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I came up with this in mind and couldn't find a way to solve it. With this progression: $$ A_0 = 0,\\ A_1 = 1,\\ A_2 = 1,\\ A_3 = 2,\\ A_4 = 2,\\ A_5 = 2,\\ ...\\ A_N $$ for the step $k$ of the sequence, there will be $k$ times the number $k-1$.

(Example for the 4th part of the sequence, there will be $4$ times the number $3$ like this: $3, 3, 3, 3$)

What will be $A_n$ for a given $n$? Why? What are these kind of progressions called?

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    This is related to the [triangle numbers](https://oeis.org/A000217) and its inverse sequence. $T_0=0, T_1=1, T_2=3, T_3=6,\dots,T_n$ can be interpreted as "*the position of the first time you see the number $n$ in your sequence*" One has $T_n = n(n+1)/2$, so if you go through the effort of finding the functional inverse of the related real function $f(x)=x(x+1)/2$, and take the floor function, one gets $A_n = \lfloor(-1+\sqrt{8n+1})/2\rfloor$ – 2017-02-16

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This appears as https://oeis.org/A003056

As mentioned in comment above, one way to derive the formula for this is to recognize its relation to the triangle numbers and find the functional inverse to that sequence, throwing in the floor function to have it map solely to integer values, yielding the expression

$$A_n = \left\lfloor \frac{-1+\sqrt{8n+1}}{2}\right\rfloor$$

For more information, visit the link at the start.

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    I get it! The binomial function that generates the triangle numbers is $X = \frac{n(n+1)}{2}$. Solving for $X$ we get this. Thanks! – 2017-02-16
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Notice $a_{\sum_{i=0}^k i=\frac {k (k+1)}2}=k $.

So $a_n=k: \frac {k (k+1)}2\le n <\frac {(k+1)(k+2)}2$

If that's not enough $k (k+1)/2=n $ means

$k^2+k-2n=0;k=\frac {-1 +\sqrt {1+8n}}{2}$

So $a_n =\lfloor \frac {-1 +\sqrt {1+8n}}{2} \rfloor $