While trying to find a lower limit for the number of twin primes I noticed the problem of having to compensate for duplicates. Once I overcame this problem the duplicates of the duplicates became a problem and I saw this was leading nowhere. So I came up with a different approach. I noticed this behavior is similar to sieving primes. After further analyzing it even seems likely that twin sieving is actually sieving primes over the results. This is also not easy to calculate so I decided to simplify to an lower limit. I shifted the result of the first sieve together so that they form one continuous sequence of sieved numbers. Then it became much easier to calculate the number because this is $\pi(\pi(n))$. The results for n up to $10^{18}$ are very promising , so my conjecture is that for large enough n: \begin{align*} \pi_2(n) &> \pi(\pi(n)) &f=\frac{\pi_2(n)}{\pi(\pi(n))} > 1\\ \end{align*} The following table shows results for $n$ up to $10^{18}\\$ $\\ \begin{array}{ | l | l | l | l |} \hline n &\pi(\pi(n)) &\pi_2(n) &f\\ \hline 1000 & 39 & 35 & 0,89744\\ \hline 10000 & 201 & 205 & 1,01990\\ \hline 100000 & 1184 & 1224 & 1,03378\\ \hline 1000000 & 7702 & 8169 & 1,06063\\ \hline 10000000 & 53911 & 58980 & 1,09403\\ \hline 100000000 & 397557 & 440312 & 1,10754\\ \hline 1000000000 & 3048955 & 3424506 & 1,12317\\ \hline 10000000000 & 24106415 & 27412679 & 1,13715\\ \hline 100000000000 & 195296943 & 224376048 & 1,14890\\ \hline 1000000000000 & 1613846646 & 1870585220 & 1,15908\\ \hline 10000000000000 & 13556756261 & 15834664872 & 1,16803\\ \hline 100000000000000 & 115465507935 & 135780321665 & 1,17594\\ \hline 1000000000000000 & 995112599484 & 1177209242304 & 1,18299\\ \hline 10000000000000000 & 8663956207026 & 10304195697298 & 1,18932\\ \hline 100000000000000000 & 76105984161825 & 90948839353159 & 1,19503\\ \hline 1000000000000000000 & 673776962356604 & 808675888577436 & 1,20021\\ \hline \end{array}\\$

For small $n$ $f<1$ because for small n the shifting actually causes less sieving. For n=3541 $\pi(\pi(n))$ and $\pi_2(n)$ are both 94, $f=1$)

After that $f>1$ because primes are far enough apart that the shifting reduces the count.

Proof: We determine the number of twin primes by first counting the number of primes that are $6n-1$ then we check the results for $6n+1$ also being a prime number. The first step is obtained by sieving for primes and dividing this by 2 which gives us: $$\frac{\pi(n)}{2}$$ There are more $6n-1$ primes so this slightly lower than the actual count. Then we start eliminating all $6n+1$ that are not prime: We need to remove all $6n+1$ that are 5 multiples: $25+5 \cdot 6x$ next all 7 multiples etc. This is the same as sieving for primes with the exception that 2 and 3 are not used. This proofs sieving for twin primes gives the same result as sieving for primes twice with the exception of 3 for the second pass (2 was necessary because we needed $\frac{\pi(n)}{2}$. Even with a crude compensation for sieving 3 the result is a much better lower limit, even much better than Hardy-Littlewood.

lower limits for the number of twin primes


  • $f$ seems to converge, it increases less with every decade. Is this true and is it possible to predict a value for $f \lim_{n \to \infty}\\ \\$

  • As $f$ only seems to grow for large $n$, this conjecture makes infinitely many twin primes very likely but is it possible to turn it into a proof?

  • 2
    $\begingroup$ Turning it into a proof would yield a result which is strictly stronger than the twin prime conjecture and strictly weaker than (but similar to) the first Hardy-Littlewood conjecture. So, if possible, it would be really hard (and quite a long shot from this). $\endgroup$
    – user228113
    Jun 27, 2017 at 13:33
  • $\begingroup$ The main error in using π(π(n)) occurs because the second pass there is no need to sieve 3. I made a crude compensation for this and the result is a much better lower limit, even much better than Hardy-Littlewood. $\endgroup$
    – pietfermat
    Jul 7, 2017 at 5:43
  • 1
    $\begingroup$ I don't see any (heuristic) good reason to relate $\pi_2(x)$ with $\pi(\pi(x))$. $\endgroup$
    – reuns
    Jul 7, 2017 at 13:30
  • $\begingroup$ I explained the reason in my proof. Try following the steps necessary to eliminate all $6n\pm1$ composites and you'll see that is the same as sieving for primes and thus they are related. The result also proves the relation why else would it be such a good approximation. I calculated the results up to $10^{16}$ even the smallest deviation from the actual sieving process would result in a very large error in this range. $\endgroup$
    – pietfermat
    Jul 7, 2017 at 13:46
  • 1
    $\begingroup$ What is the question? $\endgroup$
    – Did
    Jul 29, 2017 at 19:45

4 Answers 4


Correct me If I am wrong but I think the issue here is not using the best conjectured asymptotics. I assume you conjecture is for large $n$ :

$$ \pi_2(n) > \pi\left(\frac{\pi(n)}{1 - 1/3}\right).$$

However you compared $ \pi_2$ with

$$ 2 C_2 \frac{n}{\ln(n)^2} .$$

But if you compared with the estimates

$$ 2 C_2 \frac{n}{(\ln(n)-1)^2}, $$


$$ 2 C_2 \int_3^n (\ln(x) - 1/x)^2 d x,$$

I think the magical improvement will go away. Correct me If I am wrong.

Of course I have no proof of the infinitude of prime twins , but If the “ magical improvement “ indeed Goes away , Then i do not see How this is more helpful than the strong prime twins conjecture.

Btw error terms for the estimates ( both yours and any else's ) of about $ O( \ln(n)^2 \sqrt n ) $ are acceptable. This Also implies that such a difference between estimates of such an order is insignificant.


Infinitely many twin primes are very likely anyway, by the Hardy-Littlewood conjecture, which says that, with $C_2=0.6601618158468695739278121100145\ldots$, $$ \pi_2(n)\sim 2C_2 \frac{n}{\log^2(n)}. $$ Of course we also have $\pi(\pi(n))\sim \frac{n}{\log^2(n)}$, see here, so that showing $\pi_2(n)\sim \pi(\pi(n))$ is as hard as showing the Hardy-Littlewood conjecture. If $\pi_2(n)>\pi(\pi(n))$ is true, this will be even harder to show.

  • $\begingroup$ Comparing two sieves is easier than comparing one sieve to a (logarithmic) function. Say we are comparing sieve A and B, when A sieves more than B over the same range it is clear that: results of sieve B > results of sieve A, or am I missing something? $\endgroup$
    – pietfermat
    Jun 29, 2017 at 11:15
  • $\begingroup$ After having received a down vote, I would say, yes, you (and me) are missing something. $\endgroup$ Jul 1, 2017 at 8:55
  • $\begingroup$ Yes, I am missing an answer to my questions. $\endgroup$
    – pietfermat
    Jul 1, 2017 at 9:28
  • 2
    $\begingroup$ The answer is, to be honest, that your ideas about this are nice, but only a proof is what counts. And here you have to do something more. See also the comment by G. Sassatelli. $\endgroup$ Jul 1, 2017 at 11:39
  • 3
    $\begingroup$ And thinking outside the box is done best by those who know the box better than anyone else. Those who don't know what the known paths are, are condemned to follow them without knowing it. $\endgroup$ Jul 7, 2017 at 6:59

Another comment.

The OP used mod 6 ideas.

But if he uses mod 30 ideas he might get closer to the actual value AND the Hardy-Littlewood estimate ?

Thing is, his mod 6 idea is in the middle between them for the data given. So it can not get closer to both for those $n$ !

But for larger $n$ not in the table that might be true, that is IF the Hardy littlewood estimate is correct !



I computed $\mod 30$ and got the smaller value around $1.4$ instead of $1.5$.

I was able to show by induction that INDEED we get the prime twin constant from above as asymptotic.

It follows from sieving

$$ (1-1/2)(1-2/3)(1-2/5)...\mod 2*3*5*...$$ what works extremely well.

So $1.5, 1.4... , ...,2 C_2$ are the values of the constant factor.

So we are dealing with upper bounds in the modified ( how I described ) estimate from the OP.

In the data we seem to get to the prime twin constant from below.

Ofcourse lower bounds are unproven but the upper bound

$$4.5 n/(\ln^2(n) + o(1))$$

has been proven formally in the literature.


Just a comment I wanted to make. Please read the question and answer careful first.

The estimate from the OP uses $\pi((3/2) \pi(n))$ or $\pi(\pi((3/2) n)$, both estimates are asymptotic eventually anyways.

The constant $2 C_2$ from the k-tuple estimate is about $4/3$. Thereby estimating ( when I use $\pi$ ) about $(4/3) \pi(\pi(n)), \pi((4/3) \pi(n)),\pi(\pi((4/3) n)$ where again the estimates are asymptotics to eachother.

But the data from the OP can not even confirm that the number of twins is above $(5/4) \pi(\pi(n))$

The ratio $f$ barely gets above $1.2 ( = 6/5 )$.

Looking at the table none of those asympotics ($4/3,3/2$) get achieved yet.

Apart from focusing on the large numbers alone , I wonder if we have for all $n$

$$ \pi_2(n) < \pi(\pi((3/2) (n+2)) $$

I wonder why the limit

$$F = \lim \frac{\pi_2(n)}{\pi(\pi(n))}$$

Converges so slowly ???

Can anyone explain that ?


The other answers make this seem logical !


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