Find the sum and asymptotic expansion of $\sum_{u=1}^{[N/2]} \sum_{v=1}^{[\sqrt{N}]} \left[{|u^2-2v| \le N}\right] \left[{|-2uv| \le N}\right]$ Where $[x]$ in the summation limits is the floor function.  In the summation body this is the Iverson bracket
Find the sum and asymptotic expansion of $\sum_{u=1}^{\lfloor{N/2}\rfloor} \sum_{v=1}^{\lfloor{\sqrt{N}}\rfloor} \left[|u^2-2v| \le N\right] \left[|-2uv| \le N\right]$.
I suspect that there may be a closed form solution.  I have not made any progress to the analytic form of this sum.
I have some table values




N
Sum




1
0


10
10


10^2
100


10^3
981


10^4
10000


10^5
99928




This indicates that the sum approach $N$ as $N \rightarrow \infty$.
The second part is to find this asymptotic expansion.

Lets look at a table between perfect squares.  Say from N = 16 to 24.  Then




N
Sum
Difference




16
16
0


17
17
0


18
17
1


19
18
1


20
18
2


21
19
2


22
19
3


23
20
3


24
20
4




This pattern repeats for each perfect square sequence.
So with ${S}_{1} \left({N}\right) = \sum_{u=1}^{\lfloor{N/2}\rfloor}  \sum_{v=1}^{\lfloor{\sqrt{N}}\rfloor} \left[{|u^2-2v| \le N}\right] \left[{|-2u| v \le N}\right]$.
Then consider two consecutive perfect squares $\sqrt{N} \in \mathbb{Z}$ or $N = {n}^{2}$ to $N = \left({n + 1}\right)^{2}$.  Then from $N = {n}^{2}$ to $\left({n +1}\right)^{2} - 1$ there are $2\, n$ values.  Let $k \in \left\{{0, 1, \cdots, 2\, n}\right\}$.  Then for the values ${S}_{1} \left({{n}^{2}}\right.$, ${n}^{2} + 1$, $\cdots$, $\left.{\left({n +1}\right)^{2} - 1}\right) = N - \lfloor{k/2}\rfloor$.  The sequence of $k = 0$ to $2\, n$ is $\lfloor{k/2}\rfloor = \lfloor{(N-\lfloor{\sqrt{N}}\rfloor^{2})/2}\rfloor$.  Thus, applying induction, we can write for any perfect square interval $N \ge 9$
$${S}_{1} \left({N}\right) = N - \lfloor{\frac{N-\lfloor{\sqrt{N}}\rfloor^{2}}{2}}\rfloor$$
We add the final corrections for the values of $N \in \left\{{1, 2, 3, 5}\right\}$ resulting in
$${S}_{1} \left({N}\right) = N - \lfloor{\frac{N-\lfloor{\sqrt{N}}\rfloor^{2}}{2}}\rfloor - {\delta}_{N \in \left\{{1, 2, 3}\right\}} - {\delta}_{N=5}$$
For the asymptotic values as $N \rightarrow \infty$ we write the average value between two perfect squares of
$$\left<{\lfloor{\frac{N-\lfloor{\sqrt{N}}\rfloor^{2}}{2}}\rfloor}\right> = \frac{\text{sum of interval}}{2 \times \text{ length of the interval}}= \frac{\lfloor{\sqrt{N}}\rfloor^{2}-1}{2(\lfloor{\sqrt{N}}\rfloor+2)}\sim\frac{N-1}{2(\sqrt{N}+2)}\sim\frac{1}{2} \sqrt{N}-1+O(\frac{1}{N})$$
with the final answer
$${S}_{1}\left({N}\right) \sim N - \frac{1}{2} \sqrt{N}+1+O(\frac{1}{N})$$
