Brute Force No More

Using Hermite's Identity ,

$\lfloor nx \rfloor = \lfloor x \rfloor + \left \lfloor x + \frac{1}{n} \right \rfloor + \left \lfloor x + \frac{2}{n} \right \rfloor + \dots + \left \lfloor x + \frac{n - 1}{n} \right \rfloor$ for $x \in \mathbb{R}$ and $n \in \mathbb{N}$

$\lceil \lfloor \pi \rfloor + \lfloor \pi + \frac{1}{2016}\rfloor + \lfloor \pi + \frac{2}{2016}\rfloor + ... + \lfloor \pi + \frac{2015}{2016}\rfloor \rceil + 1$

$= \lceil \lfloor 2016\pi \rfloor \rceil + 1$

$= \lceil 6333 \rceil + 1$

$= \boxed{6334}$

First post so not sure if this is up to par.

I noted that $\lfloor \pi + \frac{x}{2016}\rfloor$ would be equal to 3 for all $x \leq 1730$ [which was found by the equation $x = 2016 * (4 - \pi)$ .

So that means within the ceiling function, there are 1731 terms (the first $\pi$ , then the 1730 floors after that) that equal 3's after the floors are applied. $3 * 1731 = 5193$

For $x > 1730$ , there are $2015-1730 = 285$ terms that equal 4's after the floors are applied. $4 * 285 = 1140$

Added together, the ceiling function is 6333. Add the 1, 6334

EDIT: counting error

For 0<=n<=1730 , $\lfloor(\pi + \frac{n}{2016})\rfloor=3$ and 1731<=n<=2015 , $\lfloor(\pi + \frac{n}{2016})\rfloor=4$

therefore answer is , $(1730-(0-1))\times3+(2015-(1731-1))\times4+1=1731\times3+285\times4+1=\boxed{6334}$