# On the prime distribution

Yong Zhao Jianqin Zhou111Corresponding author. Email<EMAIL_ADDRESS>

Department of Computer Science, Anhui University of Technology
Ma’anshan 243002, P. R. China

> In this paper, the estimation formula of the number of primes in a given
> interval is obtained by using the prime distribution property. For any prime
> pairs $p>5$ and $q>5$, construct a disjoint infinite set sequence
> $A_{1},A_{2},\ldots,A_{i}.\ldots$, such that the number of prime pairs
> ($p_{i}$ and $q_{i}$, $p_{i}-q_{i}=p-q$) in $A_{i}$ increases gradually,
> where $i>0$. So twin prime conjecture is true. We also prove that for any
> even integer $m>2700$, there exist more than 10 prime pairs $(p,q)$, such
> that $p+q=m$. Thus Goldbach conjecture is true.
>
> Keywords: Prime number; Prime distribution; Twin prime conjecture; Goldbach
> conjecture; Triple prime conjecture
>
> Mathematics Subject Classification 2010: 11N05,11P32

##  1 Introduction

Like Goldbach conjecture, twin prime conjecture is also one of the famous
unsolved problems in number theory. In 1973, Chen [1] proved that for any even
number $h$, there are infinite prime numbers $p$, so that the number of prime
factors of $p+h$ does not exceed 2. In 2008, Green and Tao [2] proved the
existence of arbitrarily long arithmetic progressions in the primes. In 2014,
Zhang [3] proved that bounded gaps between primes are all less than 70
million.

In this paper, the estimation formula of the number of primes in a given
interval is given by using the prime distribution property. For any prime
pairs $p>5$ and $q>5$, construct a disjoint infinite set sequence
$A_{1},A_{2},\ldots,A_{i}.\ldots$, such that the number of prime pairs
($p_{i}$ and $\ q_{i}$, $p_{i}-q_{i}=p-q$) in $A_{i}$ increases gradually,
where $i>0$. So the original conjecture is true. We also prove that for any
even integer $m>2700$, there exist more than 10 prime pairs $(p,q)$, such that
$p+q=m$. Thus Goldbach conjecture is true. For any triple primes $p>7,$ $q>7$
and $r>7$, construct a disjoint infinite set sequence
$A_{1},A_{2},\ldots,A_{i}.\ldots$, such that the number of triple primes
($p_{i}$ $q_{i}$ and $r_{i}$, $p_{i}-q_{i}=p-q$, $p_{i}-r_{i}=p-r$) in $A_{i}$
increases gradually, where $i>0$. So triple primes conjecture is true.

##  2 Three lemmas

In this section, three lemmas are proved.

Lemma 1. Given two coprime natural numbers $p$ and $q$. If the remainder of
natural numbers in the set $A$ with respect to $p$ is evenly distributed, then
the remainder of natural numbers in the set $\\{aq+c|a\in A\\}$ is still
evenly distributed, where $c\geq 0$ is an integer.

###### Proof.

Prove by a contradiction.

Without loss of generality, suppose that $a_{i}\equiv i\mod p,a_{i}\in A$,
$0\leq i<p$, but the remainder of natural numbers in the set
$\\{a_{i}q+c|a_{i}\in A,0\leq i<p\\}$ is not evenly distributed. Let’s say
that $a_{i}q+c$ and $a_{j}q+c$ have the same remainder about $p$,

$\Longrightarrow a_{i}q+c-(a_{j}q+c)=kp,k$ is an integer. $\Longrightarrow$
$(a_{i}-a_{j})q=kp$.

As $p$ and $q$ are coprime natural numbers, $\Longrightarrow$ $a_{i}\equiv
a_{j}\mod p$. This is a contradiction.  $\hbox to0.0pt{$\sqcap$\hss}\sqcup$

Based on Lemma 1, we prove the following Lemma 2.

Lemma 2. Given natural numbers $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$
pairwise prime, and the remainder of natural numbers in the set $A$ with
respect to $\alpha_{i},1\leq i\leq n,$ are evenly distributed, where
$|A|=\alpha_{1}\times\alpha_{2}\times\ldots\times\alpha_{n}$. We conclude that

$|\\{a|a\in A\mbox{ and }a\not\equiv 0\mod\alpha_{i},1\leq i\leq
n\\}|=(\alpha_{1}-1)\times(\alpha_{2}-1)\times\ldots\times(\alpha_{n}-1)$

###### Proof.

If $n=1$, then $|A|=\alpha_{1}$. Obviously, $|\\{a|a\in A\mbox{ and
}a\not\equiv 0\mod\alpha_{1}\\}|=\alpha_{1}-1$.

If $n=2$, then $|A|=\alpha_{1}\times\alpha_{2}$.

Without loss of generality, suppose that

$A=\left(\begin{array}[]{ccccc}1&2&\cdots&\alpha_{1}-1&\alpha_{1}\\\
\alpha_{1}+1&\alpha_{1}+2&\cdots&2\alpha_{1}-1&2\alpha_{1}\\\
\vdots&\vdots&\ddots&\vdots&\vdots\\\
(\alpha_{2}-2)\alpha_{1}+1&(\alpha_{2}-2)\alpha_{1}+2&\cdots&(\alpha_{2}-1)\alpha_{1}-1&(\alpha_{2}-1)\alpha_{1}\\\
(\alpha_{2}-1)\alpha_{1}+1&(\alpha_{2}-1)\alpha_{1}+2&\cdots&\alpha_{2}\alpha_{1}-1&\alpha_{2}\alpha_{1}\\\
\end{array}\right)$

$\Longrightarrow$

$\\{a|a\in A\mbox{ and }a\not\equiv
0\mod\alpha_{1}\\}=\left(\begin{array}[]{cccc}1&2&\cdots&\alpha_{1}-1\\\
\alpha_{1}+1&\alpha_{1}+2&\cdots&2\alpha_{1}-1\\\
\vdots&\vdots&\ddots&\vdots\\\
(\alpha_{2}-2)\alpha_{1}+1&(\alpha_{2}-2)\alpha_{1}+2&\cdots&(\alpha_{2}-1)\alpha_{1}-1\\\
(\alpha_{2}-1)\alpha_{1}+1&(\alpha_{2}-1)\alpha_{1}+2&\cdots&\alpha_{2}\alpha_{1}-1\\\
\end{array}\right)$

As $\alpha_{1}$ and $\alpha_{2}$ are coprime natural numbers, and by Lemma 1,
$\\{i\alpha_{1}+j|0\leq
i<\alpha_{2}\\}\equiv\\{0,1,2,\cdots,\alpha_{2}-1\\}\mod\alpha_{2}$, where
$0<j<\alpha_{1}$.

$\Longrightarrow|\\{a|a\in A\mbox{ and }a\not\equiv 0\mod\alpha_{i},1\leq
i\leq 2\\}|=(\alpha_{1}-1)\times(\alpha_{2}-1)$.

If $n=k+1$, then
$|A|=\alpha_{1}\times\alpha_{2}\times\ldots\times\alpha_{k}\times\alpha_{k+1}$.

As $\alpha_{1},\alpha_{2},\ldots,\alpha_{k+1}$ are pairwise primes, and by
Lemma 1, the remainder of natural numbers in the set
$\\{i\times\alpha_{k+1}|0<i\leq\alpha_{1}\times\alpha_{2}\times\ldots\times\alpha_{k}\\}$
with respect to $\alpha_{i},1\leq i\leq k$, is evenly distributed.

$\Longrightarrow$ the remainder of natural numbers in the set $\\{a|a\in
A\mbox{ and }a\not\equiv 0\mod\alpha_{k+1}\\}$ with respect to
$\alpha_{i},1\leq i\leq k$, is evenly distributed and $|\\{a|a\in A\mbox{ and
}a\not\equiv
0\mod\alpha_{k+1}\\}|=\alpha_{1}\times\alpha_{2}\times\ldots\times\alpha_{k}\times(\alpha_{k+1}-1)$.

Similarly, as $\alpha_{1},\alpha_{2},\ldots,\alpha_{k}$ are pairwise primes,
and by Lemma 1, the remainder of natural numbers in the set
$\\{i\times\alpha_{k}|0<i\leq\alpha_{1}\times\alpha_{2}\times\ldots\times\alpha_{k-1}\times(\alpha_{k+1}-1)\\}$
with respect to $\alpha_{i},1\leq i\leq k-1$, is evenly distributed.

$\Longrightarrow$ the remainder of natural numbers in the set $\\{a|a\in
A\mbox{ and }a\not\equiv 0\mod\alpha_{x},x=k,k+1\\}$ with respect to
$\alpha_{i},1\leq i\leq k-1$, is evenly distributed and $|\\{a|a\in A\mbox{
and }a\not\equiv
0\mod\alpha_{x},x=k,k+1\\}|=\alpha_{1}\times\alpha_{2}\times\ldots\times\alpha_{k-1}\times(\alpha_{k}-1)\times(\alpha_{k+1}-1)$.

$\ldots\ldots$

Finally, we have

$|\\{a|a\in A\mbox{ and }a\not\equiv 0\mod\alpha_{i},1\leq i\leq
k+1\\}|=(\alpha_{1}-1)\times(\alpha_{2}-1)\times\ldots\times(\alpha_{k+1}-1)$

$\hbox to0.0pt{$\sqcap$\hss}\sqcup$

The following examples is helpful to understand Lemma 2.

It is known that the natural numbers $2,3,5$ are mutually prime. Let
$A=\\{1,2,3,$ $\ldots,29,30\\}$. Obviously the remainder of the natural
numbers in $A$ about $2,3,5$ is evenly distributed. According to Lemma 2,
after removing all natural numbers with 0 remainder about $2,3,5$, the number
of natural numbers in $A$ becomes $(2-1)\times(3-1)\times(5-1)=8$. Namely,
$\\{1,7,11,13,17,19,23,29\\}$.

Let $A=\\{7*i+1|1\leq i\leq 30\\}$. According to Lemma 1, the remainder of
natural numbers about $2,3,5$ in $A$ is still evenly distributed. Then,
according to Lemma 2, after removing all natural numbers with 0 remainder
about $2,3,5$, the number of natural numbers in $A$ becomes
$(2-1)\times(3-1)\times(5-1)=8$.

Based on Lemma 2, the number of primes in 30 consecutive natural numbers
$\\{6,7,\ldots,$ $34,35\\}$ is
$(2-1)(3-1)(5-1)=30(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{5})=8$. As
$30=2\times 3\times 5$, the formula is accurate. These specific prime numbers
are: $7,11,13,17,19,23,29,31$.

We are concerned about the following two issues here.

1\. Number counting formula of primes $(2-1)(3-1)(5-1)$ is valid for 30
consecutive natural numbers less than 49. Because 49 is not a multiple of
$2,3,5$, but $49=7\times 7$. For example, the number of primes among 30
consecutive natural numbers $\\{20,21,\ldots,48,49\\}$ is 7. These specific
prime numbers are: $23,29,31,37,41,43,47$. That is, the actual number of
primes is less than $(2-1)(3-1)(5-1)$.

2\. Note that number counting formula of primes

$30(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{5})(1-\frac{1}{7})\approx 6.86$

is less than the actual number 8 of primes in 30 consecutive natural numbers
$\\{6,7,\ldots,$ $34,35\\}$, and also less than the actual number 7 of primes
in 30 consecutive natural numbers $\\{20,21,\ldots,48,49\\}$.

The number counting formula of primes
$30(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{5})(1-\frac{1}{7})$ is valid for
30 consecutive natural numbers less than $11^{2}$. However, it is well known
that the prime distribution is not even, and the prime numbers in the front
part of the effective range are dense and the prime numbers behind are sparse.
For example, the actual number 6 of primes in 30 consecutive natural numbers
$\\{71,72,\ldots,99,100\\}$ is less than 6.86. These specific prime numbers
are: $71,73,79,83,89,97$.

This paper mainly considers the case that the estimation formula for the
number of primes lower than the actual number of primes.

The following proves Lemma 3

Lemma 3.

$(1-\frac{d}{30\times
x+c})<(1-\frac{d}{30\times(2x+e+1)+c})(1-\frac{d}{30\times(2x+e+2)+c})$ $x\geq
0,e\geq 0,0<c<32,0<d<30x.$

###### Proof.

First prove:
$\frac{d}{30\times(2x+1)+c}+\frac{d}{30\times(2x+2)+c}<\frac{d}{30\times x+c}$

$\Longleftrightarrow(b+c)(2b+c+30+2b+c+60)<(2b+c+30)(2b+c+60),b=30x$
$\Longleftrightarrow(b+c)(4b+2c+90)<(2b+c+30)(2b+c+60)$ $\Longleftrightarrow
4b^{2}+6bc+90b+90c+2c^{2}<4b^{2}+4bc+180b+c^{2}+90c+1800$

From both sides of the inequality remove $4b^{2}+4bc+90b+90c+c^{2}$

The original inequality $\Longleftrightarrow
c^{2}+2bc<90b+1800\Longleftrightarrow 0<(90-2c)b+1800-c^{2}$.

As $c<32$, thus

$\frac{d}{30\times(2x+1)+c}+\frac{d}{30\times(2x+2)+c}<\frac{d}{30\times x+c}$

$\Longrightarrow\frac{d}{30\times(2x+e+1)+c}+\frac{d}{30\times(2x+e+2)+c}<\frac{d}{30\times
x+c}$

$\Longrightarrow(1-\frac{d}{30\times(2x+e+1)+c})(1-\frac{d}{30\times(2x+e+2)+c})$
$>1-(\frac{d}{30\times(2x+e+1)+c}+\frac{d}{30\times(2x+e+2)+c})>1-\frac{d}{30\times
x+c}$

$\hbox to0.0pt{$\sqcap$\hss}\sqcup$

##  3 Possible form of prime numbers $\\{11+30*x|x\geq 0\\}$ and
$\\{13+30*x|x\geq 0\\}$

Introduce some basic properties of prime numbers. All prime numbers are in odd
numbers with single digits of 1, 3, 7 and 9 (except 2 and 5). Now let’s take a
look at the prime number whose single digit is 1. It is easy to find that
there are only two possible forms:

$\\{11+30*x|x\geq 0\\}$ and $\\{31+30*x|x\geq 0\\}$

For prime number whose single digit is 3, there are only two possible forms:

$\\{13+30*x|x\geq 0\\}$ and $\\{23+30*x|x\geq 0\\}$

For prime number whose single digit is 7, there are only two possible forms:

$\\{7+30*x|x\geq 0\\}$ and $\\{17+30*x|x\geq 0\\}$

For prime number whose single digit is 9, there are only two possible forms:

$\\{19+30*x|x\geq 0\\}$ and $\\{29+30*x|x\geq 0\\}$

Among possible form of prime numbers $\\{11+30*x|x\geq 0\\}$, if any $11+30*x$
is not a prime number, then there are only four possible decomposition forms:

$H_{1}=[7+30a][23+30b],a\geq 0,b\geq 0$; $H_{2}=[13+30a][17+30b],a\geq 0,b\geq
0$;

$H_{3}=[11+30a][31+30b],a\geq 0,b\geq 0$; $H_{4}=[19+30a][29+30b],a\geq
0,b\geq 0$.

As $(19+30a)^{2}=30c+1$, thus $(19+30a)^{2}$ is not a possible decomposition
form of $11+30*x$.

Table 1:  The possible form of prime numbers $\\{11+30*x|0\leq x<210\\}$ 11 | 41 | 71 | 101 | 131 | 161 | 191 | 221 | 251 | 281 | 311 | 341 | 371 | 401 | 431
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
|  |  |  |  | 7*23 |  | 13*17 |  |  |  | 11*31 | 7*53 |  |
461 | 491 | 521 | 551 | 581 | 611 | 641 | 671 | 701 | 731 | 761 | 791 | 821 | 851 | 881
|  |  | 19*29 | 7*83 | 13*47 |  | 11*61 |  | 17*43 |  | 7*113 |  | 23*37 |
911 | 941 | 971 | 1001 | 1031 | 1061 | 1091 | 1121 | 1151 | 1181 | 1211 | 1241 | 1271 | 1301 | 1331
|  |  | 7 |  |  |  | 19 |  |  | 7 | 17 | 31 |  | 11
1361 | 1391 | 1421 | 1451 | 1481 | 1511 | 1541 | 1571 | 1601 | 1631 | 1661 | 1691 | 1721 | 1751 | 1781
| 13 | 7 |  |  |  | 23 |  |  | 7 | 11 | 19 |  | 17 | 13
1811 | 1841 | 1871 | 1901 | 1931 | 1961 | 1991 | 2021 | 2051 | 2081 | 2111 | 2141 | 2171 | 2201 | 2231
| 7 |  |  |  | 37 | 11 | 43 | 7 |  |  |  | 13 | 31 | 23
2261 | 2291 | 2321 | 2351 | 2381 | 2411 | 2441 | 2471 | 2501 | 2531 | 2561 | 2591 | 2621 | 2651 | 2681
7 | 29 | 11 |  |  |  |  | 7 | 41 |  | 13 |  |  | 11 | 7
2711 | 2741 | 2771 | 2801 | 2831 | 2861 | 2891 | 2921 | 2951 | 2981 | 3011 | 3041 | 3071 | 3101 | 3131
|  | 17 |  | 19 |  | 7 | 23 | 13 | 11 |  |  | 37 | 7 | 31
3161 | 3191 | 3221 | 3251 | 3281 | 3311 | 3341 | 3371 | 3401 | 3431 | 3461 | 3491 | 3521 | 3551 | 3581
29 |  |  |  | 17 | 7 | 13 |  | 19 | 47 |  |  | 7 | 53 |
3611 | 3641 | 3671 | 3701 | 3731 | 3761 | 3791 | 3821 | 3851 | 3881 | 3911 | 3941 | 3971 | 4001 | 4031
23 | 11 |  |  | 7 |  | 17 |  |  |  |  | 7 | 11 |  | 29
4061 | 4091 | 4121 | 4151 | 4181 | 4211 | 4241 | 4271 | 4301 | 4331 | 4361 | 4391 | 4421 | 4451 | 4481
31 |  | 13 | 7 | 37 |  |  |  | 11 | 61 | 7 |  |  |  |
4511 | 4541 | 4571 | 4601 | 4631 | 4661 | 4691 | 4721 | 4751 | 4781 | 4811 | 4841 | 4871 | 4901 | 4931
13 | 19 | 7 | 43 | 11 | 59 |  |  |  | 7 | 17 | 47 |  | 13 |
4961 | 4991 | 5021 | 5051 | 5081 | 5111 | 5141 | 5171 | 5201 | 5231 | 5261 | 5291 | 5321 | 5351 | 5381
11 | 7 |  |  |  | 19 | 53 |  | 7 |  |  | 11 | 17 |  |
5411 | 5441 | 5471 | 5501 | 5531 | 5561 | 5591 | 5621 | 5651 | 5681 | 5711 | 5741 | 5771 | 5801 | 5831
7 |  |  |  |  | 67 |  | 7 |  | 13 |  |  | 29 |  | 7
5861 | 5891 | 5921 | 5951 | 5981 | 6011 | 6041 | 6071 | 6101 | 6131 | 6161 | 6191 | 6221 | 6251 | 6281
| 43 | 31 | 11 |  |  | 7 | 13 |  |  | 61 | 41 |  | 7 | 11

In Table 1, the number in the cell indicates that the corresponding $11+30*x$
is not a prime number, and the number in the cell is a factor.

For example, $11+30*33=1001=11\times 7\times 13$;

$11+30*47=1421=29\times 49=7\times 203$.

Based on Lemma 2, we consider a formula for estimating the number of primes.
In Table 1, consider first 30 natural numbers $\\{11+30*x|0\leq x<30\\}$. As
30 is not the multiple of $7,11,13,17,19,23,29,31$, thus consider an
estimation formula lower than the actual number of primes:

$30(1-\frac{1}{7})(1-\frac{1}{11})(1-\frac{1}{13})(1-\frac{1}{17})(1-\frac{1}{19})(1-\frac{1}{23})(1-\frac{1}{29})(1-\frac{1}{31})\approx
17.20$

The actual number of primes in $\\{11+30*x|0\leq x<30\\}$ is 19.

If $a\in\\{11+30*x|0\leq x<30\\}$ and $a$ is a composite number, then $a$ must
has one factor in $\\{7,11,13,17,19,23\\}$, so the above estimation formula is
lower than the actual number of primes.

As $(a+30)(b+30)=30(30+a+b)+ab$, the above estimation formula is valid for
$11+30*x<53\times 37=1961=11+30\times 65$. So the above estimation formula is
applied to the front part of the effective range, which is the reason that the
above estimation formula is lower than the actual number of primes.

Table 2:  The possible form of prime numbers $\\{13+30*x|0\leq x<210\\}$ 13 | 43 | 73 | 103 | 133 | 163 | 193 | 223 | 253 | 283 | 313 | 343 | 373 | 403 | 433
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
|  |  |  | 7*19 |  |  |  | 11*23 |  |  | 7*49 |  | 13*31 |
463 | 493 | 523 | 553 | 583 | 613 | 643 | 673 | 703 | 733 | 763 | 793 | 823 | 853 | 883
| 17 |  | 7 | 11 |  |  |  | 19 |  | 7 | 13 |  |  |
913 | 943 | 973 | 1003 | 1033 | 1063 | 1093 | 1123 | 1153 | 1183 | 1213 | 1243 | 1273 | 1303 | 1333
11 | 23 | 7 | 17 |  |  |  |  |  | 7 |  | 11 | 19 |  | 31
1363 | 1393 | 1423 | 1453 | 1483 | 1513 | 1543 | 1573 | 1603 | 1633 | 1663 | 1693 | 1723 | 1753 | 1783
29 | 7 |  |  |  | 17 |  | 11 | 7 | 23 |  |  |  |  |
1813 | 1843 | 1873 | 1903 | 1933 | 1963 | 1993 | 2023 | 2053 | 2083 | 2113 | 2143 | 2173 | 2203 | 2233
7 | 19 |  | 11 |  | 13 |  | 7 |  |  |  |  | 41 |  | 7
2263 | 2293 | 2323 | 2353 | 2383 | 2413 | 2443 | 2473 | 2503 | 2533 | 2563 | 2593 | 2623 | 2653 | 2683
31 |  | 23 | 13 |  | 19 | 7 |  |  | 17 | 11 |  | 43 | 7 |
2713 | 2743 | 2773 | 2803 | 2833 | 2863 | 2893 | 2923 | 2953 | 2983 | 3013 | 3043 | 3073 | 3103 | 3133
| 13 | 47 |  |  | 7 | 11 | 37 |  | 19 | 23 | 17 | 7 | 29 | 13
3163 | 3193 | 3223 | 3253 | 3283 | 3313 | 3343 | 3373 | 3403 | 3433 | 3463 | 3493 | 3523 | 3553 | 3583
| 31 | 11 |  | 7 |  |  |  | 41 |  |  | 7 | 13 | 11 |
3613 | 3643 | 3673 | 3703 | 3733 | 3763 | 3793 | 3823 | 3853 | 3883 | 3913 | 3943 | 3973 | 4003 | 4033
|  |  | 7 |  | 53 |  |  |  | 11 | 7 |  | 29 |  | 37
4063 | 4093 | 4123 | 4153 | 4183 | 4213 | 4243 | 4273 | 4303 | 4333 | 4363 | 4393 | 4423 | 4453 | 4483
17 |  | 7 |  | 47 | 11 |  |  | 13 | 7 |  | 23 |  | 61 |
4513 | 4543 | 4573 | 4603 | 4633 | 4663 | 4693 | 4723 | 4753 | 4783 | 4813 | 4843 | 4873 | 4903 | 4933
| 7 | 17 |  | 41 |  | 13 |  | 7 |  |  | 29 | 11 |  |
4963 | 4993 | 5023 | 5053 | 5083 | 5113 | 5143 | 5173 | 5203 | 5233 | 5263 | 5293 | 5323 | 5353 | 5383
7 |  |  | 31 | 13 |  | 37 | 7 | 11 |  | 19 | 67 |  | 53 | 7
5413 | 5443 | 5473 | 5503 | 5533 | 5563 | 5593 | 5623 | 5653 | 5683 | 5713 | 5743 | 5773 | 5803 | 5833
|  | 13 |  | 11 |  | 7 |  |  |  | 29 |  | 23 | 7 | 19
5863 | 5893 | 5923 | 5953 | 5983 | 6013 | 6043 | 6073 | 6103 | 6133 | 6163 | 6193 | 6223 | 6253 | 6283
11 | 71 |  |  | 31 | 7 |  |  | 17 |  |  | 11 | 7 | 13 | 61

Among possible form of prime numbers $\\{13+30*x|x\geq 0\\}$, if any $13+30*x$
is not a prime number, then there are only four possible decomposition forms:

$H_{5}=[7+30a][19+30b],a\geq 0,b\geq 0$; $H_{6}=[13+30a][31+30b],a\geq 0,b\geq
0$;

$H_{7}=[11+30a][23+30b],a\geq 0,b\geq 0$; $H_{8}=[17+30a][29+30b],a\geq
0,b\geq 0$.

In Table 2, consider first 30 natural numbers $\\{11+30*x|0\leq x<30\\}$. As
30 is not the multiple of $7,11,13,17,19,23,29,31$, thus consider an
estimation formula lower than the actual number of primes:

$30(1-\frac{1}{7})(1-\frac{1}{11})(1-\frac{1}{13})(1-\frac{1}{17})(1-\frac{1}{19})(1-\frac{1}{23})(1-\frac{1}{29})(1-\frac{1}{31})\approx
17.20$

The actual number of primes in $\\{13+30*x|0\leq x<30\\}$ is 20.

If $a\in\\{13+30*x|0\leq x<30\\}$ and $a$ is a composite number, then $a$ must
has one factor in $\\{7,11,13,17,19\\}$, so the above estimation formula is
lower than the actual number of primes. As $19+30$ is a composite number, the
above estimation formula is valid for $13+30*x<41\times 53=2173=13+30\times
72$. So the above estimation formula is applied to the front part of the
effective range, which is the reason that the above estimation formula is
lower than the actual number of primes.

##  4 Twin prime conjecture

We further consider possible form of twin prime numbers
$\\{(11+30*x,13+30*x)|x\geq 0\\}$. From Table 1 and Table 2, (6131, 6133) is
twin prime numbers.

Consider first 30 pairs of natural numbers $\\{(11+30*x,13+30*x)|0\leq
x<30\\}$. As 30 is not the multiple of $7,11,13,17,19,23,29,31$, thus consider
an estimation formula lower than the actual number of twin prime numbers
$C(30)$:

$30(1-\frac{2}{7})(1-\frac{2}{11})(1-\frac{2}{13})(1-\frac{2}{17})(1-\frac{2}{19})(1-\frac{2}{23})(1-\frac{2}{29})(1-\frac{2}{31})\approx
9.31$

The actual number of twin prime numbers in $\\{(11+30*x,13+30*x)|0\leq
x<30\\}$ is 13.

The above estimation formula can be understood in this way (take prime number
7 as an example): for every 7 consecutive cells, there must be one cell of
$11+30*x_{1}$ can be divided by 7, and another cell of $13+30*x_{2}$ can be
divided by 7, where $x_{1}\neq x_{2}$. Therefore, one term in the above
formula is $\frac{7-2}{7}$.

When $x=19$, $11+30*19=581$ can be divided by 7, and $13+30*19=583$ can be
divided by 11. Therefore, this cell is counted twice, so the estimation
formula $C(30)$ is lower than the actual number of twin primes.

The estimation formula for the number of twin prime numbers in
$\\{(11+30*x,13+30*x)|30\leq x<90\\}$ is $C(60)$:

$60(1-\frac{2}{7})(1-\frac{2}{11})(1-\frac{2}{13})(1-\frac{2}{17})(1-\frac{2}{19})(1-\frac{2}{23})(1-\frac{2}{29})(1-\frac{2}{31})$
$(1-\frac{2}{37})(1-\frac{2}{67})(1-\frac{2}{41})(1-\frac{2}{71})(1-\frac{2}{43})(1-\frac{2}{73})(1-\frac{2}{47})(1-\frac{2}{77})(1-\frac{2}{49})(1-\frac{2}{79})(1-\frac{2}{53})(1-\frac{2}{83})(1-\frac{2}{59})(1-\frac{2}{89})(1-\frac{2}{61})(1-\frac{2}{91})$
$\approx 10.72$

The actual number of twin prime numbers in $\\{(11+30*x,13+30*x)|30\leq
x<90\\}$ is 15.

In the above formula, $77,49$ may not appear in the formula because they are
multiples of 7. For the completeness of the formula, $77,49$ are still
retained, which only make the valuation smaller.

Because $91^{2}>13+270*30$, the effective range of the above estimation
formula is: $13+30*x\leq 13+30\times 270$, where $270>90$. So the above
estimation formula is applied to the front part of the effective range, which
is the main reason that the above estimation formula is lower than the actual
number of primes.

The actual number of twin prime numbers in $\\{(11+30*x,13+30*x)|90\leq
x<210\\}$ is 24. The estimation formula for the number of twin prime numbers
in $\\{(11+30*x,13+30*x)|90\leq x<210\\}$ is $C(120)$:

$120(1-\frac{2}{7})\cdots(1-\frac{2}{31})$
$(1-\frac{2}{37})(1-\frac{2}{67})\cdots(1-\frac{2}{61})(1-\frac{2}{91})$
$(1-\frac{2}{97})(1-\frac{2}{127})(1-\frac{2}{157})(1-\frac{2}{187})\cdots(1-\frac{2}{121})(1-\frac{2}{151})(1-\frac{2}{181})(1-\frac{2}{211})$

The effective range of the above estimation formula is:
$211^{2}>30*70*210>30*450$.

More generally,

$(30*2^{k}-30)^{2}-30*(30*2^{k+1}-30)=900*(2^{2k}-2^{k+2}+2)>0$
$\Longrightarrow(30*2^{k}-30)^{2}>30*(30*2^{k+1}-30)$

where $k\geq 2$.

When $k=2$, the inequality means $90^{2}>30*210$.

When $k=3$, the inequality means $210^{2}>30*450$.

When $k=4$, the inequality means $450^{2}>30*930$.

$\cdots\cdots$

Therefore, the above estimation formula is applied to the front part of the
effective range, which is the main reason that the above estimation formula is
lower than the actual number of primes.

The actual number of twin prime numbers in $\\{(11+30*x,13+30*x)|210\leq
x<450\\}$ is 29. The estimation formula for the number of twin prime numbers
in $\\{(11+30*x,13+30*x)|210\leq x<450\\}$ is $C(240)$(only the part about
$\\{7+30*x|x\geq 0\\}$ is given here):

$(1-\frac{2}{7})$ $(1-\frac{2}{37})(1-\frac{2}{67})$
$(1-\frac{2}{97})(1-\frac{2}{127})(1-\frac{2}{157})(1-\frac{2}{187})$
$(1-\frac{2}{217})(1-\frac{2}{247})(1-\frac{2}{277})(1-\frac{2}{307})(1-\frac{2}{337})(1-\frac{2}{367})(1-\frac{2}{397})(1-\frac{2}{427})$

It is easy to obtain the following (only the part about $\\{7+30*x|x\geq 0\\}$
is given here):

$\frac{C(60)}{C(30)}=2(1-\frac{2}{37})(1-\frac{2}{67})$
$\frac{C(120)}{C(60)}=2(1-\frac{2}{97})(1-\frac{2}{127})(1-\frac{2}{157})(1-\frac{2}{187})$
$\frac{C(240)}{C(120)}=2(1-\frac{2}{217})(1-\frac{2}{247})(1-\frac{2}{277})(1-\frac{2}{307})(1-\frac{2}{337})(1-\frac{2}{367})(1-\frac{2}{397})(1-\frac{2}{427})$
$\cdots\cdots$

By Lemma 3,

$(1-\frac{2}{217})(1-\frac{2}{247})>(1-\frac{2}{97}),(1-\frac{2}{277})(1-\frac{2}{307})>(1-\frac{2}{127}),\cdots\cdots,$

Thus

$\cdots\cdots>\frac{C(240)}{C(120)}>\frac{C(120)}{C(60)}>\frac{C(60)}{C(30)}$

The complete $\frac{C(60)}{C(30)}$ is as follows,

$2(1-\frac{2}{37})(1-\frac{2}{67})(1-\frac{2}{41})(1-\frac{2}{71})(1-\frac{2}{43})(1-\frac{2}{73})(1-\frac{2}{47})(1-\frac{2}{77})(1-\frac{2}{49})(1-\frac{2}{79})(1-\frac{2}{53})(1-\frac{2}{83})(1-\frac{2}{59})(1-\frac{2}{89})(1-\frac{2}{61})(1-\frac{2}{91})$

$=2\times 0.575>1$

Therefore

$\cdots\cdots>C(240)>C(120)>C(60)>C(30)\approx 9.31$

In fact, the actual number of twin prime numbers in
$\\{(11+30*x,13+30*x)|450\leq x<930\\}$ is 71, and the actual number of twin
prime numbers in $\\{(11+30*x,13+30*x)|930\leq x<1890\\}$ is
$113,\ldots\ldots,$

Similarly, we can always find another larger interval, which has more than
9.31 twin prime numbers. Therefore, there are infinite twin primes.

Among possible form of prime numbers $\\{17+30*x|x\geq 0\\}$, if any $17+30*x$
is not a prime number, then there are only four possible decomposition forms:

$H_{9}=[7+30a][11+30b],a\geq 0,b\geq 0$; $H_{10}=[13+30a][29+30b],a\geq
0,b\geq 0$;

$H_{11}=[17+30a][31+30b],a\geq 0,b\geq 0$; $H_{12}=[19+30a][23+30b],a\geq
0,b\geq 0$.

Consider an estimation formula lower than the actual number of primes in
$\\{17+30*x|0\leq x<30\\}$:

$30(1-\frac{1}{7})(1-\frac{1}{11})(1-\frac{1}{13})(1-\frac{1}{17})(1-\frac{1}{19})(1-\frac{1}{23})(1-\frac{1}{29})(1-\frac{1}{31})\approx
17.20$

By considering an estimation formula lower than the actual number of prime
pairs in $\\{(11+30*x,17+30*x)|x\geq 0\\}$, similarly we can prove that there
are infinite prime pairs in $\\{(11+30*x,17+30*x)|x\geq 0\\}$.

To sum up, we get the following theorem.

Theorem 1. For any two prime numbers $p_{0}>5$ and $q_{0}>5$, there are
infinite prime pairs $p_{i}$ and $q_{i},i\geq 1$, such that
$p_{i}-q_{i}=p_{0}-q_{0}$.

##  5 Goldbach conjecture

Very similar to the case of twin prime conjecture, we further consider
possible form of prime pairs $\\{(11+30*x,13+30*(n-x))|x\geq 0\\}$. From Table
1 and Table 2, (131, 6163) is a pair of primes, where $n=209,x=4$.

Consider 30 pairs of natural numbers $\\{(13+30*x,13+30*(29-x))|0\leq
x<30\\}$. As 30 is not the multiple of $7,11,13,17,19,23,29,31$, thus consider
an estimation formula lower than the actual number of twin prime numbers
$C(30)$:

$30(1-\frac{2}{7})(1-\frac{2}{11})(1-\frac{2}{13})(1-\frac{2}{17})(1-\frac{2}{19})(1-\frac{2}{23})(1-\frac{2}{29})(1-\frac{2}{31})\approx
9.31$

The actual number of prime pairs in $\\{(11+30*x,13+30*(29-x))|0\leq x<30\\}$
is 11.

The above estimation formula can be understood in this way (take prime number
7 as an example): for every 7 consecutive cells, there must be one cell of
$11+30*x_{1}$ can be divided by 7, and another cell of $13+30*(29-x_{2})$ can
be divided by 7, where $x_{1}\neq x_{2}$. Therefore, one term in the above
formula is $\frac{7-2}{7}$.

When $x=11$, $11+30*11=341$ can be divided by 11, and $13+30*(29-11)=553$ can
be divided by 7. Therefore, this cell is counted twice, so the estimation
formula $C(30)$ is lower than the actual number of prime pairs.

The estimation formula for the number of prime pairs in
$\\{(11+30*x,13+30*(89-x))|30\leq x<90\\}$ is $C(60)$:

$60(1-\frac{2}{7})(1-\frac{2}{11})(1-\frac{2}{13})(1-\frac{2}{17})(1-\frac{2}{19})(1-\frac{2}{23})(1-\frac{2}{29})(1-\frac{2}{31})$
$(1-\frac{2}{37})(1-\frac{2}{67})(1-\frac{2}{41})(1-\frac{2}{71})(1-\frac{2}{43})(1-\frac{2}{73})(1-\frac{2}{47})(1-\frac{2}{77})(1-\frac{2}{49})(1-\frac{2}{79})(1-\frac{2}{53})(1-\frac{2}{83})(1-\frac{2}{59})(1-\frac{2}{89})(1-\frac{2}{61})(1-\frac{2}{91})$
$\approx 10.72$

The actual number of prime pairs in $\\{(11+30*x,13+30*(89-x))|30\leq x<90\\}$
is 16.

In the above formula, $77,49$ may not appear in the formula because they are
multiples of 7. For the completeness of the formula, $77,49$ are still
retained, which only make the valuation smaller.

The effective range of the above estimation formula is:
$91^{2}>30*270>30*210$.

The actual number of prime pairs in $\\{(11+30*x,13+30*(209-x))|90\leq
x<210\\}$ is 29. The estimation formula for the number of prime pairs in
$\\{(11+30*x,13+30*(209-x))|90\leq x<210\\}$ is $C(120)$:

$120(1-\frac{2}{7})\cdots(1-\frac{2}{31})$
$(1-\frac{2}{37})(1-\frac{2}{67})\cdots(1-\frac{2}{61})(1-\frac{2}{91})$
$(1-\frac{2}{97})(1-\frac{2}{127})(1-\frac{2}{157})(1-\frac{2}{187})\cdots(1-\frac{2}{121})(1-\frac{2}{151})(1-\frac{2}{181})(1-\frac{2}{211})$

The effective range of the above estimation formula is:
$211^{2}>30*70*210>30*450$.

More generally,

$(30*2^{k}-30)^{2}-30*(30*2^{k+1}-30)=900*(2^{2k}-2^{k+2}+2)>0$
$\Longrightarrow(30*2^{k}-30)^{2}>30*(30*2^{k+1}-30)$

where $k\geq 2$.

When $k=2$, the inequality means $90^{2}>30*210$.

When $k=3$, the inequality means $210^{2}>30*450$.

Therefore, the above estimation formula is applied to the front part of the
effective range, which is the main reason that the above estimation formula is
lower than the actual number of primes.

The actual number of prime pairs in $\\{(11+30*x,13+30*(449-x))|210\leq
x<450\\}$ is 44. The estimation formula for the number of prime pairs in
$\\{(11+30*x,13+30*(449-x))|210\leq x<450\\}$ is $C(240)$(only the part about
$\\{7+30*x|x\geq 0\\}$ is given here):

$(1-\frac{2}{7})$ $(1-\frac{2}{37})(1-\frac{2}{67})$
$(1-\frac{2}{97})(1-\frac{2}{127})(1-\frac{2}{157})(1-\frac{2}{187})$
$(1-\frac{2}{217})(1-\frac{2}{247})(1-\frac{2}{277})(1-\frac{2}{307})(1-\frac{2}{337})(1-\frac{2}{367})(1-\frac{2}{397})(1-\frac{2}{427})$

It is easy to obtain the following (only the part about $\\{7+30*x|x\geq 0\\}$
is given here):

$\frac{C(60)}{C(30)}=2(1-\frac{2}{37})(1-\frac{2}{67})$
$\frac{C(120)}{C(60)}=2(1-\frac{2}{97})(1-\frac{2}{127})(1-\frac{2}{157})(1-\frac{2}{187})$
$\frac{C(240)}{C(120)}=2(1-\frac{2}{217})(1-\frac{2}{247})(1-\frac{2}{277})(1-\frac{2}{307})(1-\frac{2}{337})(1-\frac{2}{367})(1-\frac{2}{397})(1-\frac{2}{427})$
$\cdots\cdots$

By Lemma 3,

$(1-\frac{2}{217})(1-\frac{2}{247})>(1-\frac{2}{97}),(1-\frac{2}{277})(1-\frac{2}{307})>(1-\frac{2}{127}),\cdots\cdots$

Thus

$\cdots\cdots>\frac{C(240)}{C(120)}>\frac{C(120)}{C(60)}>\frac{C(60)}{C(30)}$

The complete $\frac{C(60)}{C(30)}$ is as follows,

$2(1-\frac{2}{37})(1-\frac{2}{67})(1-\frac{2}{41})(1-\frac{2}{71})(1-\frac{2}{43})(1-\frac{2}{73})(1-\frac{2}{47})(1-\frac{2}{77})(1-\frac{2}{49})(1-\frac{2}{79})(1-\frac{2}{53})(1-\frac{2}{83})(1-\frac{2}{59})(1-\frac{2}{89})(1-\frac{2}{61})(1-\frac{2}{91})$

$=2\times 0.575>1$

Therefore

$\cdots\cdots>C(240)>C(120)>C(60)\approx 10.72$

In fact, the actual number of prime pairs in
$\\{(11+30*x,13+30*(929-x))|450\leq x<930\\}$ is 73, and the actual number of
prime pairs in $\\{(11+30*x,13+30*(1889-x))|930\leq x<1890\\}$ is
$136,\ldots\ldots$

Similarly, we can always find another larger interval, which has more than
10.72 prime pairs $(p,q)$, such that $p+q=24+30*(30*2^{k}-30-1)$, where $k>1$.

For any $24+30*a,a\geq 90$, if $30*2^{k}-30-1<a<30*2^{k+1}-30-1$, by Lemma 2,
the estimation formula for the number of prime pairs in
$\\{(11+30*x,13+30*(a-x))|0\leq x\leq a\\}$ is greater than $C(30*2^{k-1})\geq
C(60)$.

Since we can also consider the prime pairs in $\\{(7+30*x,17+30*(a-x))|0\leq
x\leq a\\}$, thus there exist much more than 10 prime pairs $(p,q)$, such that
$p+q=24+30*a$,

For example, for $24+30*99$, since $30*2^{2}-30-1<99<30*2^{2+1}-30-1$, so the
estimation formula for the number of prime pairs in
$\\{(11+30*x,13+30*(99-x))|0\leq x\leq 99\\}$ is greater than
$C(30*2^{2-1})=C(60)$. In fact, the actual number of prime pairs in
$\\{(11+30*x,13+30*(99-x))|0\leq x\leq 99\\}$ is 27, and the actual number of
prime pairs in $\\{(7+30*x,17+30*(99-x))|0\leq x\leq 99\\}$ is 32.

It is easy to show that for $a>0$,

$\\{0+30*a=7+30*(a-1)+23\\}$;

$\\{2+30*a=13+30*(a-1)+19\\}$;

$\\{4+30*a=11+30*(a-1)+23\\}$;

$\\{6+30*a=13+30*(a-1)+23\\}$;

$\\{8+30*a=19+30*(a-1)+19\\}$;

$\\{10+30*a=17+30*(a-1)+23\\}$;

$\\{12+30*a=19+30*(a-1)+23\\}$;

$\\{14+30*a=13+30*(a-1)+31\\}$;

$\\{16+30*a=17+30*(a-1)+29\\}$;

$\\{18+30*a=7+30*a+11\\}$;

$\\{20+30*a=7+30*a+13\\}$;

$\\{22+30*a=11+30*a+11\\}$;

$\\{26+30*a=13+30*a+13\\}$;

$\\{28+30*a=11+30*a+17\\}$.

Very similar to the case of $\\{24+30*a=11+30*a+13\\}$, we can prove that for
any $2l+30*a,0\leq l<15,a\geq 90$, there exist more than 10 prime pairs
$(p,q)$, such that $p+q=2l+30*a$.

To sum up, we get the following theorem.

Theorem 2. For any even integer $m>30*90$, there exist more than 10 prime
pairs $(p,q)$, such that $p+q=m$.

##  6  Triple prime conjecture

Based on Lemma 2, we consider a formula for estimating the number of primes.
Consider an estimation formula for the number of primes in first 210 natural
numbers $\\{11+210*x|0\leq x<210\\}$:

$210(1-\frac{1}{11})(1-\frac{1}{13})(1-\frac{1}{17})(1-\frac{1}{19})(1-\frac{1}{23})(1-\frac{1}{29})(1-\frac{1}{31})(1-\frac{1}{37})(1-\frac{1}{41})(1-\frac{1}{43})$
$(1-\frac{1}{47})(1-\frac{1}{53})(1-\frac{1}{59})(1-\frac{1}{61})(1-\frac{1}{67})(1-\frac{1}{71})(1-\frac{1}{73})(1-\frac{1}{79})(1-\frac{1}{83})(1-\frac{1}{89})$
$(1-\frac{1}{97})(1-\frac{1}{101})(1-\frac{1}{103})(1-\frac{1}{107})(1-\frac{1}{109})(1-\frac{1}{113})(1-\frac{1}{127})(1-\frac{1}{131})(1-\frac{1}{137})(1-\frac{1}{139})(1-\frac{1}{149})$
$(1-\frac{1}{151})(1-\frac{1}{157})(1-\frac{1}{163})(1-\frac{1}{167})(1-\frac{1}{173})(1-\frac{1}{179})(1-\frac{1}{181})(1-\frac{1}{191})(1-\frac{1}{193})(1-\frac{1}{197})(1-\frac{1}{199})(1-\frac{1}{211})$
$\approx 95.00$

The actual number of primes in $\\{11+210*x|0\leq x<210\\}$ is 98.

Similarly, consider an estimation formula for the number of primes in first
210 natural numbers $\\{17+210*x|0\leq x<210\\}$:

$210(1-\frac{1}{11})(1-\frac{1}{13})(1-\frac{1}{17})(1-\frac{1}{19})(1-\frac{1}{23})(1-\frac{1}{29})(1-\frac{1}{31})(1-\frac{1}{37})(1-\frac{1}{41})(1-\frac{1}{43})$
$(1-\frac{1}{47})(1-\frac{1}{53})(1-\frac{1}{59})(1-\frac{1}{61})(1-\frac{1}{67})(1-\frac{1}{71})(1-\frac{1}{73})(1-\frac{1}{79})(1-\frac{1}{83})(1-\frac{1}{89})$
$(1-\frac{1}{97})(1-\frac{1}{101})(1-\frac{1}{103})(1-\frac{1}{107})(1-\frac{1}{109})(1-\frac{1}{113})(1-\frac{1}{127})(1-\frac{1}{131})(1-\frac{1}{137})(1-\frac{1}{139})(1-\frac{1}{149})$
$(1-\frac{1}{151})(1-\frac{1}{157})(1-\frac{1}{163})(1-\frac{1}{167})(1-\frac{1}{173})(1-\frac{1}{179})(1-\frac{1}{181})(1-\frac{1}{191})(1-\frac{1}{193})(1-\frac{1}{197})(1-\frac{1}{199})(1-\frac{1}{211})$
$\approx 95.00$

The actual number of primes in $\\{17+210*x|0\leq x<210\\}$ is 96.

Further consider an estimation formula for the number of primes in first 210
natural numbers $\\{23+210*x|0\leq x<210\\}$:

$210(1-\frac{1}{11})(1-\frac{1}{13})(1-\frac{1}{17})(1-\frac{1}{19})(1-\frac{1}{23})(1-\frac{1}{29})(1-\frac{1}{31})(1-\frac{1}{37})(1-\frac{1}{41})(1-\frac{1}{43})$
$(1-\frac{1}{47})(1-\frac{1}{53})(1-\frac{1}{59})(1-\frac{1}{61})(1-\frac{1}{67})(1-\frac{1}{71})(1-\frac{1}{73})(1-\frac{1}{79})(1-\frac{1}{83})(1-\frac{1}{89})$
$(1-\frac{1}{97})(1-\frac{1}{101})(1-\frac{1}{103})(1-\frac{1}{107})(1-\frac{1}{109})(1-\frac{1}{113})(1-\frac{1}{127})(1-\frac{1}{131})(1-\frac{1}{137})(1-\frac{1}{139})(1-\frac{1}{149})$
$(1-\frac{1}{151})(1-\frac{1}{157})(1-\frac{1}{163})(1-\frac{1}{167})(1-\frac{1}{173})(1-\frac{1}{179})(1-\frac{1}{181})(1-\frac{1}{191})(1-\frac{1}{193})(1-\frac{1}{197})(1-\frac{1}{199})(1-\frac{1}{211})$
$\approx 95.00$

The actual number of primes in $\\{23+210*x|0\leq x<210\\}$ is 94.

We further consider possible form of triple primes
$\\{(11+210*x,17+210*x,23+210*x)|x\geq 0\\}$. For example, when $x=100$,
21011, 21017 and 21023 are all primes.

Consider an estimation formula $C(210)$ lower than the actual number of triple
primes in first 210 triplets of natural numbers
$\\{(11+210*x,17+210*x,23+210*x)|0\leq x<210\\}$:

$210(1-\frac{3}{11})(1-\frac{3}{13})(1-\frac{3}{17})(1-\frac{3}{19})(1-\frac{3}{23})(1-\frac{3}{29})(1-\frac{3}{31})(1-\frac{3}{37})(1-\frac{3}{41})(1-\frac{3}{43})$
$(1-\frac{3}{47})(1-\frac{3}{53})(1-\frac{3}{59})(1-\frac{3}{61})(1-\frac{3}{67})(1-\frac{3}{71})(1-\frac{3}{73})(1-\frac{3}{79})(1-\frac{3}{83})(1-\frac{3}{89})$
$(1-\frac{3}{97})(1-\frac{3}{101})(1-\frac{3}{103})(1-\frac{3}{107})(1-\frac{3}{109})(1-\frac{3}{113})(1-\frac{3}{127})(1-\frac{3}{131})(1-\frac{3}{137})(1-\frac{3}{139})(1-\frac{3}{149})$
$(1-\frac{3}{151})(1-\frac{3}{157})(1-\frac{3}{163})(1-\frac{3}{167})(1-\frac{3}{173})(1-\frac{3}{179})(1-\frac{3}{181})(1-\frac{3}{191})(1-\frac{3}{193})(1-\frac{3}{197})(1-\frac{3}{199})(1-\frac{3}{211})$
$\approx 17.46$

The actual number of triple primes in $\\{(11+210*x,17+210*x,23+210*x)|0\leq
x<210\\}$ is 19.

Based on Lemma 1, the above estimation formula can be understood in this way
(take prime number 11 as an example): for every 11 consecutive cells, there
must be one cell of $11+210*x_{1}$ can be divided by 11, another cell of
$17+210*x_{2}$ can be divided by 11 and the third cell of $23+210*x_{3}$ can
be divided by 11, where $x_{1},x_{2},x_{3}$ are distinct numbers. Therefore,
one term in the above formula is $\frac{11-3}{11}$.

However, when $x=23$, $11+210*23=4841$ can be divided by 47, $17+210*23=4847$
can be divided by 37, and $23+210*23=4853$ can be divided by 23. Therefore,
this cell is counted three times, so the estimation formula $C(210)$ is lower
than the actual number of triple primes.

The estimation formula for the number of triple prime numbers in
$\\{(11+210*x,17+210*x,23+210*x)|210\leq x<630\\}$ is $C(420)$:

$420(1-\frac{3}{11})(1-\frac{3}{13})\cdots(1-\frac{3}{211})$
$(1-\frac{3}{221})(1-\frac{3}{431})(1-\frac{3}{223})(1-\frac{3}{433})\cdots(1-\frac{3}{421})(1-\frac{3}{631})$
$\approx 17.66$

The actual number of triple prime numbers in
$\\{(11+210*x,17+210*x,23+210*x)|210\leq x<630\\}$ is 22.

In the above formula, $221,247,253,\cdots\cdots$ may not appear in the formula
because $221=13\times 17,247=13\times 19,253=23\times 11\cdots\cdots$. For the
completeness of the formula, $221,247,253,\cdots\cdots$ are still retained,
which only make the valuation further smaller.

The actual number of triple prime numbers in
$\\{(11+210*x,17+210*x,23+210*x)|630\leq x<1470\\}$ is 35.

The estimation formula for the number of triple prime numbers in
$\\{(11+210*x,17+210*x,23+210*x)|630\leq x<1470\\}$ is $C(840)$:

$840(1-\frac{3}{11})(1-\frac{3}{13})\cdots(1-\frac{3}{211})$
$(1-\frac{3}{221})(1-\frac{3}{431})(1-\frac{3}{223})(1-\frac{3}{433})\cdots(1-\frac{3}{421})(1-\frac{3}{631})$
$(1-\frac{3}{641})(1-\frac{3}{851})(1-\frac{3}{1061})(1-\frac{3}{1271})\cdots(1-\frac{3}{841})(1-\frac{3}{1051})(1-\frac{3}{1261})(1-\frac{3}{1471})$
$\approx 20.96$

The effective range of the above estimation formula is:
$1471^{2}>210*7*1470>210*3150$.

More generally,

$(210*2^{k}-210)^{2}-210*(210*2^{k+1}-210)=210^{2}*(2^{2k}-2^{k+2}+2)>0$
$\Longrightarrow(210*2^{k}-210)^{2}>210*(210*2^{k+1}-210)$

where $k\geq 2$.

When $k=2$, the inequality means $630^{2}>210*1470$.

When $k=3$, the inequality means $1470^{2}>210*3150$.

When $k=4$, the inequality means $3150^{2}>210*6510$.

$\cdots\cdots$

Therefore, the above estimation formula is applied to the front part of the
effective range, which is the main reason that the above estimation formula is
lower than the actual number of primes.

It is easy to obtain the following:

$\frac{C(420)}{C(210)}=2(1-\frac{3}{221})(1-\frac{3}{431})(1-\frac{3}{223})(1-\frac{3}{433})\cdots(1-\frac{3}{421})(1-\frac{3}{631})$
$\frac{C(840)}{C(420)}=2(1-\frac{3}{641})(1-\frac{3}{851})(1-\frac{3}{1061})(1-\frac{3}{1271})\cdots(1-\frac{3}{841})(1-\frac{3}{1051})(1-\frac{3}{1261})(1-\frac{3}{1471})$
$\frac{C(1680)}{C(840)}=2(1-\frac{3}{1481})(1-\frac{3}{1691})(1-\frac{3}{1901})(1-\frac{3}{2111})\cdots(1-\frac{3}{2521})(1-\frac{3}{2731})(1-\frac{3}{2941})(1-\frac{3}{3151})$
$\cdots\cdots$

Similar to the proof of Lemma 3, it is easy to prove that,

$(1-\frac{d}{210\times
x+c})<(1-\frac{d}{210\times(2x+e+1)+c})(1-\frac{d}{210\times(2x+e+2)+c})$
$x\geq 0,e\geq 0,0<c<212,0<d<210x.$

Therefore,

$(1-\frac{3}{1481})(1-\frac{3}{1691})>(1-\frac{3}{641}),(1-\frac{3}{1901})(1-\frac{3}{2111})>(1-\frac{3}{851}),\cdots\cdots,$

Thus

$\cdots\cdots>\frac{C(1680)}{C(840)}>\frac{C(840)}{C(420)}>\frac{C(420)}{C(210)}$

Since

$\frac{C(420)}{C(210)}=2(1-\frac{3}{221})(1-\frac{3}{431})(1-\frac{3}{223})(1-\frac{3}{433})\cdots(1-\frac{3}{421})(1-\frac{3}{631})$
$=2\times 0.506>1$

Therefore

$\cdots\cdots>C(1680)>C(840)>C(420)>C(210)\approx 17.46$

Similarly, we can always find another larger interval, which has more than
17.46 triple prime numbers. Therefore, there are infinite triple primes.

Very similarly, we can further consider possible form of triple primes
$\\{(17+210*x,23+210*x,29+210*x)|x\geq 0\\}$,
$\\{(31+210*x,37+210*x,43+210*x)|x\geq 0\\}$,
$\\{(47+210*x,53+210*x,59+210*x)|x\geq 0\\},\cdots\cdots,$ and prove that
there are infinite triple primes.

To sum up, we get the following theorem.

Theorem 3. For any triple primes $p>7,$ $q>7$ and $r>7$, there are infinite
triple primes $(p_{i},q_{i},r_{i})$, such that $p_{i}-q_{i}=p-q$,
$p_{i}-r_{i}=p-r$, where $i>0$.

## References

  * [1] J. Chen, On the representation of a large even integer as the sum of a prime and the product of at most two primes (in Chinese),  Science in China, Series A. 1973, (02)111-128.
  * [2] B. Green, T. Tao, The primes contain arbitrarily long arithmetic progressions, Annals of Mathematics,  167 (2) (2008)481-547.
  * [3] Y. Zhang, Bounded gaps between primes, Annals of Mathematics, 179 (2014)1121-1174.

Jianqin Zhou (1963-), Ph.D., Professor. He graduated from the Department of
mathematics of East China Normal University in 1983 and Fudan University in
1989 with a master degree in mathematics; In 2017, he obtained Ph.D. from
Department of computing in Curtin University, Australia. His main research
fields are theoretical computer science, combinatorial mathematics and
algorithms.

In 1989, he proved a conjecture in combinatorics proposed by the famous
mathematician $Paul$ $Erd\ddot{o}s$ (Please see the paper “A proof of Alavi
conjecture on integer partition”, Acta Mathematicae Sinica, 1995,
38(5)636-641). Since 1989, he has published more than 120 papers in Acta
Mathematicae Sinica, Designs, Codes and Cryptography, Acta Mathematicae
Applicatae Sinica, Combinatorica and other academic journals at home and
abroad.