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 $n$.

  • $\begingroup$ Note that both occurrences of $\frac xd$ in the first example should instead be $\lfloor\frac xd\rfloor$. $\endgroup$ Aug 7, 2023 at 6:08
  • $\begingroup$ Notice that $\Lambda(d)$ is nonzero iff $d$ is not divisible by two different primes. $\endgroup$
    – TravorLZH
    Aug 7, 2023 at 7:30
  • $\begingroup$ 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. $\endgroup$ Aug 7, 2023 at 18:15

1 Answer 1


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*}


\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.

enter image description here

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .