Looking for a conversion of sum to a single sum over d where $\sum_{n = 1}^{x} n \sum_{d \mid 2n} \frac{\Lambda \left({d}\right)}{d}$

Given the sum

$${S}_{1} \left({x}\right) = \sum_{n = 1}^{x} n \sum_{d \mid 2n} \frac{\Lambda \left({d}\right)}{d}$$

Where $$\Lambda \left({d}\right)$$ is a von Mangoldt function. Note that the number of divisors $$d$$ is over $$2n$$. Now the standard reduction if the divisors $$d$$ are over $$n$$ instead of $$2n$$ is

$$\sum_{n = 1}^{x} n \sum_{d \mid n} \frac{\Lambda \left({d}\right)}{d} = \sum_{n = 1}^{x} \sum_{n = q\, d} n\, \frac{\Lambda \left({d}\right)}{d} = \sum_{\substack{q, d \\ 1 \le q\, d \le x}} \Lambda \left({d}\right) q = \sum_{d = 1}^{x} \Lambda \left({d}\right) \sum_{q \le x/d} q = \frac{1}{2} \sum_{d = 1}^{x} \Lambda \left({d}\right) \left\lfloor\frac{x}{d}\right\rfloor \left({\left\lfloor\frac{x}{d}\right\rfloor + 1}\right)$$

How do I reduce $${S}_{1} \left({x}\right)$$ like the above example to the exact form then I generate the asymptotic expansion as $$x \rightarrow \infty$$, such as:

$${S}_{1} \left({x}\right) = \sum_{d = 1}^{x, [x/2]} \Lambda \left({d}\right)\times \text{other factors} \sim XXX$$

I have tried a number of reduction and none come out correct. This includes numerical testing as a verification. I suspect that I am missing some simple step.

We can convert $${S}_{1} \left({x}\right)$$ as

$${S}_{1} \left({x}\right) = \sum_{n = 1}^{x} n \sum_{d \mid 2\, n} \frac{\Lambda \left({d}\right)}{d} = \sum_{n = 2, \Delta n = 2}^{2\, x} \frac{n}{2} \sum_{d \mid n} \frac{\Lambda \left({d}\right)}{d}$$

Now I tried some analysis of this even sum with an approximation factor $$R = 2/3-3/4$$ of the full sum over $$n$$.

Section 3.5 p 57+ of Tom M. Apostal, Introduction to Analytic Number Theory, describes in detail the method reduction that I used in my initial example where the divisors are over $$n$$.

• Note that both occurrences of $\frac xd$ in the first example should instead be $\lfloor\frac xd\rfloor$. Aug 7, 2023 at 6:08
• Notice that $\Lambda(d)$ is nonzero iff $d$ is not divisible by two different primes. Aug 7, 2023 at 7:30
• Yes when the divisors are even, then the Mangoldt function is Log[2] or 0. However there will be divisors that are odd primes to a power. I have the summation if all divisors are even which is one component to the sums. Aug 7, 2023 at 18:15

First part is the double divisor solution: For the sum of divisors $$d \mid 2\, n$$ we have for $$n$$ being odd the sum over divisors is $$d \mid 2\, n = 2 \left({d \mid n}\right) + d \mid n$$. When $$n$$ is even and $$n = {2}^{k}\, m$$ where $$m$$ is odd, then $$d \mid 2\, n = 2 \left({d \mid n}\right) + d \mid m$$. When $$n$$ is even and $$n = {2}^{k}$$ then $$d \mid 2\, n = 2 \left({d \mid n}\right) + d \mid 1$$ where $$d$$ is the number of divisors at each stage. This is summarized as follows:

$$\begin{equation*} \sum_{d \mid 2\, n} f \left({d}\right) = \sum_{d \mid n} f \left({2\, d}\right) + \begin{cases} \sum_{d \mid n} f \left({d}\right), & \text{ if n is odd}, \\ \sum_{d \mid m} f \left({d}\right), & \text{ if n = {2}^{k}\, m for m is odd and k \ge 1}, \\ \sum_{d \mid 1} f \left({d}\right), & \text{ if n = {2}^{k} and k \ge 1}, \end{cases} \end{equation*}$$

which is further simplified as

$$\begin{equation*} \sum_{d \mid 2\, n} f \left({d}\right) = \sum_{d \mid n} f \left({2\, d}\right) + \sum_{d \mid n} f \left({d}\right) - \sum_{d \mid \lfloor{n/2}\rfloor} f \left({2\, d}\right). \end{equation*}$$

This is summarized in a double sum over a general function $$g \left({n}\right)$$ as

$$\begin{equation*} \sum_{n = 1}^{x} g \left({n}\right) \sum_{d \mid 2\, n} f \left({d}\right) = \sum_{n = 1}^{x} g \left({n}\right) \sum_{d \mid n} f \left({2\, d}\right) + \sum_{n = 1}^{x} g \left({n}\right) \sum_{d \mid n} f \left({d}\right) - \sum_{m = 1}^{\lfloor{x/2}\rfloor} g \left({2\, m}\right) \sum_{d \mid m} f \left({2\, d}\right). \end{equation*}$$

The second part is to rewrite $${S}_{1} \left({x}\right)$$ as

$$\begin{equation*} {S}_{1} \left({x}\right) = \sum_{n = 1}^{x} n \sum_{d \mid 2\, n} \frac{\Lambda \left({d}\right)}{d} = \sum_{n = 1}^{x} n \sum_{d \mid n} \frac{\Lambda \left({d}\right)}{d} + \sum_{n = 1}^{x} n \sum_{d \mid n} \frac{\Lambda \left({2\, d}\right)}{2\, d} - \sum_{m = 1}^{\lfloor{x/2}\rfloor} 2\, m \sum_{d \mid m} \frac{\Lambda \left({2\, d}\right)}{2\, d}. \end{equation*}$$

We will use the properties of the von Mangoldt function

$$\begin{equation*} \Lambda \left({2\, d}\right) = \begin{cases} \log \left({2}\right), & \text{ if } 2\, d = {2}^{k}, \\ 0, & \text{ otherwise}. \end{cases} \end{equation*}$$

Evaluating the inner sums we obtain for each of the three cases

\begin{equation*} \begin{aligned} {S}_{1}^{\prime} \left({x}\right) {}={} & \sum_{n = 1}^{x} n \sum_{d \mid n} \frac{\Lambda \left({d}\right)}{d} = \sum_{d = 1}^{x} \sum_{\substack{q, d, \\ 1 \le q\, d \le x}} q\, {\Lambda}_{k} \left({d}\right) = \sum_{d = 1}^{x} \Lambda \left({d}\right) \sum_{q \le \lfloor{x/d}\rfloor} q \\ {}={} & \frac{1}{2} \sum_{d = 1}^{x} \Lambda \left({d}\right) {\lfloor{\frac{x}{d}}\rfloor}^{2} + \frac{1}{2} \sum_{d = 1}^{x} \Lambda \left({d}\right) {\lfloor{\frac{x}{d}}\rfloor}, \end{aligned} \end{equation*}

\begin{equation*} \begin{aligned} {S}_{1}^{\prime \prime} \left({x}\right) {}={} & \sum_{n = 1}^{x} n \sum_{d \mid n} \frac{\Lambda \left({2\, d}\right)}{2\, d} = \frac{1}{2} \sum_{d = 1}^{x} \sum_{\substack{q, d, \\ 1 \le q\, d \le x}} q\, {\Lambda}_{k} \left({2\, d}\right) = \frac{1}{2} \sum_{d = 1}^{x} \Lambda \left({2\, d}\right) \sum_{q \le \lfloor{x/d}\rfloor} q \\ {}={} & \frac{1}{4} \sum_{d = 1}^{x} \Lambda \left({2\, d}\right) {\lfloor{\frac{x}{d}}\rfloor}^{2} + \frac{1}{4} \sum_{d = 1}^{x} \Lambda \left({2\, d}\right) {\lfloor{\frac{x}{d}}\rfloor}, \end{aligned} \end{equation*}

and

\begin{equation*} \begin{aligned} {S}_{1}^{\prime \prime \prime} \left({x}\right) {}={} & \sum_{m = 1}^{\lfloor{x/2}\rfloor} 2\, m \sum_{d \mid m} \frac{\Lambda \left({2\, d}\right)}{2\, d} = \sum_{d = 1}^{\lfloor{x/2}\rfloor} \sum_{\substack{q, d, \\ 1 \le q\, d \le \lfloor{x/2}\rfloor}} q\, \Lambda \left({2\, d}\right) = \sum_{d = 1}^{\lfloor{x/2}\rfloor} \Lambda \left({2\, d}\right) \sum_{q \le {\lfloor{x/\left({2\, d}\right)}}\rfloor} q \\ {}={} & \frac{1}{2} \sum_{d = 1}^{\lfloor{x/2}\rfloor} \Lambda \left({2\, d}\right) {\lfloor{\frac{x}{2\, d}}\rfloor}^{2} + \frac{1}{2} \sum_{d = 1}^{\lfloor{x/2}\rfloor} \Lambda \left({2\, d}\right) \lfloor{\frac{x}{2\, d}\rfloor}. \end{aligned} \end{equation*}

Now the evaluation/asymptotic expansions of each of these sums is a bit lengthy so I will post the final result

$$\begin{equation*} {S}_{1} \left({x}\right) \sim \frac{1}{2} \left[{- \frac{{\zeta}^{\prime} \left({2}\right)}{\zeta \left({2}\right)} + \frac{1}{3}\, \log \left({2}\right)}\right] {x}^{2} + O \left({x\, \log \left({x}\right)}\right). \end{equation*}$$

As a test the following table shows the original function definition $${S}_{1} \left({x}\right)$$, the above exact expansions, the difference between these two forms, the asymptotic expansions, its difference from the exact, and finally the asymptotic error function.

If there are any questions I would double check my part one first paragraph.