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I am trying to evaluate the sum and its asymptotic limit as $N \rightarrow \infty$ of

$$\sum_{a = 1}^{L \left({N}\right)} \left({- 1}\right)^{\left\lfloor{N/\left({2\, a + 1}\right)}\right\rfloor}$$

where the upper limit can be either $L \left({N}\right) = {L}_{1} \left({N}\right) = \left\lfloor{\frac{N}{6} - \frac{1}{2}}\right\rfloor$ or $L \left({N}\right) = {L}_{2} \left({N}\right) = \left\lfloor{\frac{1}{2} \left\lceil{\sqrt{N + 2}}\right\rceil}\right\rfloor-1$.

It looks like I should count the number of even and odd terms of $$\left({- 1}\right)^{\left\lfloor{N/\left({2\, a + 1}\right)}\right\rfloor}$$

The case for ${L}_{2} \left({N}\right)$ when $N \rightarrow \infty$ looks to be $\sim \sqrt[4]{N}$. The similar case for ${L}_{1} \left({N}\right) \sim \alpha\, N$ for some constant $\alpha$.

My literature search has not revealed any number theory equivalents nor anything for the asymptotic limits. I suspect that the sum equivalent of the sum of the number of divisors or $\sum_{n = 1}^{x} \left\lfloor{\frac{N}{n}}\right\rfloor$ has been broken down into the even and odd floor function results, but again I am not finding any literature on this subject.

These sums arise from computing the exact number of factorable quartics of the form $\left({x+a}\right)^{2} \left({x+b}\right) \left({x+c}\right)$ where $a, b, c$ are integer roots.

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2 Answers 2

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I get $ \frac{N}{2}(1-2\ln(2))+O(L(N)) $.

So we want $m \le N/(2a+1) \lt m+1 $ or $m/N \le 1/(2a+1) \lt (m+1)/N $ or $N/m \ge 2a+1 > N/(m+1) $ or $(\frac{N}{m}-1)/2 \ge a \gt (\frac{N}{m+1}-1)/2 $ or $\frac{N-m}{2m} \ge a \gt \frac{N-m-1}{2(m+1)} $.

The number of these is about

$\begin{array}\\ \frac{N-m}{2m} -\frac{N-m-1}{2(m+1)} &=\frac{(N-m)(m+1)-(N-m-1)(m)}{2m(m+1)}\\ &=\frac{(N-m)m+(N-m)-(N-m)m+m}{2m(m+1)}\\ &=\frac{N}{2m(m+1)}\\ \end{array} $

so the sum is about

$\begin{array}\\ s(N) &=\sum_{a = 1}^{L \left({N}\right)} \left({- 1}\right)^{\left\lfloor{N/\left({2\, a + 1}\right)}\right\rfloor} \\ &\approx \sum_{m=1}^{N/(2L(N))} \frac{N}{2m(m+1)}(-1)^m\\ &= \frac{N}{2}\sum_{m=1}^{f(N)} \frac{(-1)^m}{m(m+1)} \qquad f(N)=N/(2L(N))\\ &= \frac{N}{2}\sum_{m=1}^{f(N)} (-1)^m(\frac1{m}-\frac1{m+1})\\ &= \frac{N}{2}\left(\sum_{m=1}^{f(N)} \frac{(-1)^m}{m}-\sum_{m=1}^{f(N)} \frac{(-1)^m}{m+1}\right) \\ &= \frac{N}{2}\left(\sum_{m=1}^{f(N)} \frac{(-1)^m}{m}-\sum_{m=2}^{f(N)+1} \frac{(-1)^{m-1}}{m}\right)\\ &= \frac{N}{2}\left(\sum_{m=1}^{f(N)} \frac{(-1)^m}{m}+\sum_{m=2}^{f(N)+1} \frac{(-1)^{m}}{m}\right)\\ &= \frac{N}{2}\left((-1)+\sum_{m=2}^{f(N)} \frac{(-1)^m}{m}+\sum_{m=2}^{f(N)} \frac{(-1)^{m}}{m}+\frac{(-1)^{f(N)+1}}{f(N)+1}\right)\\ &= \frac{N}{2}\left((-1)+2\sum_{m=2}^{f(N)} \frac{(-1)^m}{m}+\frac{(-1)^{f(N)+1}}{f(N)+1}\right)\\ &\approx \frac{N}{2}\left((-1)+2(1-\ln(2))+\frac{(-1)^{f(N)+1}}{f(N)+1}\right)\\ &\approx \frac{N}{2}(1-2\ln(2))+\frac{(-1)^{f(N)+1}N}{2(f(N)+1)}\\ &\approx \frac{N}{2}(1-2\ln(2))+\frac{(-1)^{f(N)+1}N}{2N/(2L(N))}\\ &= \frac{N}{2}(1-2\ln(2))+(-1)^{f(N)+1}L(N)\\ &= \frac{N}{2}(1-2\ln(2))+O(L(N))\\ \end{array} $

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This is a partial answer that covers the case when $L \left({N}\right) = \lfloor{\frac{N}{6} - \frac{1}{2}}\rfloor$. First a correction, the above initial solution involving $m$ was based on the ceiling function relations, instead of the floor function relations which are:

$$\lfloor{\frac{N}{2\, a + 1}}\rfloor \Rightarrow m - 1 < \lfloor{\frac{N}{2\, a + 1}}\rfloor \le m \Rightarrow \frac{N - m}{2\, m} \le a < \frac{N - m + 1}{2 \left({m - 1}\right)} \Rightarrow \lceil{\frac{N - m}{2\, m}}\rceil \le a \le \lceil{\frac{N - m + 1}{2 \left({m - 1}\right)}}\rceil - 1$$

Note that for $m = 4$ we have

$$m = 4 \Rightarrow \lceil{\frac{N}{8} - \frac{1}{2}}\rceil \le a \le \lceil{\frac{N}{6} - \frac{1}{2}}\rceil - 1$$

which is the boundary for the ${L}_{1} \left({N}\right)$ case. The number of these for a given $a$ for $m \ge 2$ is about

$$ \#a = \lceil{\frac{N - m + 1}{2 \left({m - 1}\right)}}\rceil - \lceil{\frac{N - m}{2\, m}}\rceil \approx \frac{N - m + 1}{2 \left({m - 1}\right)} - \frac{N - m}{2\, m} = \frac{N}{2\, m \left({m - 1}\right)}. $$

We can write as follows with $f \left({N}\right) = \lfloor{N/\left({2\, {L}_{1} \left({N}\right) + 1}\right)}\rfloor$ and the the boundary in terms of Lerch transcendent as

$$ \sum_{m = 4}^{f \left({N}\right)} \left({- 1}\right)^{m + 1}\, \frac{N}{2\, m \left({m - 1}\right)} \approx - \frac{N}{2} \sum_{m = 4}^{f \left({N}\right)} \frac{\left({- 1}\right)^{m}}{m \left({m - 1}\right)} = - N \left[{ \frac{1}{3} \left({- 2 + 3\, \log \left({2}\right)}\right) N - \frac{\left({- 1}\right)^{f \left({N}\right)}}{2\, f \left({N}\right)} + \left({- 1}\right)^{f \left({N}\right)}\, \Phi \left({- 1, 1, f \left({N}\right) + 1}\right)}\right] $$

where $\Phi \left({z, s, a}\right)$ is the Lerch transcendent. Note that the expansion of the Lerch transcendent about $a \rightarrow \infty$ is

$$ \Phi \left({z, s, a}\right) \sim \frac{1}{1 - z}\, \frac{1}{{a}^{s}} + \sum_{n = 1}^{N - 1} \left({- 1}\right)^{n}\, \frac{{Li}_{- n} \left({z}\right)}{n!}\, \frac{{\left({s}\right)}_{n}}{{a}^{n + s}} + O \left({{a}^{- \left({N + a}\right)}}\right) $$

for $N \in \mathbb{N}$, $\left({s}\right)_{n}$ is the Pochhammer symbol, and where ${Li}_{s} \left({z}\right) = z\, \Phi \left({z, s, 1}\right)$ is the polylogarithm.

Now taking the limit as $f \left({N}\right) \rightarrow \infty$ in the summation gives

$$ \sum_{m = 4}^{f \left({N}\right)} \left({- 1}\right)^{m + 1}\, \frac{N}{2\, m \left({m - 1}\right)} \sim - \frac{1}{3} \left({- 2 + 3\, \log \left({2}\right)}\right) N + O \left({\frac{1}{f \left({N}\right)}}\right) \sim - \frac{1}{3} \left({- 2 + 3\, \log \left({2}\right)}\right) N + O \left({1}\right) \sim - 0.026480513 \cdots N $$

which is borne out by numerical testing.

Note that as $N \rightarrow \infty$ we actually have $f \left({N}\right) \rightarrow 3$. However this is a reciprocal term, so the resultant total limit is still correct.

Now the case where $L \left({N}\right) = {L}_{2} \left({N}\right)$ does not work in this case from what I can see because we do not have a clear value of the initial limit of $m$ due to the square root of this limit. In the case for $L \left({N}\right) = {L}_{1} \left({N}\right)$ we have $m$ start at $4$ instead of $1$.

In the case of $L \left({N}\right) = {L}_{2} \left({N}\right)$ what value of $m$ in the sum do I start with? In fact numerical testing shows values that oscillates in sign roughly around the value of ${N}^{1/4}$ with large variations. When I take the average of these values say from $N-1000$ to $N+1000$ I still get large variations.

A table for evaluations of the primary sum for the limit ${L}_{2} \left({N}\right)$ is

$$\begin{array}{cccc} N & {L}_{2} \left({N}\right) & \text{Accumulative Sum} & \lfloor{{N}^{1/4}}\rfloor \\ 1 & 0 & 0 & 1 \\ 10 & 1 & -1 & 1 \\ {10}^{2} & 4 & 0 & 3 \\ {10}^{3} & 15 & 5 & 5 \\ {10}^{4} & 49 & 5 & 10 \\ {10}^{5} & 157 & 13 & 17 \\ {10}^{6} & 499 & -13 & 31 \\ {10}^{7} & 1{,}580 & 14 & 56 \\ {10}^{8} & 4{,}999 & 235 & 100 \\ {10}^{9} & 15{,}810 & 148 & 177 \\ {10}^{10} & 49{,}999 & -335 & 316 \\ {10}^{11} & 158{,}113 & 313 & 562 \\ {10}^{12} & 499{,}999 & 235 & 1{,}000 \\ {10}^{13} & 1{,}581{,}138 & 3{,}450 & 1{,}778 \\ {10}^{14} & 4{,}999{,}999 & -5{,}241 & 3{,}162 \\ {10}^{15} & 15{,}811{,}387 & 2{,}835 & 5{,}623 \\ {10}^{16} & 49{,}999{,}999 & 6{,}439 & 10{,}000 \\ {10}^{17} & 158{,}113{,}882 & -6{,}920 & 17{,}782 \end{array}$$

A plot of this sum for $N \in \left\{{{10}^{6}, {10}^{6} + 1{,}000}\right\}$ for $L \left({N}\right) = {L}_{2} \left({N}\right)$ is shown below

Plot of sum

So defining an average of this sum for $L \left({N}\right) = {L}_{2} \left({N}\right)$ may not be well defined in that it converges to a meaningful value.

Now if we use the accumulative average then we see a possible convergent answer shown below

Accumulative average of the sum for L2[N]

Then we have the approximation

$$\frac{1}{N} \sum_{n = 1}^{N} \sum_{a = 1}^{{L}_{2} \left({n}\right)} \left({- 1}\right)^{\lfloor{n/\left({2\, a + 1}\right)}\rfloor} \sim 0.11 \sim O \left({1}\right)$$

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