A Sieve for Twin Primes

TL;DR

This paper introduces a new sieve algorithm for twin primes, provides heuristic estimates for their counts, and compares these estimates with actual data, offering a simpler correction factor than previous methods.

Contribution

The paper presents a novel sieve algorithm for twin primes and introduces a simpler heuristic correction factor for estimating their distribution.

Findings

The sieve algorithm effectively generates twin primes.

Heuristic estimates closely match actual counts up to 8009.

A new correction factor simplifies twin prime distribution estimates.

Abstract

We present an algorithm analogous to the sieve of Eratosthenes that produces the list of twin primes. Next, we count the number of twin primes resulting from the construction with two different heuristic arguments. The first method is essentially the same as the one in Hardy and Wright. However, the second method is novel. It results in the same asymptotic formula but it uses a simpler correction factor than theirs. Though we have no theory for the accuracy of our estimates, we compute them both without and with the correction factor and they turn out to be close to the actual counts up to 8009.

Tables

Table 1. Table 1: The Set T 5 .

5	7
11	13
17	19
23	25
29	31

Table 2. Table 2: The Set T 7 .

11	13	17	19	29	31
41	43	47	49	59	61
71	73	77	79	89	91
101	103	107	109	119	121
131	133	137	139	149	151
161	163	167	169	179	181
191	193	197	199	209	211

Table 3. Table 3: The Set T 11 .

11	13	17	19	29	31	41	43	59	61	71	73	101	103	107	109	137	139	149	151	167	169	179	181	191	193	197	199	209	211
221	223	227	229	239	241	251	253	269	271	281	283	311	313	317	319	347	349	359	361	377	379	389	391	401	403	407	409	419	421
431	433	437	439	449	451	461	463	479	481	491	493	521	523	527	529	557	559	569	571	587	589	599	601	611	613	617	619	629	631
641	643	647	649	659	661	671	673	689	691	701	703	731	733	737	739	767	769	779	781	797	799	809	811	821	823	827	829	839	841
851	853	857	859	869	871	881	883	899	901	911	913	941	943	947	949	977	979	989	991	1007	1009	1019	1021	1031	1033	1037	1039	1049	1051
1061	1063	1067	1069	1079	1081	1091	1093	1109	1111	1121	1123	1151	1153	1157	1159	1187	1189	1199	1201	1217	1219	1229	1231	1241	1243	1247	1249	1259	1261
1271	1273	1277	1279	1289	1291	1301	1303	1319	1321	1331	1333	1361	1363	1367	1369	1397	1399	1409	1411	1427	1429	1439	1441	1451	1453	1457	1459	1469	1471
1481	1483	1487	1489	1499	1501	1511	1513	1529	1531	1541	1543	1571	1573	1577	1579	1607	1609	1619	1621	1637	1639	1649	1651	1661	1663	1667	1669	1679	1681
1691	1693	1697	1699	1709	1711	1721	1723	1739	1741	1751	1753	1781	1783	1787	1789	1817	1819	1829	1831	1847	1849	1859	1861	1871	1873	1877	1879	1889	1891
1901	1903	1907	1909	1919	1921	1931	1933	1949	1951	1961	1963	1991	1993	1997	1999	2027	2029	2039	2041	2057	2059	2069	2071	2081	2083	2087	2089	2099	2101
2111	2113	2117	2119	2129	2131	2141	2143	2159	2161	2171	2173	2201	2203	2207	2209	2237	2239	2249	2251	2267	2269	2279	2281	2291	2293	2297	2299	2309	2311

Table 4. Table 4: Count of Twin Primes between p and p 2

p	Actual	Equation 7	r	Equation 15
101	404	394	1.03975	410
199	1150	1143	1.01694	1162
307	2288	2332	0.99588	2323
401	3578	3618	0.99050	3683
503	5170	5263	0.98667	5193
601	6974	7103	0.98036	6964
701	8946	9186	0.97882	8992
797	11128	11426	0.97493	11140
907	13674	14223	0.97287	13837
1009	16556	17053	0.97038	16548
1999	53556	55038	0.96144	52916
3001	107610	111342	0.95734	106592
4001	176914	184081	0.95390	175595
5003	261086	272412	0.95202	259343
6007	358978	375972	0.95005	357192
7001	469528	492326	0.94945	467437
8009	594636	625062	0.94773	592388

Equations

$$d_{p_{n-1}}=\frac{2{\prod }_{k=2}^{n-1}\left(p_k-2\right)}{p_{n-1}\#}=\prod _{k=2}^{n-1}\frac{p_k-2}{p_k}.$$

$$N_p\left(5\right)\approx 2\cdot \frac{\pi \left(p^2\right)}{\phi \left(30\right)}+1\approx \frac{2}{\left(3-1\right)\left(5-1\right)}\pi \left(p^2\right)=\frac{1}{4}\pi \left(p^2\right)$$

$$N_p\left(7\right)\approx 2\left(5-2\right)\frac{\pi \left(p^2\right)}{\phi \left(210\right)}=\frac{2\left(5-2\right)}{\left(3-1\right)\left(5-1\right)\left(7-1\right)}\pi \left(p^2\right)=\frac{1}{8}\pi \left(p^2\right)$$

$$N_p\left(11\right)$$

$$=\frac{\left(5-2\right)\left(7-2\right)}{\left(5-1\right)\left(7-1\right)\left(11-1\right)}\pi \left(p^2\right)=\frac{1}{16}\pi \left(p^2\right)$$

$$N_p\left(p_k\right)$$

$$=\prod _{i=3}^{k-1}\frac{p_i-2}{p_i-1}\cdot \frac{1}{p_k-1}\pi \left(p^2\right)=\prod _{i=3}^{k-1}\left(1-\frac{1}{p_i-1}\right)\cdot \frac{1}{p_k-1}\pi \left(p^2\right)$$

$$=\left[\prod _{i=3}^{k-1}\left(1-\frac{1}{p_i-1}\right)-\prod _{i=3}^{k-1}\left(1-\frac{1}{p_i-1}\right)\cdot \left(1-\frac{1}{p_k-1}\right)\right]\pi \left(p^2\right)$$

$$=\left[\prod _{i=3}^{k-1}\left(1-\frac{1}{p_i-1}\right)-\prod _{i=3}^k\left(1-\frac{1}{p_i-1}\right)\right]\pi \left(p^2\right).$$

$$\sum _{k=3}^n{N}_{p_n}\left(p_k\right)\approx \left[1-\prod _{i=3}^n\left(1-\frac{1}{p_i-1}\right)\right]\pi \left(p_n^2\right),$$

$$\prod _{i=3}^n\left(1-\frac{1}{p_i-1}\right)\pi \left(p_n^2\right)=\prod _{i=3}^n\left(\frac{p_i-2}{p_i-1}\right)\pi \left(p_n^2\right).$$

$$\pi \left(p_n^2\right)=r{p}_n^2\prod _{i=1}^n\frac{p_i-1}{p_i}.$$

$$r=\frac{\pi \left(p_n^2\right)}{p_n^2}\prod _{i=1}^n\frac{p_i}{p_i-1}.$$

$$r=\frac{\pi \left(x\right)}{x}\prod _{p\le \sqrt{x}}\frac{p}{p-1}.$$

$$\prod _{p\le \sqrt{x}}\frac{p-1}{p}\sim \frac{2e^{-\gamma }}{\log x},$$

$$\pi \left(x\right)\sim \frac{x}{\log x}.$$

$$r\sim \frac{1}{\log x}\cdot \frac{\log x}{2e^{-\gamma }}=\frac{e^{\gamma }}{2}.$$

$${\pi }_2\left(x\right)$$

$$=\frac{{\left(\pi \left(x\right)\right)}^2}{x}\cdot 4\prod _{3\le p\le \sqrt{x}}\frac{p\left(p-2\right)}{{\left(p-1\right)}^2}.$$

$${\pi }_2\left(x\right)\sim \frac{4x}{{\log }^{2}x}\prod _{3\le p}\frac{p\left(p-2\right)}{{\left(p-1\right)}^2},$$

Full text

Next, we count the number of twin primes resulting from the construction with two different heuristic arguments. The first method is essentially the same as the one in [HW]. However, the second method is novel. It results in the same asymptotic formula but it uses a simpler correction factor than [HW].

Though we have no theory for the accuracy of our estimates, we compute them both without and with the correction factor and they turn out to be close to the actual counts up to $8009^{2}$.

A famous unproved problem in number theory is the twin prime conjecture, which says that the number of twin  primes, that is, primes that differ from each other by 2, is infinite. A stronger form, still not completely proved, due to Hardy and Wright (See [HW], p. 372.) gives an asymptotic formula for the number of twin primes under a real number $x$ as $x\rightarrow\infty.$

In this paper we present a new approach to the problem by proposing a new kind of sieve, which produces the twin primes, and we use it to count their number under $x$ with two different heuristic arguments. The first method is essentially the same as the one in [HW]. However, the second method is novel. It is based on Dirichlet’s theorem on primes in arithmetic progressions (see e.g. [A], p. 146), and results in the same asymptotic formula as the one given in [HW], but it uses a simpler correction factor than theirs. Though we have no theory for the accuracy of our estimates, we compute them, both without and with the correction factor, and they turn out to be close to the actual counts up to $8009^{2}$.

1 The Double Sieve.

The following property of twin primes is well known:

Lemma 1

All twin primes greater than $3$ are of the form $6n+1$ and $6n-1$ for the same $n\in\mathbb{N}.$

Proof. Clearly, the numbers greater than $3$ congruent to 0,2,3, and $4\operatorname{mod}6$ are composite, and so all primes greater than $3$ must be of the form $6n\pm 1.$ Now for the same $n$ the two numbers $6n\pm 1$ differ by 2, but for $m\neq n,$ $6m\pm 1$ and $6n\pm 1$ differ by more than 2.

Since we want to study twin primes and those are of the form $6n\pm 1$, we sift the set of such numbers. For the sake of convenience we call the two numbers $6n\pm 1$, for any positive integer $n,$ twins of each other, regardless of whether they are primes or not.

For any prime $p\geq 5,$ let $L_{p}$ denote the set of all positive integers of the form $6n\pm 1$ obtained by sifting out those with prime factors less than $p$ together with their twins.

We construct these sets successively as follows.

First, $L_{5}=$ $\left\{6n\pm 1|n\in\mathbb{N}\right\},$ because this is the set of all positive numbers with prime factors $\geq 5.$ However, with an eye on the upcoming construction of $L_{7},$ we write $L_{5}$ as a table $T_{5}$ of two columns and five rows (see Table 1.), with the numbers $6n-1$ in the first column and the numbers $6n+1$ in the second column for $n=1,\ldots,5,$ and consider the rest of $L_{5}$ as the numbers of this table $+30k$ for $k=1,2,\ldots.$ Here $30=5\#.$111$n\#$ $=\prod_{p\leq n}p=n$-primorial

To construct $L_{7}$, we proceed as follows: In $T_{5}$, we delete the multiples of 5, that is, 5 and 25 (shown in dark orange) and the twins of these multiples, that is, 7 and 23.(shown in light orange), and in $L_{5}$ we delete all the numbers in the residue classes of these four numbers $\operatorname{mod}30.$ Notice that, apart from the number 5, among the deleted residue classes only those of 7 and 23 contain primes, and, except for 7, these are nontwin primes, because their twins are multiples of 5.

$T_{5}$ has 5 rows and 2 columns, and in each column we delete 2 numbers. So, we have $2\left(5-2\right)=6$ undeleted numbers left. We write a table $T_{7}$ for $L_{7}$ (see Table 2) consisting of six columns headed by the six numbers that remained undeleted in $L_{5},$ and seven rows that are obtained by continuing the residue classes $\operatorname{mod}30$ of the headers of the columns. $L_{7}$ is the set of the numbers of this table $+210k$ for $k=1,2,\ldots.$ Here $210=7\#.$

To construct $L_{11}$, we proceed similarly: In $L_{7}$, we delete the multiples of 7 (shown in dark orange in $T_{7}$) and the twins of these multiples (shown in light orange), plus all the numbers in the residue classes of these twelve numbers $\operatorname{mod}7\#=\operatorname{mod}210.$ Thus the deleted numbers include all the nontwin primes that are twins of multiples of 7 and no other primes. $T_{7}$ has 7 rows and 6 columns, and, by the Chinese remainder theorem, in each column we have a multiple of 7 and a twin of a multiple of 7, which we delete. So, in $T_{7}$ we have $2\left(5-2\right)\left(7-2\right)=30$ undeleted numbers left.

From the undeleted numbers in $T_{7}$, we form the first row of $T_{11},$ shown in Table 3. Below this row we write 10 more rows by adding $210k$ for $k=1,2,\ldots,10$ to each header. $T_{11}$ is the first complete set of residues $\operatorname{mod}11\#=\operatorname{mod}2310$ of $L_{11}.$

We continue building $L_{p}$ for all successive primes in a similar fashion.

Lemma 2

For $p>3,$ every number $q<p^{2}-2$ in $L_{p}$ must be a twin prime, and every twin prime $\geq p$ is in $L_{p}.$

Proof. Consider $L_{p}$ for some given $p.$ Then $p^{2}$ is the smallest composite number with all factors $\geq p,$ and so no composite $q<p^{2}$ can be in $L_{p}.$

If $q<p^{2}-2$ is a nontwin prime $>3,$ then its twin is composite and $<p^{2},$ hence $q$ cannot be in $L_{p}.$

So, if there are numbers in $L_{p}$ under $p^{2}-2,$ then they must be twin primes.

Furthermore, if $q$ is a twin prime $\geq p,$ then it is a member of $L_{p}.$ This is so, because in the construction of $L_{p}$ we have sifted out all multiples of the primes $<p$ and their twins, but no twin prime $\geq p$ is one of those.

Thus, our construction provides a sieve for the twin primes analogous to the sieve of Eratosthenes, as illustrated in Tables 1-3, which start with the twin primes between $p$ and $p^{2}-2,$ for $p=5,7,11.$ We have no proof, however, that the construction will give twin primes for every $p,$ no matter how large. We have proved only that if there are numbers between $p$ and $p^{2}-2$ in $L_{p},$ then they are the twin primes of that interval.

Based on the construction above, in the next two sections we shall estimate the number of twin primes in $L_{p}$ under $p^{2}-2$ in two ways. First, by counting the number of deletions in each step of the construction and computing the density of the undeleted numbers left that way.

The second way of counting the number of twin primes in $L_{p}$ under $p^{2}-2,$ is by counting the number of primes deleted under $p^{2}$ in the steps of the construction of $L_{p}$ and subtracting that from the total number $\pi\left(p^{2}\right)$ of primes under $p^{2}.$

2 Counting all deletions under $p^{2}.$

In $T_{5}$ we delete $2$ entries in each column, and so we keep $2\left(5-2\right)=6$ entries out of $5\#=30$ numbers. Hence the density of the undeleted numbers in $T_{5}$ is $d_{5}=2\left(5-2\right)/5\#=1/5.$ These numbers become the entries of $T_{7,}$ and so $d_{5}$ is also the density of $T_{7}$ and of $L_{7}.$

In $T_{7}$ we delete $2$ entries out of 7 in each column, and so the density of the undeleted numbers in $T_{7}$ becomes $d_{7}=\left(7-2\right)d_{5}/7=2\left(5-2\right)\left(7-2\right)/7\#=1/7,$ which is then also the density of $T_{11}$ and of $L_{11}.$ Continuing in this way, and writing $p_{n}$ for the $n$th prime, we obtain the density of $T_{p_{n}}$ and $L_{p_{n}},$ for $n=3,4,\ldots$ as

[TABLE]

This density is essentially the same222[HW] gives the number of pairs, we give the density of individual primes. as the one on p. 372 in [HW] and has the same shortcoming, namely that it is the density of $L_{p_{n}}$ over the huge interval $\left[1,p_{n}\#\right],$ but we need the density of $L_{p_{n}}$ over the much shorter interval $\left[p_{n},p_{n}^{2}\right],$ where it is the same as the density of the twin primes there. As detailed in [HW], the corresponding ratio of the number of all primes relative to $p_{n}\#$ in the short interval to that in the long interval is known to be about $e^{\gamma}/2,$ by Mertens’ theorem. (Thm. 429, lc. Here $\gamma$ is Euler’s constant.) Thus it is conjectured there that for twin primes the corresponding correction factor should be $\left(e^{\gamma}/2\right)^{2}.$ This amounts to assuming the statistical independence of the two numbers in a twin pair. (See [P].)

We try to avoid this problem by counting just the primes deleted under $p_{n}^{2}$ in the construction, since those are all nontwin primes. We do this in the next section. Unfortunately, however, we have no theory to determine the accuracy of our approximation. Empirically it turns out to be very good, though, and with a plausible correction factor, which has no square as the one in the first method, it becomes the same as the one above.

3 Counting deleted primes.

Let $N_{p}\left(p_{k}\right)=$ number of primes under $p^{2}$ deleted from $L_{p_{k}}$ when we build $L_{p_{k+1}}$ from $L_{p_{k}}$ for $3<p_{k}<p.$

In $L_{5}$ there are 2 columns in the first period, and in each column there is one entry that is the twin of a multiple of 5, namely 23 in the first column and 7 in the second. Thus, the primes that we delete from $L_{5}$ are the primes under $p^{2}$ of the two residue classes $7\operatorname{mod}30$ and $23\operatorname{mod}30$ plus the single prime $5.$ According to Dirichlet’s theorem on primes in arithmetic progressions (see e.g. [A], p. 146), each residue class that is relatively prime to the modulus has an approximately equal share of the $\pi\left(p^{2}\right)$ primes under $p^{2},$ that is, about $\pi\left(p^{2}\right)/\varphi\left(30\right)$ primes. Thus, for large $p$ we delete

[TABLE]

primes from $L_{5}$ under $p^{2}$. This number is just an approximation, because the fraction is correct only over integer multiples of the period and Dirichlet’s theorem is only asymptotically true. The same considerations apply to the estimates below as well.

Similarly, from $L_{7}$ under $p^{2}$ we delete about

[TABLE]

primes, because $L_{7}$ has a period of 210, the first period has $2\left(5-2\right)=6$ columns, and in each column we delete one multiple of $7$ and one twin of such a multiple, which is prime to 210. (All entries of $L_{7}$ are prime to $2,3,$ and $5,$ and the twin of a multiple of 7 in $L_{7}$ is prime also to 7, and so to $7\#.$)

From $L_{11}$ we delete about

[TABLE]

primes under $p^{2}$.

Similarly, for general $p_{k}<p,$ with $k\geq 4,$ the number of primes deleted from $L_{p_{k}}$ under $p^{2}$  is

[TABLE]

Thus, if we sum these expressions over $k,$ we get a telescoping sum and the total number of primes deleted under $p_{n}^{2}$ is

[TABLE]

and so the number of twin primes left in $L_{p_{n}}$ under $p_{n}^{2},$ that is, in the interval $\left[p_{n},p_{n}^{2}\right],$ is about

[TABLE]

The second expression in Equation 7 can also be interpreted as the ratio of the number of twins relatively prime to $p_{n}\#$ to that of all numbers relatively prime to $p_{n}\#$ in the interval $[5,p_{n}\#],$ times the number of primes in $[1,p_{n}^{2}],$ (which has the same order of magnitude as that in $\left[p_{n},p_{n}^{2}\right]).$ Thus, while we did not assume that the density of twin primes is the same in the short interval as in the long one, we obtained a result that is equivalent to the ratio of the densities of the twin relative primes and of all relative primes being the same in the two intervals. (Clearly, in $\left[p_{n},p_{n}^{2}\right]$ the numbers relatively prime to $p_{n}\#$ are the same as the primes there.)

Apparently, we can improve the estimate above by applying the same correction factor $r$ that appears when comparing the numbers of all relative primes in the two intervals:

[TABLE]

Here $p_{n}^{2}$ is the length of the short interval and the product is the density of all relative primes in the long interval. Thus, $p_{n}^{2}\prod_{i=1}^{n}\frac{p_{i}-1}{p_{i}}$ would be the number of primes in the interval$\ [1,p_{n}^{2}],$ if the density there were the same.

Hence

[TABLE]

Denoting $p_{n}^{2}$ by $x,$ we can also write

[TABLE]

By Mertens’ theorem,

[TABLE]

and by the prime number theorem,

[TABLE]

Thus,

[TABLE]

Applying the correction factor $r$ to the estimate in Equation 7, we get

[TABLE]

In Table 4 we compare the estimate in Equation 7 and this estimate to the actual count of twin primes in the interval $\left[p,p^{2}\right]$. As one can see, the numbers from Equation 7 are pretty close to the actual counts, but those from Equation 15 are much closer. (See also Fig. 1.)

If we apply the asymptotic estimate $\pi\left(x\right)\sim x/\log x$ for the number of all primes under $x,$ then from Equation 15 we obtain

[TABLE]

as an asymptotic estimate for the number of twin primes under $x.$ This result equals the estimate in Equation 22.20.1 in [HW], where the number of pairs is counted, accounting for the factor of 2 in place of our 4 for the count of the individual primes.