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The answer to the first question is: yes. The function $f(x)$ jumps $0$ or $1$ at every integer (because $0\leq\alpha\leq 1$), and also $f(0)=0$, hence $A_1(n)=f(n+1)$. Therefore $$\frac{A_1(n)}{n}=\frac{[(n+1)\alpha]}{n}=\frac{n\alpha+O(1)}{n}=\alpha+O\left(\frac{1}{n}\right)=\alpha+o(1). $$

Regarding the second question: the limit exists and equals $1-N\alpha$. Indeed, $f(k+N)=[k\alpha+N\alpha]$, hence $f(k)=f(k+N)$ means that the fractional part of $k\alpha$ is less than $1-N\alpha$. Here we used that $0<N\alpha<1$. As $\alpha$ is irrational, the fractional parts of $k\alpha$ are equidistributed in $(0,1)$ by Weyl's theorem, hence the density of $k$'s with $f(k)=f(k+N)$ equals the length of the interval $(0,1-N\alpha)$ which is $1-N\alpha$.

More generally, if $1\leq M\leq N$ is fixed, then the density of $k$'s with $f(k)=f(k+M)$ equals $1-M\alpha$. In particular, the case $M=1$ yields an alternate proof for the first paragraph, since $1-(1-\alpha)=\alpha$.

The answer to the first question is: yes. The function $f(x)$ jumps $0$ or $1$ at every integer (because $0\leq\alpha\leq 1$), and also $f(0)=0$, hence $A_1(n)=f(n+1)$. Therefore $$\frac{A_1(n)}{n}=\frac{[(n+1)\alpha]}{n}=\frac{n\alpha+O(1)}{n}=\alpha+O\left(\frac{1}{n}\right)=\alpha+o(1). $$

Regarding the second question: the limit exists and equals $1-N\alpha$. Indeed, $f(k+N)=[k\alpha+N\alpha]$, hence $f(k)=f(k+N)$ means that the fractional part of $k\alpha$ is less than $1-N\alpha$. Here we used that $0<N\alpha<1$. As $\alpha$ is irrational, the fractional parts of $k\alpha$ are equidistributed in $(0,1)$ by Weyl's theorem, hence the density of $k$'s with $f(k)=f(k+N)$ equals the length of the interval $(0,1-N\alpha)$ which is $1-N\alpha$.

More generally, if $1\leq M\leq N$ is fixed, then the density of $k$'s with $f(k)=f(k+M)$ equals $1-M\alpha$. In particular, the case $M=1$ yields an alternative proof for the first paragraph, since $1-(1-\alpha)=\alpha$.

The answer to the first question is: yes. The function $f(x)$ jumps $0$ or $1$ at every integer (because $0\leq\alpha\leq 1$), and also $f(0)=0$, hence $A_1(n)=f(n+1)$. Therefore $$\frac{A_1(n)}{n}=\frac{[(n+1)\alpha]}{n}=\frac{n\alpha+O(1)}{n}=\alpha+O\left(\frac{1}{n}\right)=\alpha+o(1). $$

Regarding the second question: the limit exists and equals $1-N\alpha$. Indeed, $f(k+N)=[k\alpha+N\alpha]$, hence $f(k)=f(k+N)$ means that the fractional part of $k\alpha$ is less than $1-N\alpha$. Here we used that $0<N\alpha<1$. As $\alpha$ is irrational, the fractional parts of $k\alpha$ are equidistributed in $(0,1)$ by Weyl's theorem, hence the density of $k$'s with $f(k)=f(k+N)$ equals the length of the interval $(0,1-N\alpha)$ which is $1-N\alpha$.

The answer to the first question is: yes. The function $f(x)$ jumps $0$ or $1$ at every integer (because $0\leq\alpha\leq 1$), and also $f(0)=0$, hence $A_1(n)=f(n+1)$. Therefore $$\frac{A_1(n)}{n}=\frac{[(n+1)\alpha]}{n}=\frac{n\alpha+O(1)}{n}=\alpha+O\left(\frac{1}{n}\right)=\alpha+o(1). $$

Regarding the second question: the limit exists and equals $1-N\alpha$. Indeed, $f(k+N)=[k\alpha+N\alpha]$, hence $f(k)=f(k+N)$ means that the fractional part of $k\alpha$ is less than $1-N\alpha$. Here we used that $0<N\alpha<1$. As $\alpha$ is irrational, the fractional parts of $k\alpha$ are equidistributed in $(0,1)$ by Weyl's theorem, hence the density of $k$'s with $f(k)=f(k+N)$ equals the length of the interval $(0,1-N\alpha)$ which is $1-N\alpha$.

More generally, if $1\leq M\leq N$ is fixed, then the density of $k$'s with $f(k)=f(k+M)$ equals $1-M\alpha$. In particular, the case $M=1$ yields an alternate proof for the first paragraph, since $1-(1-\alpha)=\alpha$.

The answer to the first question is yes. The function $f(x)$ jumps $0$ or $1$ at every integer (because $0\leq\alpha\leq 1$), and also $f(0)=0$, hence $A_1(n)=f(n+1)$. Therefore $$\frac{A_1(n)}{n}=\frac{[(n+1)\alpha]}{n}=\frac{n\alpha+O(1)}{n}=\alpha+O\left(\frac{1}{n}\right)=\alpha+o(1). $$

For your second question: the limit exists and equals $1-N\alpha$. Indeed, $f(k+N)=[k\alpha+N\alpha]$, hence $f(k)=f(k+N)$ means that the fractional part of $k\alpha$ is less than $1-N\alpha$. Here we used that $0<N\alpha<1$. As $\alpha$ is irrational, the fractional parts of $k\alpha$ are equidistributed in $(0,1)$ by Weyl's theorem, hence the density of $k$'s with $f(k)=f(k+N)$ equals the length of the interval $(0,1-N\alpha)$ which is $1-N\alpha$.

The answer to the first question is: yes. The function $f(x)$ jumps $0$ or $1$ at every integer (because $0\leq\alpha\leq 1$), and also $f(0)=0$, hence $A_1(n)=f(n+1)$. Therefore $$\frac{A_1(n)}{n}=\frac{[(n+1)\alpha]}{n}=\frac{n\alpha+O(1)}{n}=\alpha+O\left(\frac{1}{n}\right)=\alpha+o(1). $$

Regarding the second question: the limit exists and equals $1-N\alpha$. Indeed, $f(k+N)=[k\alpha+N\alpha]$, hence $f(k)=f(k+N)$ means that the fractional part of $k\alpha$ is less than $1-N\alpha$. Here we used that $0<N\alpha<1$. As $\alpha$ is irrational, the fractional parts of $k\alpha$ are equidistributed in $(0,1)$ by Weyl's theorem, hence the density of $k$'s with $f(k)=f(k+N)$ equals the length of the interval $(0,1-N\alpha)$ which is $1-N\alpha$.