Problem with my floor... Let the set of numbers $A,n$ be of defined as $A_n = \text{the }n\text{'th value of }x\text{ such that }2\not\mid\lfloor{ x^2 - \sqrt{x}}\rfloor$, and $n$ is a positive integer.
So as the first 10 sums are:
$$\begin{align}
x&&\lfloor{ x^2 - \sqrt{x}}\rfloor\\
1 &&0\\
2 &&2\\
3 &&7\\
4 &&14\\
5 &&22\\
6 &&33\\
7 &&46\\
8 &&61\\
9 &&78\\
10 &&96\\
\end{align}$$
The terms of $A$ is $A_1 = 3, A_2 =6,A_3 = 8,$ and so on.
I'm trying to determine whether the following is correct:
$$2 n+\lfloor{\frac{1}{2}+\frac{1}{2} \sqrt{8 n - 7}}\rfloor = A_n$$ 
$A_n = $ Complement of the Connell sequence
This is a real pain, as I have little to no experience in working with the floor function. How can this be (dis)proven?
Any help would be appreciated.
 A: I'll just start writing down my thoughts. I don't know where they'll end...
You want $\lfloor{ x^2 - \sqrt{x}}\rfloor$ to be odd. Because you only look at integers $x$, we know that $\lfloor{ x^2 - \sqrt{x}}\rfloor=x^2+\lfloor -\sqrt x\rfloor=x^2-\lceil\sqrt x\rceil$. Now, you want $\lceil \sqrt x\rceil$ to have the opposite parity of $x$ (because $x^2$ has the same parity as $x$. You know that $\lceil\sqrt x\rceil$ is odd when $(2k)^2<x\leq(2k+1)^2$ for some integer $k$. It is even when $(2k+1)^2< x\leq(2k+2)^2$.  
Now, we find the following values:
$$
3,6,8,11,13,15,18,20,22,24,27,29,31,33,35,38,\dots
$$
As you can see, you get all even or odd values between consecutive squares, and when crossing a square, you change the parity. Also, the differences are always $2$, except when crossing a square; then they are $3$.
Your formula gives these numbers too, so it seems to be correct. (Now, I assume you came this far yourself...)
We know every group of numbers of the same parity is one larger than its predecessor. We want to solve $\frac{k(k+1)}2=n$ to find the number of the group $n$ is in. This is what we want to add to $2n$ to make a correction for the steps of size $3$. Solving it gives $\lfloor\frac{1}{2} \left(\sqrt{8 n+1}-1\right)\rfloor$. We still have to add $1$ to correct for the offset. (this can/should be proven/explained more carefully) Now, we obtain exactly the formula you gave yourself. This is (almost) enough explanation to make it a proof IMO.
