Why does the limit behavior of this function take over at 35? 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
$$\frac{10}{13}N < \left\lceil \frac{3}{4}N \right\rceil$$.

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}
$
