Find the 2038th non-perfect cube.

**
Clarification:
**
Non-perfect cubes are positive integers which cannot be expressed as
$k^3$
, where
$k$
is a positive integer. For example: 2, 3, 4, 5, 6, 7, 9,... and 2 is the first non-perfect cube.

You may want to try 1973rd non-perfect square .

The answer is 2050.

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Let $t(n)$ be the $n$ th non-perfect integer. Then the sequence of non-perfect integers up to $t(n)=k^3 +1$ , where $k^3$ is the $k$ th perfect cube is as follows:

$\begin{aligned} \{ t(n) \} & = \underbrace{ {\color{#D61F06}\cancel 1}, 2, 3, 4, 5, 6, 7, {\color{#D61F06}\cancel 8}, 9, 10, \cdots {\color{#D61F06}\cancel {k^3}}, k^3+1}_{k^3+1-k \text{ terms}} \\ \implies n & = k^3 - k + 1 \\ t(k^3-k+1) & = k^3 + 1 \end{aligned}$

Now the largest perfect cube smaller than 2038 is $\lfloor \sqrt[3] {2038}\rfloor =12$ . Then, we have:

$\begin{aligned} t(k^3-k+1) & = k^3+1 & \small \color{#3D99F6} \text {Putting }k=12 \\ t(12^3-12+1) & = 12^3+1 \\ t(1717) & = 1729 \\ t(1717+321) & = 1717+321 \\ t(2038) & = \boxed{2050} \end{aligned}$

Inspired by Chew Seong-Cheong's solution to 1973rd non-perfect square