sqrt3}\right)=\frac{-1}3(\omega+2).\end{aligned}$$
Hence, $n\,{\rm mod}\,3=1+\frac13\big((\omega-1)\omega^n-(\omega+2)\omega^{2n}\big)$.
$$\begin{aligned}\sum_{0\leq k\lt m}\left\lfloor x+\frac km\right\rfloor&=\sum_{j,k\geq0}\left[1\leq j\leq x+\frac km\right][k\lt m]\\&=\sum_{j,k\geq0}\big[1\leq j\leq\lceil x\rceil\big]\big[m(j-x)\leq k\lt m\big]\\&=\sum_k\Big[m\big(\lceil x\rceil-x\big)\leq k\lt m\Big]+\sum_{j,k}\big[1\leq j\lt\lceil x\rceil\big][0\leq k\lt m]\\&=\lceil m\rceil-\Big\lceil m\big(\lceil x\rceil-x\big)\Big\rceil+m\big(\lceil x\rceil-1\big)\\&=-\lceil-mx\rceil=\lfloor mx\rfloor.\end{aligned}$$
$$\begin{aligned}&\big\lfloor\log_b\lfloor x\rfloor\big\rfloor=m\\&\iff m\leq\log_b\lfloor x\rfloor\lt m+1\\&\iff b^m\leq\lfloor x\rfloor\lt b^{m+1};\\\\&\lfloor\log_bx\rfloor=m\\&\iff m\leq\log_bx\lt m+1\\&\iff b^m\leq x\lt b^{m+1}.\end{aligned}$$
If and only if $b^m,b^{m+1}$ are integers, we have $b^m\leq\lfloor x\rfloor\lt b^{m+1}$$\iff$$b^m\leq x\lt b^{m+1}$. Since $m\geq0$, this happens only when $b$ is an integer.
$$\begin{aligned}\sum_kkx[\alpha\leq kx\leq\beta]&=x\sum_kk\left[\frac\alpha x\leq k\leq\frac\beta x\right]\\&=\frac12x\big(\left\lfloor\beta/x\right\rfloor+\left\lceil\alpha/x\right\rceil\big)\big(\left\lfloor\beta/x\right\rfloor-\left\lceil\alpha/x\right\rceil+1\big).\end{aligned}$$
(Note: ‘$\lg x$’ is binary logarithm, and ‘$\log x$’ is common logarithm.)
$$\begin{aligned}&\sum_{k,m}\big[10^k\leq2^m\lt2\cdot10^k\big][0\leq m\leq M]\\&\quad=\sum_{k,m}\big[\lg10^k\leq m\lt\lg2\cdot10^k\big][0\leq m\leq M]\\&\quad=\sum_{k,m}\Big[\left\lceil\lg10^k\right\rceil\leq m\lt\left\lceil\lg10^k\right\rceil+1\Big][0\leq m\leq M]\\&\quad=\sum_{k,m}\big[m=\lceil k\lg10\rceil\big][-1\lt m\leq M]\\&\quad=\sum_k\big[-\log2\lt k\leq M\log2\big]\\&\quad=1+\lfloor M\log2\rfloor.\end{aligned}$$
An integer $n$ can always be decomposed into $2^{k-1}q$, where $k,q$ are integers and $q$ is odd. Hence, only the $k$-th term in $S_{n-1}$ is one less than $S_n$. In other words, we have the recurrence
$$\begin{aligned}S_0&=0;\\S_n&=S_{n-1}+1,\quad\text{for $n\geq1$.}\end{aligned}$$
This implies $S_n=n$.
Likewise, only the $k$-th term in $T_{n-1}$ is $2^kq=2n$ less than $T_n$. We have the recurrence
$$\begin{aligned}T_0&=0;\\T_n&=T_{n-1}+2n,\quad\text{for $n\geq1$.}\end{aligned}$$
This implies $T_n=n(n+1)$.
The $n$-th element of the sequence equals to $m$, which gives
$$\begin{aligned}&\frac{m(m-1)}2\lt n\leq\frac{m(m+1)}2\\&\iff m^2-m\lt 2n\leq m^2+m\\&\iff m^2-m+\frac14\leq 2n\lt m^2+m+\frac14\\&\iff m-\frac12\leq\sqrt{2n}\lt m+\frac12\\&\iff m=\left\lfloor\sqrt{2n}+\frac12\right\rfloor.\end{aligned}$$
The number of times a non-negative integer occurs in ${\rm Spec}\big(\alpha/(\alpha+1)\big)$ is exactly one more than the number of times it occurs in ${\rm Spec}(\alpha)$. That’s because
$$\begin{aligned}&N\big(\alpha/(\alpha+1),n\big)-N(\alpha,n)\\&\quad=\left\lceil(\alpha+1)\frac{n+1}\alpha\right\rceil-1-\left\lceil\frac{n+1}\alpha\right\rceil+1\\&\quad=n+1.\end{aligned}$$
If we could find an $m$ such that $K_m\leq m$, we could violate the stated inequality. But the existence of such an $m=n^\prime+1$ requires
$$K_{\lfloor n^\prime/2\rfloor}\leq\lfloor n^\prime/2\rfloor\quad\text{or}\quad K_{\lfloor n^\prime/3\rfloor}\leq\lfloor n^\prime/3\rfloor.$$
This goes down further and further, implying that $K_0\leq0$, which contradicts $K_0=1$. Hence, there’s no such an $m$, and the inequality that $K_n>n$ stands.
$D_n^{(q)}$ is an auxiliary Josephus number that satisfies the recurrence
$$\begin{aligned}D_0^{(q)}&=1;\\D_n^{(q)}&=\left\lceil\frac q{q-1}D_{n-1}^{(q)}\right\rceil,\quad\text{for $n\geq0$.}\end{aligned}$$
Show that the auxiliary Josephus numbers satisfy
$$\left(\frac q{q-1}\right)^n\leq D_n^{(q)}\leq q\left(\frac q{q-1}\right)^n,\quad\text{for $n\geq0$.}$$
The left part is easy to prove. First, the basis $1\leq D_0^{(q)}$ stands when $n=0$; then
$$\begin{aligned}&\left(\frac q{q-1}\right)^{n-1}\leq D_{n-1}^{(q)}\\&\iff\left(\frac q{q-1}\right)^n\leq\frac q{q-1}D_{n-1}^{(q)}\\&\implies\left(\frac q{q-1}\right)^n\leq\left\lceil\frac q{q-1}D_{n-1}^{(q)}\right\rceil\\&\iff\left(\frac q{q-1}\right)^n\leq D_n^{(q)}.\end{aligned}$$
But the right part is a little tricky. Let’s try to prove it directly, we would have
$$\begin{aligned}&D_{n-1}^{(q)}\leq q\left(\frac q{q-1}\right)^n\\&\iff\frac q{q-1}D_{n-1}^{(q)}\leq q\left(\frac q{q-1}\right)^{n+1}\end{aligned}$$
but no more, because there’s no “space” left on the right-hand side for us to add a pair of ceiling on the left-hand side.
To proceed, we would have to subtract something from the right-hand side at the beginning. For example, we could subtract $q-1$ from the right-hand side and then prove the stronger hypothesis:
$$D_n^{(q)}\leq q\left(\frac q{q-1}\right)^n-q+1=(q-1)\left(\left(\frac q{q-1}\right)^{n+1}-1\right),\quad\text{for $n\geq0$.}$$
First, the basis $D_0^{(q)}\leq1$ stands when $n=0$; then
$$\begin{aligned}&D_{n-1}^{(q)}\leq(q-1)\left(\left(\frac q{q-1}\right)^n-1\right)\\&\iff\frac q{q-1}D_{n-1}^{(q)}\leq(q-1)\left(\left(\frac q{q-1}\right)^{n+1}-1\right)-1\\&\implies\left\lceil\frac q{q-1}D_{n-1}^{(q)}\right\rceil\lt(q-1)\left(\left(\frac q{q-1}\right)^{n+1}-1\right)\\&\iff D_n^{(q)}\lt(q-1)\left(\left(\frac q{q-1}\right)^{n+1}-1\right).\end{aligned}$$
Since the hypothesis is stronger than original, in other words, $(q-1)\left(\left(\frac q{q-1}\right)^{n+1}-1\right)\lt q\left(\frac q{q-1}\right)^n$, we have $D_n^{(q)}\leq q\left(\frac q{q-1}\right)^n$.
Any positive integer could be decomposed into $2^mb+a$, where integer $m\geq1$, positive integer $b$ is odd, and $a$ is $0$ or $1$. If $D_n^{(3)}=2^mb-a$, we can deduce that
$$\begin{aligned}&D_n^{(3)}=2^mb-a\\&\iff D_{n+m}^{(3)}=3^mb-a\\&\iff D_{n+m}^{(3)}=(3^mb-1)+(1-a)\\&\iff D_{n+m}^{(3)}=2^{m^\prime}b^\prime+(1-a).\end{aligned}$$
Then, we have a new version of the three original variables. Since $m^\prime\geq1$, we could push $D_n^{(3)}$ further and further with larger $n$’s. While we’re proceeding with $D_n^{(3)}$, $a$ alters between $1$ and $0$ every iteration; since $2^mb$ is even, this causes $D_n^{(3)}$ to change between odd and even every iteration. Thus, infinitely many of the numbers $D_n^{(3)}$ are even, and that infinitely many are odd.
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