Find the sum (if possible) and asymptotic expansion (necessary) to as many terms as possible as $N \rightarrow \infty$ for
$$ S \left({N}\right) = \sum_{a = \lceil\sqrt{N + 2}\rceil}^{\lfloor 2N/3 \rfloor} \left\{{\frac{{a}^{2}-N}{2\, a}}\right\} $$
where $\left\{{...}\right\}$ is the fractional part. Note that the fractional part is defined to be in the interval [0,1). However in Mathematica a negative result can occur, so the use of absolute value is required.
This sum arises from counting the number of reducible quartics of the form $\left({x + a}\right)^{2} \left({x + b}\right) \left({x + c}\right) = {x}^{4} + {a}_{3}\, {x}^{3} + {a}_{2}\, {x}^{2} + {a}_{1}\, x + {a}_{0}$ where $- N \le {a}_{3}, {a}_{2}, {a}_{1}, {a}_{0} \le + N$ for positive integer $N \ge 1$.
Numerical testing ($N \rightarrow {10}^{7}$) shows that $S \left({N}\right) \sim 0.3285 \cdots N$. So I would expect that the next ordered term would be $\sqrt{N}$ or so.
I have tried letting $N = 2\, a\, u + v$ where $u = \lfloor{N/\left({2\, a}\right)}\rfloor$ and $v = N \left({\textrm{mod}\ 2\,a}\right) \in \left\{{0}\right.$, $1$, $\cdots$, $\left.{2\, a - 1}\right\}$ by the Quotient Remainder Theorem. Then we have
$$\left\{{\frac{{a}^{2}-N}{2\,a}}\right\} = \left\{{\frac{a}{2}-\frac{N}{2\,a}}\right\} = \left\{{\frac{a}{2}-\frac{v}{2\,a}}\right\}. $$
where the fractional part of $a/2 = 0$ if $a$ is even and $a/2 = 1/2$ if $a$ is odd. I do not get sums that work out properly.
Looking for suggestions, references, etc. on how to solve this problem.
From the properties of the floor function we can write
$$ \left\{{x \pm y}\right\} = x \pm y - \left({\lfloor{x}\rfloor \pm \lfloor{y}\rfloor}\right) - \lfloor{\left\{{x}\right\} \pm \left\{{y}\right\}}\rfloor. $$
Thus we write
$$S \left({N}\right) = \sum_{a = \lceil{\sqrt{N + 2}}\rceil}^{\lfloor{2N/3}\rfloor} \frac{{a}^{2} - N}{2 a} - \sum_{a = \lceil{\sqrt{N + 2}}\rceil}^{\lfloor{2N/3}\rfloor} \lfloor{\frac{a}{2}}\rfloor + \sum_{a = \lceil{\sqrt{N + 2}}\rceil}^{\lfloor{2N/3}\rfloor} \lfloor{\frac{N}{2 a}}\rfloor- \sum_{a = \lceil{\sqrt{N + 2}}\rceil}^{\lfloor{2N/3}\rfloor} \lfloor{\left\{{\frac{a}{2}}\right\} - \left\{{\frac{N}{2\, a}}\right\}}\rfloor$$
We can solve the other sums except for the last one. The sums bounds are $$-1 \le \lfloor{\left\{{\frac{a}{2}}\right\} - \left\{{\frac{N}{2\, a}}\right\}}\rfloor \le 0$$ therefore
$$\sum_{a = \lceil{\sqrt{N + 2}}\rceil}^{\lfloor{2N/3}\rfloor} \lfloor{\left\{{\frac{a}{2}}\right\} - \left\{{\frac{N}{2\, a}}\right\}}\rfloor \le \sum_{a = \lceil\sqrt{N + 2}\rceil}^{\lfloor 2N/3 \rfloor} 1 = \lfloor \frac{2N}{3} \rfloor - \lceil\sqrt{N + 2}\rceil + 1 \sim O \left({\frac{2}{3} N}\right) \sim O \left({N}\right)$$
This is verified from the numerical testing. So considering this can I obtain an more precise value for the constant times $N$ and the possible next term?
Using the summation over even series the corrected results from solution #1 is
$$\frac{1}{12} \left({1 + 3\, \gamma - 6\, \log \left({2}\right) + 3\, \log \left({3}\right)}\right) N + O \left({\sqrt{N}}\right) \sim 0.155717 \cdots N + O \left({\sqrt{N}}\right)$$
The numerical results for $N = 10^8$ is $0.15569 \cdots N$. Which we see is getting close to the above asymptotic expansion.
For the odd sum we have
$$\begin{aligned} \sum_{a = \lceil{\sqrt{N + 2}}\rceil \wedge \text{ odd}}^{\lfloor{2N/3}\rfloor} \left\{{\frac{{a}^{2} - N}{2\, a}}\right\} {}={} & \sum_{a = 2\, \lfloor{\lceil{\sqrt{N + 2}}\rceil/2}\rfloor + 1 \wedge {\Delta}\, a = 2}^{2 \lfloor{\left({\lfloor{2N/3}\rfloor - 1}\right)/2}\rfloor + 1} \left\{{\frac{{a}^{2} - N}{2\, a}}\right\} \\ {}={} & \sum_{a = \lfloor{\lceil{\sqrt{N + 2}}\rceil/2}\rfloor}^{\lfloor{\left({\lfloor{2N/3}\rfloor - 1}\right)/2}\rfloor} \left\{{\frac{4\, {a}^{2} + 4\, a + 1 - N}{2 \left({2\, a + 1}\right)}}\right\} \\ {}={} & \sum_{a = \lfloor{\lceil{\sqrt{N + 2}}\rceil/2}\rfloor}^{\lfloor{\left({\lfloor{2N/3}\rfloor - 1}\right)/2}\rfloor} \left\{{\frac{1}{2} - \frac{N}{2 \left({2\, a + 1}\right)}}\right\} \\ {}={} & \sum_{a = 1}^{\lfloor{\left({\lfloor{2N/3}\rfloor - 1}\right)/2}\rfloor} \left\{{\frac{1}{2} - \frac{N}{2 \left({2\, a + 1}\right)}}\right\} - \sum_{a = 1}^{\lfloor{\lceil{\sqrt{N + 2}}\rceil/2}\rfloor - 1} \left\{{\frac{1}{2} - \frac{N}{2 \left({2\, a + 1}\right)}}\right\} \\ {}\sim{} & \sum_{a = 1}^{\lfloor{\left({\lfloor{2N/3}\rfloor - 1}\right)/2}\rfloor} \left\{{\frac{1}{2} - \frac{N}{2 \left({2\, a + 1}\right)}}\right\} + O \left({\sqrt{N}}\right). \end{aligned} $$
The summation argument is positive if $\frac{1}{2} - \frac{N}{2 \left({2\, a + 1}\right)} \ge 0 \Rightarrow a \ge \lceil{\frac{N - 1}{2}}\rceil$. However, the upper limit of the sum is $\lfloor{\frac{1}{2} \left({\lfloor{\frac{2N}{3}}\rfloor - 1}\right)}\rfloor = \lfloor{\frac{N}{3}}\rfloor - \frac{1}{2} \left[{\left({- 1}\right)^{\lfloor{2\, N/3}\rfloor} + 1}\right] \sim \frac{N}{3}$ see Sloane Integer Sequence A011655 for details, which is less than the positive limit requirement hence the argument of the fractional part is negative. Using the same argument as for the even summation and that $- N/\left({4\, a + 2}\right)$ is an integer is at most $\tau \left({N}\right)$ we can write
$$\begin{aligned} \sum_{a = 1}^{\lfloor{\left({\lfloor{2N/3}\rfloor - 1}\right)/2}\rfloor} \left\{{\frac{1}{2} - \frac{N}{2 \left({2\, a + 1}\right)}}\right\} {}\sim{} & \sum_{a = 1}^{\lfloor{N/3}\rfloor} \left({1 - \left\{{\frac{N}{2 \left({2\, a + 1}\right)} - \frac{1}{2}}\right\}}\right) + O \left({\sqrt{N}}\right) \\ {}\sim{} & \lfloor{\frac{N}{3}}\rfloor - \sum_{a = 1}^{\lfloor{N/3}\rfloor} \left({\left({\frac{N}{2 \left({2\, a + 1}\right)} - \frac{1}{2}}\right) - \lfloor{\frac{N}{2 \left({2\, a + 1}\right)} - \frac{1}{2}}\rfloor}\right) + O \left({\sqrt{N}}\right) \\ {}\sim{} & \lfloor{\frac{N}{3}}\rfloor - \left({\frac{1}{2}\, \lfloor{\frac{N}{3}}\rfloor + \frac{1}{2} \left({1 - \log \left({2}\right)}\right) N - \frac{1}{4}\, N\, {H}_{1/2 + \lfloor{N/3}\rfloor}}\right) \\ & {}+{} \sum_{a = 1}^{\lfloor{N/3}\rfloor} \lfloor{\frac{N}{2 \left({2\, a + 1}\right)} - \frac{1}{2}}\rfloor + O \left({\sqrt{N}}\right) \\ {}\sim{} & - \frac{1}{4}\, N\, \log \left({N}\right) + \frac{1}{4} \left({4 - \gamma - 2\, \log \left({2}\right) + \log \left({3}\right)}\right) N \\ & {}+{} \sum_{a = 1}^{\lfloor{N/3}\rfloor} \lfloor{\frac{N}{2 \left({2\, a + 1}\right)} - \frac{1}{2}}\rfloor + O \left({\sqrt{N}}\right). \end{aligned}$$
The next reduction is when the argument of the floor function in the remaining sum is less than one. This condition is $\frac{N}{2 \left({2\, a + 1}\right)} - \frac{1}{2} < 1 \Rightarrow a \ge \lfloor{\frac{N + 3}{6}}\rfloor < \lfloor{\frac{N}{3}}\rfloor$. However, the upper limit of the sum now becomes $\sum_{a = 1}^{\lfloor{N/3}\rfloor} \lfloor{\frac{N}{2 \left({2\, a + 1}\right)} - \frac{1}{2}}\rfloor = \sum_{a = 1}^{\lfloor{\left({N - 3}\right)/6}\rfloor} \lfloor{\frac{N}{2 \left({2\, a + 1}\right)} - \frac{1}{2}}\rfloor$.
This is as far as I have gotten concerning the odd summation.
Here's half a solution—an asymptotic formula for the terms in the sum with $a$ even. Hopefully the method can be carried through with minor modifications for $a$ odd.
We start with \begin{align*} \sum_{\substack{a = \lceil\sqrt{N + 2}\rceil \\ a\text{ even}}}^{\lfloor 2N/3 \rfloor} \biggl\{ \frac{{a}^{2}-N}{2a} \biggr\} &= \sum_{\substack{a=1 \\ a\text{ even}}}^{\lfloor 2N/3 \rfloor} \biggl\{ \frac{{a}^{2}-N}{2a} \biggr\} + O(\sqrt N) \\ &= \sum_{\substack{1\le a\le 2N/3 \\ a\text{ even}}} \biggl\{ \frac{-N}{2a} \biggr\} + O(\sqrt N) = \sum_{1\le b\le N/3} \biggl\{ \frac{-N}{b} \biggr\} + O(\sqrt N) \end{align*} after setting $b=2a$. Note that $\{-y\}=1-\{y\}$ unless $y$ is an integer, while the number of $b$ such that $\frac{-N}b$ is an integer is at most $\tau(N)$, the number of divisors of $N$, which is certainly $O(\sqrt N)$. Therefore \begin{align*} \sum_{b\le N/3} \biggl\{ \frac{-N}{b} \biggr\} &= \sum_{b\le N/3} \biggl( 1 - \biggl\{ \frac{N}{b} \biggr\} \biggr) + O(\sqrt N) \\ &= \frac N3 - \sum_{b\le N/3} \biggl( \frac{N}{b} - \biggl\lfloor \frac{N}{b} \biggr\rfloor \biggr) + O(\sqrt N) \\ &= \frac N3 - N\biggl( \log \frac N3 + \gamma + O\biggl( \frac1N \biggr) \biggr) + \sum_{b\le N/3} \biggl\lfloor \frac{N}{b} \biggr\rfloor + O(\sqrt N) \\ &= -N\log N + N(\tfrac13 + \log3 - \gamma) + \sum_{b\le N/3} \biggl\lfloor \frac{N}{b} \biggr\rfloor + O(\sqrt N), \end{align*} where $\gamma\approx 0.577216$ is Euler's constant. On the other hand, \begin{align*} \sum_{b\le N/3} \biggl\lfloor \frac{N}{b} \biggr\rfloor &= \sum_{b\le N/3} \sum_{\substack{d\le N \\ b\mid d}} 1 = \sum_{d\le N} \sum_{\substack{b\le N/3 \\ b\mid d}} 1 \\ &= \sum_{d\le N} \bigl( \tau(d) - [1\text{ if } d>\tfrac N3] - [1\text{ if } d>\tfrac{2N}3\text{ and $d$ is even}] \bigr) \\ &= \sum_{d\le N} \tau(d) - \frac{2N}3 - \frac N6 + O(1) \\ &= N\log N + (2\gamma-1)N - \frac{2N}3 - \frac N6 + O(\sqrt N) \\ &= N\log N + N\bigl( 2\gamma-\tfrac{11}6 \bigr) + O(\sqrt N). \end{align*} Consequently, \begin{align*} \sum_{\substack{a = \lceil\sqrt{N + 2}\rceil \\ a\text{ even}}}^{\lfloor 2N/3 \rfloor} \biggl\{ \frac{{a}^{2}-N}{2a} \biggr\} &= -N\log N + N(\tfrac13 + \log3 - \gamma) \\ &\qquad{}+ N\log N + N\bigl( 2\gamma-\tfrac{11}6 \bigr) + O(\sqrt N) \\ &= N(\log3 + \gamma - \tfrac32) + O(\sqrt N). \end{align*} Here the constant $\log3 + \gamma - \tfrac32 \approx 0.175828$ is borne out by numerical experiments.