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If we define $$\pi^2(N)=\vert [p: p\leq N, p\in\mathbb{P}, p-2\in\mathbb{P}]\vert$$ where $\mathbb{P}$ is the set of all primes (as the number of twin primes less than $N$), and we define $$\pi^D(N)=\vert [p: p\leq N, p\in\mathbb{P}, Dp-2\in\mathbb{P}]\vert$$

Can anyone think of how to show that for $D$ prime and large enough with respect to $N$ and $N$ large enough, we have $\pi^2(N)\geq \pi^D(N)$. It seems heuristically true, simply because the probability, given a prime $p$, that $Dp-2$ is also prime is less than the probability that $p-2$ (which is smaller than $Dp-2$ and thus within a denser area of primes) is also prime.
Sieve methods allow us to conclude that the upper bounds of both sets are multiples of each other, but not much else. Any ideas?

flag	asked Mar 22 2012 at 16:47 Alex Botros 225●1●5

I see that $p-2$ is more often prime than $2p-2$, but why is $p-2$ more often prime than $3p-2$? – Woett Mar 22 2012 at 23:25

That's the interesting thing: computationally $3p-2$ is far more often than $p-2$ or any other $Dp-2$, and it was shown that there are infinitely many prime tuples $p,3p-2$. It seems, however, than if $N$ is sufficiently large, and $D$ is bigger than $N^{3/8}$ say, then the above holds. I just don't know why. – Alex Botros Mar 23 2012 at 0:02

Ah, I get it. But it was shown that there are infinitely many prime tuples $p, 3p-2$? Really? – Woett Mar 23 2012 at 1:23

3p-2 is never a multiple of 3. That is a big advantage over p+2. I'd expect close to twice as many 3p+2 are prime (up to x) as p+2. – Aaron Meyerowitz Mar 23 2012 at 4:02

1

I would be interested in a reference for this claim that there is an infinite number of prime pairs $(p, 3p -2).$ – Igor Rivin Mar 23 2012 at 4:29

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Aaron Meyerowitz · Accepted Answer · 2012-03-25 15:18:51Z UTC

What you are calling $\pi^2(N)$ should really be $\pi^1(N).$ It is slightly unfortunate that this is the number of twin primes up to $N$ which is commonly written as $\pi_2(N),$ but that is just notation. There is no reason to think that $D$ being prime will make a difference. It will be clearer to let $D$ be any nonegative integer and

$$\pi^D(N)=\begin{cases} \vert [p: p\leq N, p\in\mathbb{P}, Dp-2\in\mathbb{P}]\vert& \text{if $D$ is odd},\\ \vert [p: p\leq N, p\in\mathbb{P}, Dp-1\in\mathbb{P}]\vert& \text{if $D$ is even}. \end{cases}$$

For "large" $N$ relative to $D$ there will be hardly any difference between the density of primes near $N$ and near $DN:$ $$\frac1{\ln(N)}-\frac1{\ln(DN)}<\frac{\ln(D)}{(\ln{N})^2}.$$ What should really matter is the set of odd primes dividing $D$ (the prime $2$ is built in.)

Recall that the number of primes up to $N$ is $\pi(N)\sim \frac{N}{\ln{N}}.$ It is quite true that no one (yet) can prove that $\lim_{N \to \infty}\pi^1(N)=\infty$, but it is known how fast it goes there. Less humorously: it is widely believed that the number of twin prime pairs up to $N$ is $\pi^1(N)=\pi_2(N)=C_2\frac{N}{(\ln{N})^2}$ for the constant $C_2 = \prod_{p\ge 3} \frac{p(p-2)}{(p-1)^2} \approx 0.66016.$

We can discuss the present question with a high confidence that computational results will support us. In the spirit of the amazing paper Heuristic Reasoning in the Theory of Numbers (read it!) one might define $R_D(N)=\frac{\pi^D(N)}{\pi^1(N)}$ and experiment to make conjectures. I will predict that for large enough $N$ $R_2,R_4,R_8 \sim 1$ , $R_3,R_6,R_9,R_{12},R_{18} \sim 2$ and in general $R_D$ will be very close to the $R_d$ from Polya's paper. This will take very large $N$ for larger $D.$ The same phenomenon should be easier to observe for $\pi^D(M,N)$ with the condition on $p$ changed to $M \le p \le N.$

Here is a very small experiment. The first $2000$ primes following $10,000,000$ run from $M=10000019$ to $N=10032181.$ The formula number of twin primes predicts $\pi_2(N)-\pi_2(M)=143.27.$ Here is a sorted list of $[D,\pi^D(M,N)]$ for selected values.

[3, 329], [6, 305], [9, 296], [24, 282], [12, 280], [18, 276], [27, 259], [36, 242], [5, 218], [10, 188], [25, 182], [20, 176], [1, 169], [2, 166], [40, 163], [16, 152], [4, 140], [8, 137], [32, 135]

The values are grouped as predicted. The values for $1,2,4,8,16,32$ were supposed to all be around $143.27.$ The evidence given is fair but maybe not overwhelming. Here are the ratios $\frac{\pi^D}{143.27}$

[3, 2.295882763], [6, 2.128401954], [9, 2.065596650], [24, 1.967899512], [12, 1.953942777], [18, 1.926029309], [27, 1.807397069], [36, 1.688764829], [5, 1.521284020], [10, 1.311933008], [25, 1.270062805], [20, 1.228192603], [1, 1.179344033], [2, 1.158408932], [40, 1.137473831], [16, 1.060711793], [4, .9769713887], [8, .9560362875], [32, .9420795534]

I leave more extensive calculations to the motivated reader.

Woett · Answer 2 · 2012-03-23 02:17:58Z UTC

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If $D$ is not too large, then it is surely the case that $\pi^D(N)$ goes to infinity when $N$ goes to infinity. So if you want to show that $\pi^2(N) \ge \pi^D(N)$, for all large $N$, then you also need to show that $\pi^2(N)$ goes to infinity with $N$. And this would resolve the twin prime conjecture. So no, I don't think that anyone can think of a way to show $\pi^2(N) \ge \pi^D(N)$ in that case (at this moment).

edited Mar 23 2012 at 2:17

answered Mar 23 2012 at 1:38

Woett
1,021●4●16

	There are asymptotics for the number of twin primes modulo standard conjectures, so presumably one can do same for $\pi^D,$ and see what the truth is, whether or not one can actually prove it. – Igor Rivin Mar 23 2012 at 4:28
	It may not be the case that $\pi^D(N)$ goes to infinity with $N$. Consider Brun's sieve: if $\omega(p)$ is the standard sieving function for which $$A_d=A\frac{\omega(d)}{d}+R_d$$ Then notice that the only difference between $\omega(d)$ in the case of $p,Dp-2$ tuples and $p,p-2$ tuples, is the fact that in $Dp-2$ tuples, $\omega(d)$ is also equal to one if $D\vert d$. That's the only difference between the two. Thus, Brun's "main" terms in his upper bound can be shown to be about equal for large enough $D$. THe only problem lies in coming up with a reasonable way to deal with the remainde – Alex Botros Mar 23 2012 at 14:26
	Let me clarify: When I say it may not be the case that $\pi^D(N)$ go to infinity, what I mean is: there may be a way to show a very close relationship between $\pi^D$ and $\pi^2$. THe fact is, we can show that their difference is bounded above by some multiple of $N/log^2(N)$. That's a pretty good start – Alex Botros Mar 23 2012 at 14:29
	I see what you're saying, but, to be honest, that's not a good start. Because unless you prove that the difference between $p^2(N)$ and $p^D(N)$ is bounded by a finite amount (which doesn't sound very reasonable), you either show no asymptotics at all, or you prove that their difference can be arbitrarily large, which is impossible without resolving the twin prime conjecture. – Woett Mar 23 2012 at 15:17
	Naively reasoning, the number of twin primes should be something like $\fracN{\ln^2N}$ or maybe around half that since we can only have the smaller prime in a pair be $2 \mod 3$. The more refined predicted value is below. – Aaron Meyerowitz Mar 25 2012 at 15:13

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Twin primes and D primes

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