How To find the 1991-th number of this series $$ 2,3,5,6,7,10,11,... $$
When I remove the perfect cube numbers and square numbers I get this series 
What is the 1991th term of this series 
Thanks for all answers in advance.
 A: HINTS:
Start with the list $$1,2,3,\cdots,1991$$ Now in this sequence there are $44$ squares since $45^2>1991$ and $12$ cubes since $13^3>1991$. 
Also, there are $3$ numbers that are both squares and cubes, namely $1^6$, $2^6$ and $3^6$.
All this accounts to $53$ different numbers. 
Hence in your sequence there are only $1991-53$ numbers left.
Can you continue?
A: Given any positive integer $N$, 


*

*The number of perfect squares $\le N$ is $\lfloor N^{1/2} \rfloor$.

*The number of cubes $\le N$ is $\lfloor N^{1/3}\rfloor$.

*These two sets of squares and cubes can overlap. The overlaps are sixth power of some other numbers. The number of overlaps is $\lfloor N^{1/6}\rfloor$.


This means for numbers $\le N$, there are
$$\mathcal{N}(N) \stackrel{def}{=} N - \lfloor N^{1/2} \rfloor - \lfloor N^{1/3}\rfloor + \lfloor N^{1/6}\rfloor$$
numbers which are neither prefect squares or cubes. The $n^{th}$ entry of the sequence is the smallest $N$ such that $n = \mathcal{N}(N)$.
By direct computation, we have $\mathcal{N}(1991) = 1938$. This suggests the number we seek is around $1991 + (1991-1938) = 2044$. By direction computation again, we have $\mathcal{N}(2044) = 1990$ and $\mathcal{N}(2045) = 1991$. This means the $1991^{th}$ entry of the sequence is $2045$.
