Why does the limit behavior of this function take over at 35?

Question

I've been working with this function on an semi-related question:

$$f(N)=\left\lfloor \frac{10N}{\lceil \frac{3}{4} N \rceil} \right\rfloor$$

It's clear that $10\leq f(N) \leq 13$, and that $\displaystyle\lim_{n\rightarrow \infty} f(N)=13$.

It's also true that $f(N)=13\,, \forall N\geq 36$.

Why is $35$ the last integer at which $f$ deviates from $13$?

Should be something simple, but I got stuck at

$$\frac{10}{13}N < \left\lceil \frac{3}{4}N \right\rceil$$.

Answer 1

Suppose $N=4n$. Then $f(4n)=\left\lfloor\frac{40n}{3n}\right\rfloor=\left\lfloor\frac{40}{3}\right\rfloor=13+\left\lfloor\frac23\right\rfloor=13.$

Suppose $N=4n+1.$ Then $f(4n+1)=\left\lfloor\frac{40n+10}{3n+1}\right\rfloor=\left\lfloor\frac{40n+10}{3n+1}\right\rfloor=13+\left\lfloor\frac{n-3}{3n+1}\right\rfloor$ which is 13 for $n\ge3.$

Check the last two cases and you should be able to solve the problem.

Answer 2

Let $N = 4n+k$ where $0 \le k \le 3$. I will show that $f(N) = 13+\left\lfloor \frac{1-\frac{3k}{n}}{3+\frac{k}{n}} \right\rfloor $. Therefore, if $n \ge 9$, $f(N) = 13$.

Since $\lceil \frac{a}{b} \rceil = \lfloor \frac{a+b-1}{b} \rfloor $ if $a$ and $b$ are integers with $a \ge 0$ and $b \ge 1$,

$\begin{array}\\ f(N) &=\left\lfloor \frac{10N}{\lceil \frac{3}{4} N \rceil} \right\rfloor\\ &=\left\lfloor \frac{10(4n+k)}{\lceil \frac{3(4n+k)}{4} \rceil} \right\rfloor\\ &=\left\lfloor \frac{40n+10k}{\lceil \frac{12n+3k}{4} \rceil} \right\rfloor\\ &=\left\lfloor \frac{40n+10k}{\lceil 3n+\frac{3k}{4} \rceil} \right\rfloor\\ &=\left\lfloor \frac{40n+10k}{3n+\lceil \frac{3k}{4} \rceil} \right\rfloor\\ &=\left\lfloor \frac{40n+10k}{3n+\lfloor \frac{3k+3}{4} \rfloor} \right\rfloor\\ &=\left\lfloor \frac{40+\frac{10k}{n}}{3+\frac1{n}\lfloor \frac{3k+3}{4} \rfloor} \right\rfloor\\ &=\left\lfloor 13+\frac{40+\frac{10k}{n}-13(3+\frac1{n}\lfloor \frac{3k+3}{4} \rfloor)}{3+\frac1{n}\lfloor \frac{3k+3}{4} \rfloor} \right\rfloor\\ &=13+\left\lfloor \frac{1+\frac{10k}{n}-13(\frac1{n}\lfloor \frac{3k+3}{4} \rfloor)}{3+\frac1{n}\lfloor \frac{3k+3}{4} \rfloor} \right\rfloor\\ &=13+\left\lfloor \frac{1+\frac1{n}(10k-13(\lfloor \frac{3k+3}{4} \rfloor)}{3+\frac1{n}\lfloor \frac{3k+3}{4} \rfloor} \right\rfloor\\ \end{array} $

For $k=0, 1, 2, 3$, $\lfloor \frac{3k+3}{4} \rfloor) =0, 1, 2, 3 =k $ so $10k-13(\lfloor \frac{3k+3}{4} \rfloor) =10k-13k =-3k $.

Therefore

$\begin{array}\\ f(N) &=13+\left\lfloor \frac{1+\frac1{n}(10k-13(\lfloor \frac{3k+3}{4} \rfloor)}{3+\frac1{n}\lfloor \frac{3k+3}{4} \rfloor} \right\rfloor\\ &=13+\left\lfloor \frac{1+\frac1{n}(-3k)}{3+\frac{k}{n}} \right\rfloor\\ &=13+\left\lfloor \frac{1-\frac{3k}{n}}{3+\frac{k}{n}} \right\rfloor\\ \end{array} $