I'd recommend finding the formula first

If the first few terms are listed, a pattern can be noticed (besides the obvious one that each term n occurs n times):

${ a }_{ 1 }=1\\ { a }_{ 2 }=2\\ { a }_{ 3 }=2\\ { a }_{ 4 }=3\\ { a }_{ 5 }=3\\ { a }_{ 6 }=3\\ { a }_{ 7 }=4\\ { a }_{ 8 }=4\\ { a }_{ 9 }=4\\ { a }_{ 10 }=4\\$

Seeing how there seems to be a pattern where each term stops repeating (1,3,6,10,...). We can define a second sequence, ${ b }_{ n }$ , that tells us the nth term where n stops repeating; ${ b }_{ n }=\sum _{ k=1 }^{ n }{ k } =\frac { n(n+1) }{ 2 }$ . (For example, ${ b }_{ 3 }=6$ and ${ b }_{ 4 }=10$ ).

Now, if we were to take the inverse of ${ b }_{ n }$ , it would actually output instead what term n is in the midst of being repeated at the nth term, so given some manipulation:

$n=\frac { { a }_{ n }({ a }_{ n }+1) }{ 2 } \\ 2n={ a }_{ n }({ a }_{ n }+1)\\ \left\lfloor \sqrt { 2n } \right\rfloor =\left\lfloor { a }_{ n }\sqrt { 1+\frac { 1 }{ { a }_{ n } } } \right\rfloor \\ { a }_{ n }=\left\lfloor \sqrt { 2n } \right\rfloor$

(Upon checking a table for accuracy, it was off by .5, so then a translation was made to correct the error, although personally I'm not sure where the error came up.)

${ a }_{ n }=\left\lfloor \sqrt { 2n } +\frac { 1 }{ 2 } \right\rfloor \\ { a }_{ 2016 }=63$ .