• Notice how $n^{th}$ term reacts with sequence reduction.
• If $n(A)$ denotes the number of element of a set $A$ then $n(A \cup B)=n(A)+n(B)-n(A \cap B)$

In this sequence,

$n$th term would be = $n+\lfloor \sqrt[2]{n} \rfloor+\lfloor \sqrt[3]{n} \rfloor - \lfloor \sqrt[6]{n} \rfloor+k$ where k is the number of square and cube from $n$ to $\lfloor \sqrt[2]{n} \rfloor+\lfloor \sqrt[3]{n} \rfloor - \lfloor \sqrt[6]{n} \rfloor$

Because, suppose there is a sequence $A=(a_1,a_2,a_3,...,a_n)$ and you remove $x$ terms and make it $B=(b_1,b_2,..b_{n-x})$

Now consider a term $a_i$ in $A$ which is also in $B$.

$a_i$ is $i^{th}$ position in $A$ and number of terms removed before $a_i$ is $y$

Now $i^{th}$ position in B would be $a_{i+y}$

Note that:

From $1$ to $n$ there are total $\lfloor \sqrt[2]{n} \rfloor+\lfloor \sqrt[3]{n} \rfloor-\lfloor \sqrt[6]{n}$

As there are $\lfloor \sqrt[2]{n} \rfloor$ square numbers,$\lfloor \sqrt[3]{n} \rfloor$ cubic and $\lfloor \sqrt[6]{n} \rfloor$ numbers both square and cubic from $1$ to $n$. (From set union theorem)

For this reason, as $1,2,3,...,n$ has been reduced by squares and cubic numbers,

$n^{th}$ term would be $n+\lfloor \sqrt[2]{n} \rfloor+\lfloor \sqrt[3]{n} \rfloor - \lfloor \sqrt[6]{n} \rfloor$

So, $2019^{th}$ term $=2019+\lfloor \sqrt[2]{2019} \rfloor+\lfloor \sqrt[3]{2019} \rfloor - \lfloor \sqrt[6]{2019} \rfloor+k$

$2019^{th}$ term $=2019+44+12-3+k=2072+k$

Here $k=1$($2025$ is a square)

Hence the $2019^{th}$ term would be $2073$.