How To find the 1991-th number of this series

Question

$$ 2,3,5,6,7,10,11,... $$

When I remove the perfect cube numbers and square numbers I get this series What is the 1991^th term of this series

Thanks for all answers in advance.

Answer 1

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?

Answer 2

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 $2045$.