How to prove that this function all over the positive integers gives us this sequence? On the first hand, I have this sequence : $0,1,1,2,2,2,3,3,3,3,...$ which is the sequence where an $n$ positive integer appears $n+1$ times consecutively.
On the other hand, I have this function : $a_n=\lfloor\frac{\sqrt {1+8n}-1}{2}\rfloor$ where $n\ge0$
$a_0=0$ ; $a_1=1$ ; $a_2=1$; $a_3=2$
This function seems to be a formula for this sequence.
However, if it is the case, how can we prove it ?
And if it isn't, what is the explanation ?
To begin with, i did something :
$\frac{\sqrt {1+8n}-1}{2}=t+b$ where $t\in \mathbb N$ and $b\in [0,1[$
After some simplification, i get this :
$8n= 4t^2+4b^2+8tb+12t+12b+8$
After this, i don't know how to continue...

The first entry of $k$ in your sequence comes in position $T_k=\frac 12k(k+1)$
If you use the quadratic equation, this gives $k=\frac {-1+\sqrt{1+8k}}2$
