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 $n$.
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.