Is there a closed form for this sequence $a_n = m, \ \binom{m}{2}\le n < \binom{m+1}{2}$?

Question

I was playing around with the sequence: $$ 2,2, 3,3,3,4,4,4,4,\dots $$ If I denote the first element of said sequence by $a_1$ I realized the sequence can be written as $$ a_n = m, \quad \binom{m}{2}\le n < \binom{m+1}{2} $$ for $m \ge 2$. Although the above does work, I would like to find a closed for of a sequence that gives $a_n$ in terms of common functions like the floor function. Does anyone know how I could do this?

I managed to re-arrange the condition $ \binom{m}{2}\le n < \binom{m+1}{2}$ into $$ 0\le \frac{n}{m} + \frac{1-m}{2} < 1 $$ with hopes of using something like the floor function on it, but I couldn't seem to make it work.

Answer 1

We have $$ \begin{array}{l} \left( \begin{array}{c} m \\ 2 \\ \end{array} \right) = \frac{{m\left( {m - 1} \right)}}{2} \le n < \left( \begin{array}{c} m + 1 \\ 2 \\ \end{array} \right) = \frac{{m\left( {m + 1} \right)}}{2}\quad \Rightarrow \\ \Rightarrow \quad m^2 - m - 2n \le 0 < m^2 + m - 2n\quad \Rightarrow \\ \Rightarrow \quad \frac{{\sqrt {1 + 8n} - 1}}{2} = \frac{{\sqrt {1 + 8n} + 1}}{2} - 1 < m \le \frac{{\sqrt {1 + 8n} + 1}}{2} \\ \end{array} $$ and since $$ x - 1 < \left\lfloor x \right\rfloor \le x $$ then $$ a_n = \left\lfloor {\frac{{\sqrt {1 + 8n} + 1}}{2}} \right\rfloor $$ where $a_0=1, a_1=2, a_2=2, \cdots$

Answer 2

The sequence is listed on the OEIS https://oeis.org/A002024, and lists the function $$ \operatorname{floor}\left(\frac{\left(1+\sqrt{1+8x}\right)}{2}\right).$$

It is the inverse of the triangular numbers, which is simply the sum of numbers $1$ to $n$. The sequence $2, 2, 3, 3, 3,...$ follows after observing this fact. (It is simply the floor of the inverse.)

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