How do you derive the continuous analog of the discrete sequence $1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ...$? I was wondering what the rate of growth of the sequence $$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, ...$$ was, and found the related question, Formula for the $n$th term of $1, 2, 2, 3, 3, 3, 4, 4 ,4, 4, 5, ...$, in which one of the answers given is $$a_n = \operatorname{round}{\sqrt {2n}}$$ so the "continuous analog" (not sure if there's a better term) is $\sqrt{2n}$, whose rate of growth turns out to be its reciprocal, $\sqrt{2n}^{-1}$.
But there's no explanation in the aforementioned answer. What are the steps behind going from a discrete sequence to the function that approximates it?
Edit
By trial-and-error, I would have eventually figured it out—even easier if I used hindsight (knowing that the rate of growth is $\sqrt{}$). But how would we characterize a sequence's growth to begin with, to even come up with a guess whose growth matches?
 A: This has already been answered in various forms by others earlier but here's my attempt:
$$\begin{align}n&=1,\ \ 2,\ \ 3,\ \ 4,\ \ 5,\ \ 6,\ \ 7,\ \ 8,\ \ 9,\  10,\  11,\ 12,...\\
m=f(n)&=1,\ \ 2, \ \ 2, \ \ 3,\ \ 3,\ \ 3,\ \ 4,\ \ 4,\ \ 4,\ \ \  4, \ \ \ 5,\ \ \ 5,... \\
&\quad \  *\qquad *\qquad \quad\ \ *\qquad \quad \quad  \ \ \ \ * \end{align}$$
At positions marked * which occur just before $m$ steps up, 
$$\begin{align}
n&=\lbrace 1, \ 3, \ 6, 10, 15, ...\rbrace\\
m&=\lbrace 1,\ 2,\  3,  \ \ 4, \ \ 5, ... \rbrace\\
\Rightarrow n&=\sum_{i=1}^m i = \frac {m(m+1)}2\\
m^2+m-2n&=0 \\
m=f(n)&=\frac{\sqrt{1+8n}-1}2\end{align}$$
As $m$ steps up 1 each time, for all values of $n$, $m=f(n)$ can be given as 
$$m=f(n)=\Biggl\lceil{\frac{\sqrt{1+8n}-1}2} \Biggr\rceil $$
Simplifying further by approximation gives
$$\begin{align}
m=f(n)&=\Biggl\lceil{}\frac{2\sqrt{2n(1+\frac1{8n})}-1}2 \Biggr\rceil\\
&=\Biggl\lceil{}\sqrt{2n(1+\frac1{8n})}-\frac 12 \Biggr\rceil\\
&\approx \Biggl\lceil{}\sqrt{2n}-\frac 12 \Biggr\rceil 
\qquad \text{for moderate $n$ and larger }\\
&\approx \text{round}(\sqrt{2n})
\end{align}$$
A: I actually see it as the other way around:  the continuous function is used to generate the discrete function.  The sequence in the linked answer is discretized from the continuous function.
As for how to find the continuous function, one way is simply by inspection.  The sum of the first $n$ odd positive integers is $n^2$, so in between each square we have an increasing number of integers: $1,3,5,7,\dots$.  Let's round $\sqrt{n}$ up to the nearest integer, starting with $n=1$:
$$1, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 5, \dots$$
Not quite.  Getting rid of every other one might gets us there.  So $\sqrt{2n}$ starting with $n=1$:
$$2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, \dots$$
Almost, except we lost the $1$ at the beginning.  Maybe start at $n=0$ instead?
$$1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, \dots$$
There it is.
A: Observe that the sequence increases by one unit after $1, 1+2, 1+2+3, 1+2+3+4...$ terms, and more generally, it equals $k$ from the term $\frac{k(k-1)}2\approx\frac{k^2}2$ onwards. (In this particular case, contrary to habit, it turns out to be easier to express the index of the term as a function of the value of the term.)
This shows you the growth rate, $$T_{\frac{k^2}2}=k,$$ or by inversion, $$T_n=\sqrt{2n}.$$
