# 2017th edition

We have a peculiar series in which a lead number $$(say \space a_{1})$$ is written first and then the number of consecutive numbers $$(say \space a_{2}, a_{3}, \dots)$$ that follow the lead number (including the lead number) is equal to the lead number itself i.e. $$\underbrace{a_{1}, a_{2}, a_{3}, \dots, a_{a-2}, a_{a-1}, a_{a}}$$. And then we continue the series with the number $$a_{2}$$ and follow the same procedure.

If the series starts from 1 as $$\{$$1, $$\,$$ 2, 3, $$\,$$ 3, 4, 5, $$\,$$ 4, 5, 6, 7,  ...$$\}$$, find the smallest $$n$$, where the $$n$$th term $$a_n=2017$$.

$\underbrace{\color{blue}{1}}_\color{blue}{1}, \underbrace{\color{blue}{2}, 3}_\color{blue}{2}, \underbrace{\color{blue}{3}, 4,5}_\color{blue}{3}, \underbrace{\color{blue}{4}, 5,6,7}_\color{blue}{4},...$

The series above comprises segments of a leader integer (blue) followed by consecutive follower integers such that, if a segment starts with a leader integer $$\color{blue}{k}$$, it is followed by $$k-1$$ follower integers: $$k+1$$, $$k+2$$, $$k+3$$,... $$k+k-1$$, before the next segment, which starts with leader integer $$\color{blue}{k+1}$$.

Find the smallest $$n$$ such that the $$n$$th term of the series is 2017.

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