Find the $2011^{th}$ term of the sequence $2,3,5,6,7,8,10,11,...$ 
Find the $2011^{th}$ term of the sequence $2,3,5,6,7,8,10,11,...$
  (a) $2056$
  (b) $2011$
  (c) $2013$
  (d) $2060$

My Approach:
Let $2,3,5,6,7,8$ be one set of numbers.
In every 6 terms we reach 8 terms forward. From 1 to 9, 6 terms are present but the jump is of 8.
$2011/6=335+1/6$.
Hence 335 such whole sets and 1st term of the 336th set would give us the 2011th term.
335 full sets starting from 1, each of a jump 8 will reach $2+(335)(8)=2682$. Now the first term of the next set is $2683$.   
But this doesn't match with any of the options given. Have I identified the sequence wrongly or is there a flaw in my logic?
 A: If $a_n$ is the given sequence then $a_n-n=$ the number of squares less than or equal to $a_n$ which equals $\left\lfloor \sqrt{a_n}\right\rfloor $
Therefore we need to solve for $m$
$$m-\left\lfloor \sqrt{m}\right\rfloor =2011$$
This leads to the inequalities
$$2010<m-\sqrt{m}\leq 2011$$
and so
$$2055.34<m\leq 2056.35$$
and 
$$m=2056$$
A: Hint: One way to continue this sequence is to eliminate all squares: $$1=1^2, \quad 4=2^2, \quad 9=3^2, \quad \dots,\quad 45^2=2025$$
A: Just skipping the squares leads us to 2056. So the answer to the question is 
(a) 2056
A: Compare the two sequences:
$$\color{red}1,2,3,\color{red}{2^2},5,6,7,8,\color{red}{3^2},10,...,\underbrace{\color{red}{44^2}}_{1936},1937,1938,...,2010,\color{blue}{2011},2012,...,\underbrace{\color{green}{45^2}}_{2025},...,\underbrace{\color{green}{46^2}}_{2116},...\\
2,3,5,6,7,8,10,...,2011,\underbrace{2012}_{\color{red}1},\underbrace{2013}_{\color{red}{2^2}},\underbrace{2014}_{\color{red}{3^2}},...,\underbrace{2055}_{\color{red}{44^2}},\underbrace{2056}_{\color{green}{45^2}}\\
\color{blue}{2011}+\color{green}{45}=2056.$$
Note: $2011<\color{green}{2025=45^2}<2011+\color{red}{44}<\color{green}{2116=46^2}$, therefore $2025=\color{green}{45^2}$ is skipped too.
