# "Squeezing" the primes?

The logical idea here is to map a curve that encodes the primes into the region $$(0,1)^2$$ and analyze the distribution there more easily and achieve tight bounds.

To assess the distribution of primes, define a curve by the set of points: $$P_t=\bigg(1-\frac{\pi(t)}{t},1-\big(1-t^{-1}\big)\bigg)$$

for $$t\ge10$$ Here $$\pi(t)$$ is the prime counting function.

I'd like to show the following bounds on $$P_t$$ are tight $$\forall t$$ and $$\forall x$$ and as $$t \to \infty$$ and as $$x \to 1^{-}$$:

$$\varphi^2(x)\le P_t\le\varphi(x)$$

where $$\varphi(x)=e^{\frac{1}{\log x}}$$ for $$x\in(0,1).$$

In other words I want to show that the $$P_t$$ curve is squeezed between the $$\varphi(x)$$ and $$\varphi^2(x)$$ curves.

Can this be shown?

I did numerical analysis and the bounds are good for at least $$k=15.$$ Curious if they hold for all $$k$$. I also did some analysis on the inequality itself and reached the conclusion that, yes, the inequality holds everywhere. Still on the fence about tightness.

• You’re comparing an interval to a number and the bounds in x tend to 0 without regard to k. If you fix your mistakes perhaps we could understand your intent better. Commented Sep 1, 2023 at 15:40
• @Charles $P_k$ defines a parametric plot of points. I'm trying to bound all these points between $\varphi(x)$ and $\varphi^2(x)$. Commented Sep 1, 2023 at 15:49
• Here's a diagram in case that helps convey what I'm trying to do: desmos.com/calculator/fq583mmtr6 I've used the logarithmic integral representation here as opposed to the prime counting function. Commented Sep 1, 2023 at 15:54
• I have a formula for a pseudo-CDF (Cumulative Distribution Function) related to the primes. You could evaluate it at $1/x$ to compress it into the interval $0<x<1$, but it only converges when the series evaluation limit is much larger than $x$, so you always end up with an area near zero where the series evaluation doesn't converge. However you can always make this area of non-convergence smaller by making the evaluation limit larger. I can post this alternative approach as an answer if you're interested. Commented Sep 2, 2023 at 22:15
• @StevenClark Yes I'm interested Commented Sep 2, 2023 at 22:17

$$x=1 - \frac{\operatorname{li}(10^s)}{10^s} \approx 1 - \frac{1}{s\log 10}$$ for $$1\le s\le8$$ and then noting that $$\exp\left(\frac{1}{\log x}\right) <10^{-s}< \exp\left(\frac{2}{\log x}\right)$$ or in other words $$\frac{1}{\log x} <-s\log 10< \frac{2}{\log x}.$$

Let's define $$\varepsilon=\varepsilon(s)=\frac{\operatorname{li}(10^s)}{10^s}$$ for conciseness, so $$x=1-\varepsilon.$$

For first intuition, we have $$s\approx\frac{1}{\varepsilon\log 10}$$ and so this is roughly

$$\frac{1}{\log(1-\varepsilon)} <-\frac{1}{\varepsilon}< \frac{2}{\log(1-\varepsilon)}$$

or $$\frac{-1}{\log(1-\varepsilon)} >\frac{1}{\varepsilon}> \frac{-2}{\log(1-\varepsilon)}$$

which is, to the second order, $$\frac{-1}{-\varepsilon-\varepsilon^2/2} >\frac{1}{\varepsilon}> \frac{-2}{-\varepsilon-\varepsilon^2/2}$$

which simplifies to $$0 < \varepsilon < 2.$$

This could be made asymptotically precise and even effective with more work.

This answer describes an alternate approach to "squeezing the primes" (more accurately prime-powers) based on my related Math Overflow question on the inverse Mellin transform

$$p(x)=\mathcal{M}_s^{-1}\left[-\xi(s)\,\frac{\zeta'(s)}{s\,\zeta(s)^2}\right]\left(\frac{1}{x}\right)=\frac{1}{2 \pi i} \int\limits_{\alpha-i\, \infty}^{\alpha-i\, \infty} \left(\pi^{-\frac{s}{2}}\, (1-s)\, \Gamma\left(\frac{s}{2}+1\right)\, \frac{\zeta'(s)}{\zeta(s)}\right)\, \frac{x^s}{s}\, ds$$ $$=\underset{K\to\infty}{\text{lim}}\left(\sum\limits_{k=1}^K \Lambda(k) \left(\frac{2 \pi k^2}{x^2}-1\right) e^{-\frac{\pi k^2}{x^2}}\right)\tag{1}$$

where $$\xi(s)$$ is the Riemann Xi function, $$\Lambda(k)$$ is the von Mangoldt function, and the related Mellin transform

$$P(s)=s\, \mathcal{M}_x[p(x)](-s)=s \int\limits_0^\infty p(x)\, x^{-s-1}\, ds$$ $$=\pi^{-\frac{s}{2}}\, (1-s)\, \Gamma\left(\frac{s}{2}+1\right)\,\frac{\zeta'(s)}{\zeta(s)}\,,\quad\Re(s)>1\tag{2}$$

converges for $$\Re(s)>1$$.

The residue formula for $$p(x)$$, which I believe converges for $$x>0$$, is

$$p_o(x)=\log(2 \pi)+p_{-2n}(x)+p_\rho(x)\tag{3}$$

where $$p_{-2n}(x)$$ and $$p_\rho(x)$$ defined in formulas (4) and (5) below are sums over the trivial and non-trivial zeta zeros respectively.

$$p_{-2n}(x)=\sum\limits_{n=1}^{\infty} \frac{x^{-2 n} (-\pi)^n \left(\zeta'(-2 n)+n (2 n+1) \left(\zeta''(-2 n)+\zeta'(-2 n) (\psi^{(0)}(n)+2 \log(x)-\log(\pi))\right)\right)}{2 n\, n!\, \zeta'(-2 n)}\tag{4}$$

$$p_\rho(x)=\sum\limits_{\rho} \left(\pi^{-\frac{\rho}{2}} \, \left(1-\rho\right)\, \Gamma\left(\frac{\rho}{2}+1\right)\, \frac{x^{\rho}}{\rho}\right)\tag{5}$$

Figure (1) below illustrates formula (1) above for $$p(x)$$ (blue curve) seems to converge to $$\log(2 \pi)$$ (orange horizontal dashed line) as $$x\to\infty$$ which is consistent with the leading term in the explicit formula for $$p_o(x)$$ defined in formula (3) above.

Figure (1): Illustration of formula (1) for $$p(x)$$

The function $$p(x)$$ defined in formula (1) above is an even function of $$x$$, but Figure (1) above originally seemed to suggest to me the function

$$f(x)=\left\{\begin{array}{cc} 0 & x<0 \\ \frac{p(x)}{\log(2 \pi)} & x\ge 0 \\ \end{array}\right.\tag{6}$$

could potentially qualify as a CDF (Cumulative Distribution Function) related to the distribution of the primes. But I now suspect the oscillation in $$p_\rho(x)$$ (defined in formula (5) above) illustrated in figure (2) below causes $$f'(x)$$ to have an infinite number of $$0$$ crossings which also perhaps eventually leads to $$f(x)$$ having an infinite number of $$1$$ crossings.

Figure (2) below illustrates formula (5) above for $$p_\rho(x)$$ evaluated over the first $$100$$ pairs of non-trivial zeta-zeros which is largely dominated by the contribution of the first pair of non-trivial zeta-zeros. Note the oscillation in $$p_\rho(x)$$ seems to grow in magnitude as $$x$$ increases.

Figure (2): Illustration of formula (5) for $$p_\rho(x)$$

I've been struggling to define an accurate and simple asymptotic for $$p(x)$$ defined in formula (1) above, but the only accurate asymptotic I've come up with is $$\log(2 \pi)+p_{-2n}(x)$$ which is based on the residue formula for $$p(x)$$ illustrated in formula (3) above.

I'm investigating a potential pseudo-closed-form (meaning non-elementary functions are acceptable) for the function $$p_{-2n}(x)$$ defined in formula (4) above (which was the subject of my related Math StackExchange question) and have made some progress. I believe $$p_{-2n}(x)$$ defined in formula (4) above can also be evaluated as illustrated in formula (7) below where I use the series representation in formula (10) below to evaluate $$g(x)$$ and the pseudo-closed-forms in formulas (8) and (9b) below to evaluate $$r(x)$$ and $$h(x)$$, but this is not yet a closed-form much less simple.

$$p_{-2n}(x)=r(x)+h(x)+g(x)\tag{7}$$

$$r(x)=-\frac{1}{2 x^4} \left(3 \pi x^2 \text{1F1}^{(1,0,0)}\left(1,2,-\frac{\pi }{x^2}\right)-\pi ^2 \text{1F1}^{(1,0,0)}\left(2,3,-\frac{\pi }{x^2}\right)+x^4 \left(E_1\left(\frac{\pi }{x^2}\right)+\log \left(\frac{1}{x^2}\right)+2 \log (x)-2\right)+e^{-\frac{\pi }{x^2}} x^2 \left(x^2 (\gamma +2+\log (\pi ))+\left(4 \pi -2 x^2\right) \log (x)-2 \pi (\gamma -1+\log (\pi ))\right)\right)\tag{8}$$

$$h(x)=\sum\limits_{n=1}^\infty \frac{(-1)^n\, (2 n+1)}{n!}\, \pi^n \left(\gamma+\log(2 \pi)-H_{2 n}\right)\, x^{-2 n}\tag{9a}$$ $$=\frac{1}{2 x^2} e^{-\frac{\pi }{x^2}} \left(x^2 \, _1F_1^{(1,0,0)}\left(0,\frac{1}{2},\frac{\pi }{x^2}\right)-2 \pi \, _1F_1^{(1,0,0)}\left(0,\frac{3}{2},\frac{\pi }{x^2}\right)+e^{\frac{\pi }{x^2}} \left(2 \pi \, _1F_1^{(1,0,0)}\left(2,2,-\frac{\pi }{x^2}\right)-x^2 \, _1F_1^{(1,0,0)}\left(1,1,-\frac{\pi }{x^2}\right)\right)-2 \left(e^{\frac{\pi }{x^2}}-1\right) x^2 (\gamma +\log (2 \pi ))+\pi (-4 \gamma +6-4 \log (2 \pi ))\right)\tag{9b}.$$

$$g(x)=\underset{N\to\infty}{\text{lim}}\left(\sum\limits_{n=1}^N \frac{(-1)^{n+1}\, (2 n+1)}{n!}\, \pi^n\, \frac{\zeta'(2 n+1)}{\zeta(2 n+1)} x^{-2 n}\right)\tag{10}$$

Figure (3) below illustrates formula (6) for $$f\left(\frac{1}{x}\right)$$ above using formula (1) above for $$p(x)$$ evaluated at $$K=1000$$.

Figure (3): Illustration of formula (6) for $$f\left(\frac{1}{x}\right)$$ using formula (1) for $$p(x)$$

Figure (4) below illustrates formula (6) for $$f\left(\frac{1}{x}\right)$$ above using formula (3) above for $$p_o(x)$$ where formula (4) is evaluated over the first $$100$$ trivial zeta-zeros and formula (5) is evaluated over the first $$100$$ pairs of non-trivial zeta-zeros.

Figure (4): Illustration of formula (6) for $$f\left(\frac{1}{x}\right)$$ using formula (3) for $$p_o(x)$$

Figure (5) below illustrates formula (1) above for $$p(x)$$ evaluated at $$K=100$$ in blue and $$K\in\{9,11,13,16,17\}$$ in orange through brown where the blue horizontal dashed line is at $$\log(2 \pi)$$. Note the contribution of each prime-power comes along just in time to temporarily "boost support" of formula (1) for $$p(x)$$ until the contribution of the next prime-power. Omitting the contribution of a single prime-power or adding the contribution of a non-prime-power will cause a significant perturbation in the evaluation of formula (1) for $$p(x)$$.

Figure (5): Illustration of formula (1) for $$p(x)$$ evaluated at $$K=100$$ in blue and $$K\in\{9,11,13,16,17\}$$ in orange through brown

I'll note that formula (1) for $$p(x)$$ only converges when $$K\gg x$$, and when holding $$K$$ constant formula (1) for $$p(x)$$ converges to $$-\psi(K)$$ as $$x\to\infty$$ where

$$\psi(x)=\sum\limits_{n\le x} \Lambda(n)\tag{11}$$

is the second Chebyshev function.