Peculiar Sequence of Positive Integers!

We arrange the terms of the sequence $(x_n)$ in a triangular array according to the given rule so that the $1^{st}$ level consists of a single 1 and for all $k \geq 2$ , the $k^{th}$ level consists of the $k$ consecutive even (odd) integers that follow the last odd (even) integer in the $(k-1)^{th}$ level:

For any given integer $n$ , let $l_n$ denote the unique integer such that:

${l_n \choose 2} < n \leq {l_n + 1 \choose 2}$

that is,

$\dfrac{(l_n-1)l_n}{2} < n \leq \dfrac{l_n(l_n+1)}{2}\quad \quad \text{ ....(1)}$

Note that $l_n$ is simply the number of the level which $x_n$ is on. We claim that:

$x_n = 2n - l_n \quad \quad \text{ .... (2) }$

When $n = 1$ , clearly $l_1 = 1$ and thus $2n-l_n = 1 = x_1$ . Suppose (2) holds for some $n \geq 1$ . There are two possible cases:

Case-1 : If $n+1 \leq {l_n+1 \choose 2}$ ; that is, if $x_{n+1}$ and $x_n$ are on the same level, then $l_{n+1} = l_n$ and hence

$x_{n+1} = x_n + 2 = 2n - l_n + 2 = 2(n+1) - l_{n+1}$

Case-2 : If $n+1 > {l_n + 1 \choose 2}$ then $x_n$ is the last number on the $l_n^{th}$ level and $x_{n+1}$ is the first number on the $l_{n+1}^{th}$ level. Thus

$l_{n+1} = l_n + 1$

and

$x_{n+1} = x_n + 1 = 2n - l_n + 1 = 2(n+1) - l_{n+1}$

Hence, by induction, (2) is established. From (1) we get

$l_n^2 - l_n <2n \leq (l_n + 1)^2 - (l_n + 1)$

Solving the inequalities, we easily obtain:

$l_n < \dfrac{1 + \sqrt{1+8n}}{2} \leq l_n + 1$

Hence,

$l_n = \left \lceil \dfrac{1 + \sqrt{1+8n}}{2} \right \rceil - 1 \quad \quad \text{ ....(3)}$

From (2) and (3) we conclude that:

$\large{x_n = 2n + 1 - \left \lceil \dfrac{1 + \sqrt{1+8n}}{2} \right \rceil}$

Substituting $n = 2015$ , we obtain $x_{2015} = \boxed{3967}$ .

Let list the terms of the sequence, call it $S$ , in rows of odd and even numbers as follows:

$\begin{array} {lc} k = 1 & x_{\color{#D61F06}{1}} = 1 = \color{#3D99F6}{1^2} \\ k= 2 & x_2 = 2 \quad x_{\color{#D61F06}{3}} = 4 = \color{#3D99F6}{2^2} \\ k=3 & x_4 = 5 \quad x_5 = 7 \quad x_{\color{#D61F06}{6}} = 9 = \color{#3D99F6} {3^2} \\ k = 4 & x_7 = 10 \quad x_8 = 12 \quad x_9 = 14 \quad x_{\color{#D61F06}{10}} = 16 = \color{#3D99F6}{4^2} \\ ... & ... \\ k = r & ... \quad ... \quad ... \quad ... \quad ... \quad ... \quad ... \quad ... \quad x_{\color{#D61F06}{T_r}} = \color{#3D99F6}{r^2} \end{array}$

We note that the last term of the $r^{th}$ row is the $\color{#D61F06}{T_r}^{th}$ term, where $T_r = \dfrac{r(r+1)}{2}$ is the $r^{th}$ triangular number and that $x_{\color{#D61F06}{T_r}} = \color{#3D99F6}{r^2}$ .

Since $T_{63} = 2016 \quad \Rightarrow x_{2016} = 2016^2$ and $x_{2015} = 2016^2 - 2 = \boxed{3967}$

In general:

$\begin{aligned} x_n & = \left \lfloor \sqrt{2n} \right \rfloor ^2 - 2 \left(\dfrac{\left \lfloor \sqrt{2n} \right \rfloor \left( \left \lfloor \sqrt{2n} \right \rfloor + 1 \right)}{2} - n \right) \\ \Rightarrow x_n & = 2n - \left \lfloor \sqrt{2n} \right \rfloor \\ \Rightarrow x_{2015} & = 4030 - \left \lfloor \sqrt{4030} \right \rfloor = 4030 - 63 = \boxed{3967} \end{aligned}$