2016th edition

Lets divide the sequence into segments as follows:-
$\color{#3D99F6}{1} \ | \ \color{#3D99F6}{4},5 \ | \ \color{#3D99F6}{9},10,11 \ | \ \color{#3D99F6}{16},17,18,19 \cdots$ .

So, $1$ ,is the first term in the sequence.
$4$ is the third term.
$9$ is the sixth term.
$16$ is the tenth term.

So, $n^2$ is the ${\left(\left(\displaystyle \sum_{i = 1}^n i\right) - (n - 1)\right)}^{\text{th}}$ term of sequence.

So, let $x^2$ be the first square in the ${2016}^{\text{th}}$ segment.
Then $\displaystyle \sum_{i = 1}^x i = 2016\\ \dfrac{x(x + 1)}{2} = 2016\\ x^2 + x - 4032 = 0\\ x^2 + 64x - 63x - 4032= 0\\ (x + 64)(x - 63) = 0\\ x = -64 ; x = 63$ .

But, we neglect $x = -64$ , because $\displaystyle \sum_{i = 1}^{-64} i$ does not exist.
So, $x = 63$ .
So, the perfect square of the ${2016}^{\text{th}}$ segment is ${63}^2 = 3969$ .
So, the ${2016}^{\text{th}}$ term of the sequence is $3979 + 62 = \color{#3D99F6}{\boxed{4031}}$