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Consider the function $f(x)$ defined in (2) below related to the fundamental prime counting function $\pi(x)$ defined in (1) below. Note $b(n)$ is the Möbius transform of $a(n)$.

(1) $\quad \pi(x)=\sum\limits_{n=1}^{x}a(n)\,,\quad a(n)=\begin{array}{cc} \{ & \begin{array}{cc} 1 & n\in\mathbb{P} \\ 0 & n\notin\mathbb{P} \\ \end{array} \\ \end{array}\qquad$ (see https://oeis.org/A010051)

(2) $\quad f(x)=\sum\limits_{n=1}^{x}b(n)\,,\quad b(n)=\sum\limits_{d|n} a(d)\,\mu\left(\frac{n}{d}\right)\qquad\,\,$ (see https://oeis.org/A143519)


The following plot illustrates $f(x)$ defined in formula (2) above.


Illustration of f(x)

Figure (1): Illustration of $f(x)$ defined in formula (2)


The integer zeros of $f(x)$ for $x\le 10,000$ are listed in (3) below.

(3) $\quad${1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109}


The zero crossings of $f(x)$ for $x\le 10,000$ where $f(x)$ doesn't settle at zero are listed in (4) below.

(4) $\quad${14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682}


Question (1): Does $f(x)$ have a finite number of integer zeros, and if so what is the largest integer zero of $f(x)$?

Question (2): Does $f(x)$ have an finite number of zero crossings, and if so what is the largest zero crossing of $f(x)$?

Question (3): What is the asymptotic for the long term growth of $f(x)$? What are the associated error bounds predicted by the Prime Number Theorem and the Riemann Hypothesis?


The Dirichlet transforms of $a(n)$ and $b(n)$ are defined in (5) and (6) below where $P(s)$ is the prime zeta function.

(5) $\quad\sum\limits_{n=1}^\infty\frac{a(n)}{n^s}=P(s),\quad\Re(s)>1$

(6) $\quad\sum\limits_{n=1}^\infty\frac{b(n)}{n^s}=\frac{P(s)}{\zeta(s)},\quad\Re(s)>1$


The following figure illustrates the Dirichlet series for $\frac{P(s)}{\zeta(s)}$ defined in (6) above in orange where formula (6) is evaluated over the first $10,000$ terms. The underlying blue reference function is $\frac{P(s)}{\zeta(s)}$.


Illustration of formula (6) for P(s)/zeta(s)

Figure (2): Illustration of formula (6) for $\frac{P(s)}{\zeta(s)}$ (orange curve) and reference function (blue curve)


The following four figures illustrate formula (6) for $\frac{P(s)}{\zeta(s)}$ evaluated along the line $s=1+i\,t$ in orange where formula (6) is evaluated over the first $1,000$ terms. The underlying blue reference function is $\frac{P(s)}{\zeta(s)}$. The red discrete portions of the plots illustrate the evaluation of formula (6) for $\frac{P(1+i\,t)}{\zeta(1+i\,t)}$ where $t$ equals the imaginary part of a non-trivial zeta zero.


Illustration of formula (6) for Abs(P(1+i t)/zeta(1+i t))

Figure (3): Illustration of formula (6) for $\left|\frac{P(1+i\,t)}{\zeta(1+i\,t)}\right|$


Illustration of formula (6) for Re(P(1+i t)/zeta(1+i t))

Figure (4): Illustration of formula (6) for $\Re\left(\frac{P(1+i\,t)}{\zeta(1+i\,t)}\right)$


Illustration of formula (6) for Im(P(1+i t)/zeta(1+i t))

Figure (5): Illustration of formula (6) for $\Im\left(\frac{P(1+i\,t)}{\zeta(1+i\,t)}\right)$


Illustration of formula (6) for Arg(P(1+i t)/zeta(1+i t))

Figure (6): Illustration of formula (6) for $Arg\left(\frac{P(1+i\,t)}{\zeta(1+i\,t)}\right)$


Question (4): What is the range of convergence of the Dirichlet series for $\frac{P(s)}{\zeta(s)}$ defined in (6) above? Does it converge only for $\Re(s)>1$, or does it also converge for $\Re(s)=1\land\Im(s)\ne 0$?


Note $\frac{P(s)}{\zeta(s)}$ has a pole at each non-trivial zeta zero.


Question (5): Are there explicit formulas for $f(x)$ and $\frac{P(s)}{\zeta(s)}$ expressed in terms of the non-trivial zeta zeros?

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  • $\begingroup$ It's worth pointing out that the characteristic function of the primes, alternately denoted by $\chi_{\mathbb{P}}(n)$, can be expressed as a convolution of the prime $\omega(n)$ function and $\mu(n)$: $$\chi_{\mathbb{P}}(n) = (\omega \ast \mu)(n)$$ (cf. the exercises in Apostol's book). This should demystify some of the notation and secrecy used in stating some of the questions. $\endgroup$ – mds Jun 3 at 3:56
  • $\begingroup$ @mds Thanks for your comment. I was not familiar with this relationship. I updated the question above in an attempt to make it easier to understand without needing to consult the OEIS, but I still included OEIS references for those who might be interested in more information. I think the definition I used for the characteristic function of the primes in the updated question above is easier to comprehend at a glance than the Dirichlet convolution in your comment, but I appreciate your comment as I learned about a new relationship. $\endgroup$ – Steven Clark Jun 3 at 15:58
  • $\begingroup$ It's not an obvious, or well-known relation. I found it myself by scanning through exercises in the back of a textbook which stated this property much more subtly. The point here is that it gives a nice easy to see way to express the Dirichlet series over the function in terms of standard functions. $\endgroup$ – mds Jun 3 at 16:51
  • $\begingroup$ I also believe I misspoke about the convergence of $P(s)$ when $\Re(s) = 1$ but $\Im(s) \neq 0$. If you search around Math Stack Exchange for the prime zeta function, there is apparently a way to show convergence at these points. The bound I gave only holds when the imaginary part is zero. $\endgroup$ – mds Jun 3 at 16:52
  • $\begingroup$ @mds I understand your point about expressing Dirichlet series coefficients in terms of standard functions. Thanks also for the update on the convergence. I was still wondering if that part of your answer was correct. $\endgroup$ – Steven Clark Jun 3 at 18:30
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Proof of (4): We have that $f(x) := \sum_{n \leq x} f_0(n)$ is the summatory function of the sequence $f_0(n) := \sum_{d|n} \mu(n/d) (\pi(d) - \pi(d-1))$. So the Dirichlet series of $f_0$ is given by $$D_{f_0}(s) = \sum_{n \geq 1} \frac{f_0(n)}{n^s} = \frac{1}{\zeta(s)} \times \sum_{p\mathrm{\ prime}} p^{-s} = \frac{P(s)}{\zeta(s)},$$ where $P(s)$ is the prime zeta function.

Partial solution to (5): Now by Perron's formula, we obtain that for integers $x \geq 1$, $$f(x) - \frac{1}{2}f_0(x) = {\sum_{n \leq x}}^{\prime} f_0(n) = \frac{1}{2\pi\imath} \int_{c-\imath\infty}^{c+\imath\infty} D_{f_0}(s) \frac{x^s}{s} ds$$, where $c > \sigma_{f_0}$ makes the Dirichlet series absolutely convergent. The integrand of the previous equation has poles at 1) $s := 0$; 2) $s := \frac{1}{k}$ for all integers $k \geq 1$ (by the representation of $P(s)$ by a series over $\log \zeta(s)$); and 3) at all of the zeros of $\zeta(s)$. Thus, if we were able to apply the residue theorem (see my request for an answer), AND there were only singularities of the integrand at the negative even integers (corresponding to the trivial zeros of $\zeta(s)$), then we might get a formula that looks something like this: $$\require{cancel}{\sum_{n \leq x}}^{\prime} f_0(n) = -\lim_{s \rightarrow 0} 2 P(s) + \sum_{\rho} \frac{P(\rho) x^{\rho}}{\rho \cdot \zeta^{\prime}(\rho)} - \sum_{n \geq 1} \cancel{\frac{P^{\ast}(-2n)}{2n \cdot x^{2n} \zeta^{\prime}(-2n)}} + \sum_{k \geq 1} \operatorname{Res}_{s=\frac{1}{k}}\left[\frac{P(s)}{\zeta(s)}\right].$$

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  • $\begingroup$ By proof of (4), I believe you meant proof of formula (4) versus question (4)? With respect to partial solution to (5), I'm wondering if the relationship $\log\zeta(s)=s\int\limits_0^\infty\frac{\pi(x)}{x\,\left(x^s-1\right)}\,dx\,,\Re(s)>1$ can somehow be inverted. $\endgroup$ – Steven Clark Jun 2 at 16:01
  • $\begingroup$ Question 4 follows from Merten's theorem: $\sum_{p \leq x} p^{-1} = \log\log(x) + B + O\left(\frac{1}{\log x}\right) \rightarrow \infty$ as $x \rightarrow \infty$. So the answer to question (4) is "no". $\endgroup$ – mds Jun 2 at 16:04
  • $\begingroup$ RE: Inversion Problem: I believe it can. It should be common knowledge (i.e., the proof is explicitly given on the MathWorld page for the prime zeta function) that $$P(s) = \sum_{k \geq 1} \frac{\mu(k)}{k} \log \zeta(ks).$$ The integral representation you have written just looks like a manipulation of the formula $$P(s) = s \cdot \int_1^{\infty} \frac{\pi(x)}{x^{s+1}} dx.$$ The problem is that this series for $P(s)$ makes $s := 1/k$ a pole of the function for all integers $k \geq 1$. Due to density of these singularities around zero, zero is a natural boundary, so no continuation methods. $\endgroup$ – mds Jun 2 at 16:18
  • $\begingroup$ My thinking was for the relationship I illustrated, $\log(\zeta(s))$ avoids the problem where you're stuck. In a couple of my other questions (math.stackexchange.com/q/3196256 and math.stackexchange.com/q/3200866) I illustrate explicit formulas for $\pi(x)$ derived from explicit formulas for the Mertens function $M(x)$, $\Pi(x)$, and $\psi(x)$, but these are not the kind of formulas I was looking for here, and I suspect they're probably not useful with respect to your investigation either. $\endgroup$ – Steven Clark Jun 2 at 17:04
  • $\begingroup$ In my other linked post, I'm trying to evaluate a symmetric (in the real part) contour integral over a function that cannot be continued at or left of zero. The function $\log \zeta(s)$ you are suggesting seems to inherit a whole different set (say) of complications: at the zeros of $\zeta(s)$ we have definite poles. $\endgroup$ – mds Jun 2 at 17:20

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