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

• Please show your effort too. – Kavita Juneja Apr 24 '18 at 9:46
• I would say that you have to figure out how many perfect squares and cubes there are that are less than a certain number. Any ideas? – Matti P. Apr 24 '18 at 9:47
• i have a paper that i worked on how can i upload pictures here ? – user555654 Apr 24 '18 at 9:48
• Please do not use pictures of your post. Pictures may not be legible, cannot be searched and are not view-able to some, such as those who use screen readers. – GNUSupporter 8964民主女神 地下教會 Apr 24 '18 at 9:49
• @MattiP. i found out that it goes upto 44^2 and 12^3 and if more it goes above the 1991 th term – user555654 Apr 24 '18 at 9:50

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?

• thanks for the answer i might be able to continue on this – user555654 Apr 24 '18 at 9:51
• after I talked to my friends and they told me that they already have worked on this and they havent been able to reach the answer ps:(im a languages guy with interest in mathematics in an high school my friends doesnt know english that well ) – user555654 Apr 24 '18 at 9:59

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 $2045$.

• thank you so much this helped my class tremendously :D – user555654 Apr 25 '18 at 13:57