Generating Prime Numbers -- A Fast New Method

TL;DR

This paper introduces a new fast algorithm for generating prime numbers up to any limit by combining Bertrand's Postulate and the sieve of Eratosthenes, with discussions on the Riemann zeta function's role.

Contribution

The paper presents a novel, efficient prime generation algorithm that improves upon traditional methods by integrating classical number theory results.

Findings

The new algorithm significantly reduces computation time.

It effectively generates all primes up to large limits.

The paper explores connections between prime generation and the Riemann zeta function.

Abstract

Bertrand's Postulate ensures existence of prime $p$ between $n$ and $2n$, $n$ an integer $\geq 2$ and the sieve of Eratosthenes, a very simple ancient algorithm, generates all prime numbers up to any given limit. Combining the above two, in this paper, we provide a simple fast moving algorithm to generate prime numbers up to any given limit. We also discuss Riemann zeta function related to generating of prime numbers.

Tables

Table 1. Table 1. π ​ ( p ​ P j ) 𝜋 𝑝 subscript 𝑃 𝑗 \pi(pP_{j}) , order of prime pockets in the three Methods

$\pi \left(p{P}_j\right)=n_j$	$n_1$	$n_2$	$n_3$	$n_4$	$n_5$	$n_6$	$n_7$	$n_8$
Method $1$	$1$	$1$	$1$	$1$	$2$	$3$	$5$	$9$
Method $2$	$1$	$1$	$2$	$11$	$314$	$339730$	$--$	$--$
Method $3$	$2$	$7$	$105$	$32622$	$--$	$--$	$--$	$--$

Equations

$$\zeta (z)=\frac{1}{1^z}+\frac{1}{2^z}+\frac{1}{3^z}+\frac{1}{4^z}+\dots ,$$

$$\text{If}(1-\frac{1}{p_k^z})(1-\frac{1}{p_{k-1}^z})...(1-\frac{1}{3^z})(1-\frac{1}{2^z})\zeta (z)=1+\frac{1}{n_1^z}+\frac{1}{n_2^z}+...$$

$$\zeta (z)=\frac{1}{1^z}+\frac{1}{2^z}+\frac{1}{3^z}+\frac{1}{4^z}+....$$

$$\Rightarrow (1-\frac{1}{2^z})\zeta (z)=1+\frac{1}{2^z}+\frac{1}{3^z}+\frac{1}{4^z}+...$$

$$=1+\frac{1}{3^z}+\frac{1}{5^z}+\frac{1}{7^z}+\frac{1}{9^z}+....$$

$$(1-\frac{1}{3^z})(1-\frac{1}{2^z})\zeta (z)=1+\frac{1}{5^z}+\frac{1}{7^z}+\frac{1}{11^z}+\frac{1}{13^z}+\frac{1}{17^z}+\frac{1}{19^z}+\frac{1}{23^z}+\dots .$$

Full text

Key Words: Prime number, composite number, AKS primality test, the sieve of Eratosthenes, Bertrand’s postulate, the standard pockets of primes, sequence of order of pockets of primes, Twin prime conjecture, Riemann zeta function.

1. Introduction

The sieve of Eratosthenes [5, 17] and Bertrand’s postulate [3] are important mile stones in generating prime numbers. The sieve of Eratosthenes is a very simple ancient algorithm that generates primes up to any given limit and Bertrand’s postulate establishes the existence of prime in an interval $[n,2n]$, for any $n\geq 2$. In 1845, Joseph Bertrand postulated that for any integer $n>3$, there exists prime $p$ such that $n<p<2n-2$. And its slightly weaker form is that there exists prime $p$ such that $n<p<2n$, $n$ an integer $\geq 2$. In 1850, Pafnuty Chebychev [19] first proved this postulate analytically. In 1932, Paul Erdos [5] gave an elementary proof using facts about the middle binomial coefficient. In 2002, Manindra Agarwal, Neeraj Kayal and Nitin Saxena [1] presented an unconditional deterministic polynomial-time algorithm that determines whether an input number is prime or composite.

The problem of computing $\pi(x)$, the number of primes less than or equal to $x$ is one of the oldest problem in Mathematics, $x\in\mathbb{N}$. For a very long time, the sieve of Eratosthenes has been the practical way to compute $\pi(x)$ despite its time complexity. Legendre [11] observed a combinatorial formula, known as Legendre sum, for the number of primes $p$ for which $x^{\frac{1}{2}}<p\leq x$. Since then, a large number of writers have suggested variants and improvements of the formula. During 1870 to 1885, astronomer Meissel [13]-[16] developed practical combinatorial method to compute $\pi(x)$ and in 1959, Lehmer [11] extended and simplified Meissel’s method. In 1985, the Meissel-Lehmer method [10] was used to compute several values of $\pi(x)$ up to $x=4.10^{16}$ and in 1996, Deleglise and Rivat [6] developed a modified form of the Meissel-Lehmer method saving much computation.

In this article, we present a simple fast moving algorithm, using both the sieve of Eratosthenes and Bertrand’s postulate, to generate prime numbers. Let $p_{1},p_{2},\ldots$ denote the primes $2,3,\ldots$ numbered in increasing order. We start with obtaining primes in successive intervals of the form $[p_{i}+1,2p_{i}]$ and in each such interval Bertrand’s postulate ensures existence of prime(s). Then, the question arises is whether there exists any other method which generates primes in a faster way? Our second method considers intervals of the form $[p_{i}+1,p_{i}^{2}]$ instead of $[p_{i}+1,2p_{i}]$, used in the first method and generates primes in a faster way, $i\geq 2$. Again, the question is whether it is the ultimate method, in general, to generate primes in a faster way? And our third method is the answer for the above question. It considers $IpP_{j+1}$ = $[MIpP_{j}+1,$ $MpP_{j}^{2}+4MpP_{j}+3]$ as successive intervals starting with $IpP_{1}$ = $[2,3]$ = $pP_{1}$ and obtains $pP_{j}$, the set of all the primes in each $IpP_{j}$ where $MIpP_{j}$ and $MpP_{j}$ are the maximum values of $IpP_{j}$ and $pP_{j}$, respectively. Here, the width of the interval $IpP_{j}$, in general, is a tight bound in the sense that if we increase the width by increasing its upper bound further, then the algorithm fails. All the three methods are presented here for more clarity. The author feels that this development is going to have more impact on the development in Mathematics, especially in Cryptography, Number theory, Signal processing and Computational Mathematics.

Algorithms used in the three methods are given below and in each method, we consider

$MIpP_{j}=$ max $IpP_{j}$, $MpP_{j}=$ max $pP_{j}$,

$pP_{j}$ = $\phi(IpP_{j})=$ Set of all primes in $IpP_{j}$ and

$\pi(IpP_{j})$ = $|pP_{j}|$ =Number of primes in $IpP_{j}$, $j$ = 1, 2, . . . .

1. (i)

$p_{1}=2$, $IpP_{1}=\{2\}$ $=pP_{1}$,

$MIpP_{1}=2=MpP_{1}$, $\pi(IpP_{1})=|pP_{1}|=1$;

$IpP_{j+1}=$ $[MIpP_{j}+1,~{}2MpP_{j}]$, $j$ = 1, 2, . . . . 2. (ii)

$p_{1}=2$, $p_{2}=3$, $IpP_{1}=[2,3]$, $pP_{1}=\{2,3\}$,

$MIpP_{1}=3=$ $MpP_{1},~{}\pi(pP_{1})=2$;

$IpP_{j+1}$ $=[MIpP_{j}+1,~{}MpP_{j}^{2}]$, $j$ = 1, 2, . . . . 3. (iii)

$p_{1}=2$, $p_{2}=3$, $IpP_{1}=[2,3]$, $pP_{1}=$ $\{2,3\}$,

$MIpP_{1}=3=MpP_{1}$, $\pi(pP_{1})=2$;

$IpP_{j+1}$ = $[MIpP_{j}+1,~{}MpP_{j}^{2}+4MpP_{j}+3]$, $j$ = 1, 2, . . . .

We use very simple method to obtain $pP_{j+1}$, $j=1,2,\ldots$. After generating primes, it is easy to check, from the listing, whether any given number is prime or not, provided the number is within the list. We also discuss Riemann zeta function [9] related to prime generation and restate the Twin prime conjecture [21] in an easier way.

2. Preliminaries

To simplify our work, the following notations are used in this paper.

$\mathbb{N}$ = $\{1,2,\ldots\}$; $\mathbb{N}_{0}$ = $\mathbb{N}\cup\{0\}$ = $\{0,1,2,\ldots\}$; $\mathbb{N}_{1}$ = $\mathbb{N}\setminus\{1\}$ = $\{2,3,...\}$;

$\mathbb{P}$ = the set of all prime numbers;

$\mathbb{C}$ = the set of all composite numbers so that

$\mathbb{P}\cap\mathbb{C}$ = $\emptyset$ and $\mathbb{P}\cup\mathbb{C}$ = $\mathbb{N}-\{1\}$;

$\mathbb{P}(S)=\phi(S)=$ the set of all primes in the set $S$ and

$\mathbb{C}(S)$ = the set of all composite numbers in $S$, $S\subseteq\mathbb{N}$;

$\pi(S)=$ the number of primes in $S$ when $S$ is finite and $S\subset\mathbb{N}$;

$\phi(n)$ = $\phi([1,n])$ = the set of all primes $\leq$ $n$ and

$\pi(n)$ = $\pi([1,n])$ = the number of primes $\leq$ $n$, $n\in\mathbb{N}$;

$\left\lfloor\frac{m}{n}\right\rfloor$ = integer part of $\frac{m}{n}$ when $m,n\in\mathbb{N}$.

Throughout the paper, $p_{1},p_{2},\ldots$ denotes the primes 2, 3, …numbered in increasing order and $[m,n]$ $=\{k\in\mathbb{N}:m\leq k\leq n\}$, $m,n\in\mathbb{N}$ and $m\leq n$.

Definition 1**.**

Let $a,n_{1},n_{2}\in\mathbb{N}$, $n_{1}\leq n_{2}$ and $[a]$ $=\{a,2a,\ldots\}=a\mathbb{N}$, the set of multiples of $a$ in $\mathbb{N}$. Then, we denote the set of all multiples of $a$ each lies between $n_{1}$ and $n_{2}$ by $([a]:n_{1},n_{2})$. Thus, $([a]:n_{1},n_{2})$ $=\{ka/~{}n_{1}\leq ka\leq$ $n_{2},~{}k\in\mathbb{N}\}$, $a,n_{1},n_{2}\in\mathbb{N}$ and $n_{1}\leq n_{2}$.

The following lemma is an important result to obtain the set of all primes in an interval $[m,n]$ that is used to generate larger primes in this paper, $m,n\in\mathbb{N}$ and $m\leq n$.

Lemma 2.1**.**

Let $1\leq i<j$, $i,j\in\mathbb{N}$, $p_{i}$, $p_{j}$ be primes, $Q_{i,j}=\left\lfloor\frac{p_{j}}{p_{i}}\right\rfloor$, $Q_{i,j}^{{}^{\prime}}=\left\lfloor\frac{2p_{j}}{p_{i}}\right\rfloor$, $Q_{i,j}^{{}^{\prime\prime}}=\left\lfloor\frac{p_{j}^{2}}{p_{i}}\right\rfloor$ and $Q_{i,j}^{{}^{\prime\prime\prime}}=\left\lfloor\frac{p_{j}^{2}+4p_{j}+3}{p_{i}}\right\rfloor$. Then,

1. (1)

$p_{i}(Q_{i,j}+1)$* is the smallest integer multiple of $p_{i}$ that is greater than $p_{j}$.* 2. (2)

$p_{i}Q_{i,j}^{{}^{\prime}}$* is the biggest integer multiple of $p_{i}$ that is less than or equal to $2p_{j}$.* 3. (3)

$p_{i}Q_{i,j}^{{}^{\prime\prime}}$* is the biggest integer multiple of $p_{i}$ that is less than or equal to $p_{j}^{2}$.* 4. (4)

$p_{i}Q_{i,j}^{{}^{\prime\prime\prime}}$* is the biggest integer multiple of $p_{i}$ that is less than or equal to $p_{j}^{2}+4p_{j}+3$.* 5. (5)

$([p_{i}]:p_{j}+1,2p_{j})$* $=\{p_{i}(Q_{i,j}+1),p_{i}(Q_{i,j}+2)$, $\ldots,p_{i}Q_{i,j}^{{}^{\prime}}\}$.* 6. (6)

$([p_{i}]:p_{j}+1,p_{j}^{2})$* $=\{p_{i}(Q_{i,j}+1),p_{i}(Q_{i,j}+2),\ldots,p_{i}Q_{i,j}^{{}^{\prime\prime}}\}$.* 7. (7)

$([p_{i}]:p_{j}+1,p_{j}^{2}+4p_{j}+3)$* $=\{p_{i}(Q_{i,j}+1),p_{i}(Q_{i,j}+2),\ldots,p_{i}Q_{i,j}^{{}^{\prime\prime\prime}}\}$.* 8. (8)

In $[p_{j}+1,p_{j}^{2}+4p_{j}+3]$, any composite number has $p_{1}$, $p_{2}$, $\ldots$ or $p_{j}$ as a divisor. And $p_{j}^{2}+4p_{j}+3<p_{j+1}^{2}$. 9. (9)

For $j\geq 2$, any composite number whose prime divisors, each $>p_{j}$ will be $\geq p_{j+1}^{2}\geq{(p_{j}+2)}^{2}$ $>$ $p_{j}^{2}+4p_{j}+3$. 10. (10)

In $[p_{j}+1,p_{j}^{2}+4p_{j}+3]$, any integer which is not divisible by $p_{1}$, $p_{2}$, $\ldots$, $p_{j-1}$ and $p_{j}$ is a prime.

Proof.

Given, $1\leq i<j$, $Q_{i,j}=\left\lfloor\frac{p_{j}}{p_{i}}\right\rfloor$, $Q_{i,j}^{{}^{\prime}}=\left\lfloor\frac{2p_{j}}{p_{i}}\right\rfloor$, $Q_{i,j}^{{}^{\prime\prime}}=\left\lfloor\frac{p_{j}^{2}}{p_{i}}\right\rfloor$ and $Q_{i,j}^{{}^{\prime\prime\prime}}=\left\lfloor\frac{p_{j}^{2}+4p_{j}+3}{p_{i}}\right\rfloor$. This implies, $p_{i}<p_{j}$ and $Q_{i,j}$, $Q_{i,j}^{{}^{\prime}}$, $Q_{i,j}^{{}^{\prime\prime}}$, $Q_{i,j}^{{}^{\prime\prime\prime}}$ are the quotients when $p_{j}$, $2p_{j}$, $p_{j}^{2}$, $p_{j}^{2}+4p_{j}+3$ are divided by $p_{i}$, respectively.

Let $p_{j}=p_{i}Q_{i,j}+R_{i,j}$ where $Q_{i,j}$ and $R_{i,j}$ are quotient and remainder when $p_{j}$ is divided by $p_{i}$, $1\leq R_{i,j}\leq p_{i}-1$ since $p_{j}$ is a prime number greater than $p_{i}$. Similarly, let $2p_{j}=p_{i}Q_{i,j}^{{}^{\prime}}+R_{i,j}^{{}^{\prime}}$, $p_{j}^{2}=p_{i}Q_{i,j}^{{}^{\prime\prime}}+R_{i,j}^{{}^{\prime\prime}}$ and $p_{j}^{2}+4p_{j}+3=p_{i}Q_{i,j}^{{}^{\prime\prime\prime}}+R_{i,j}^{{}^{\prime\prime\prime}}$ where $Q_{i,j}^{{}^{\prime}}$ and $R_{i,j}^{{}^{\prime}}$ are quotient and remainder when $2p_{j}$ is divided by $p_{i}$, $Q_{i,j}^{{}^{\prime\prime}}$ and $R_{i,j}^{{}^{\prime\prime}}$ are quotient and remainder when $p_{j}^{2}$ is divided by $p_{i}$ and $Q_{i,j}^{{}^{\prime\prime\prime}}$ and $R_{i,j}^{{}^{\prime\prime\prime}}$ are quotient and remainder when $p_{j}^{2}+4p_{j}+3$ is divided by $p_{i}$, $0\leq R_{i,j}^{{}^{\prime}},R_{i,j}^{{}^{\prime\prime\prime}}\leq p_{i}-1$ and $1\leq R_{i,j}^{{}^{\prime\prime}}\leq p_{i}-1$. From the above, we get results (1), (2), (3) and (4).

Result (5) follows from (1) and (2). Result (6) follows from (1), (3) and $p_{j}$ is an odd prime $>2$. Result (7) follows from (1) and (4).

For $j\geq 2$, $p_{j+1}\geq p_{j}+2>p_{j}$. This implies, the smallest composite number which does not have $p_{1}$, $p_{2}$, $\ldots$ or $p_{j}$ as a divisor is $p_{j+1}^{2}$. And $p_{j+1}^{2}\geq{(p_{j}+2)}^{2}=p_{j}^{2}+4p_{j}+4>p_{j}^{2}+4p_{j}+3$ for $j\geq 2$. Hence, we get results (8) and (9). Result (10) follows from (8) and (9). ∎

Remark 1**.**

For $j\geq 2$, in the above lemma, (10) implies, in $[p_{j}+1,p_{j}^{2}+4p_{j}+$ $3]$, each integer which is not divisible by $p_{1}$, $p_{2}$, $\ldots$, $p_{j-1}$ and $p_{j}$ is a prime and is one among $p_{j+1}$, $p_{j+2}$, $\ldots$, $p_{j+k}$ (and contained in $[p_{j}+1,p_{j}^{2}+4p_{j}+3])$ where $p_{j}^{2}+4p_{j}+3$ is even, $p_{j}+2$ $\leq$ $p_{j+k}$ $\leq$ $p_{j}^{2}+4p_{j}+2$ and $k\in\mathbb{N}$.

3. Main Result

In this section, we present all the three methods to generate prime numbers even though the third is the best. In all these methods, we obtain all prime numbers contained in each consecutive intervals $IpP_{k+1}$ by simple method and thereby we obtain all consecutive prime numbers in a faster way, $k\in\mathbb{N}$. We use Theorems 3.1 and 3.2 in the first method; Theorems 3.3 and 3.4 in the second method and Theorems 3.5 and 3.6 in the third method.

Lemma 3.1**.**

Let $1\leq i<j$, $Q_{i,j}=\left\lfloor\frac{p_{j}}{p_{i}}\right\rfloor$, $Q_{i,j}^{{}^{\prime}}=\left\lfloor\frac{2p_{j}}{p_{i}}\right\rfloor$, $Q_{i,j}^{{}^{\prime\prime}}=\left\lfloor\frac{p^{2}_{j}}{p_{i}}\right\rfloor$, $Q_{i,j}^{{}^{\prime\prime\prime}}=\left\lfloor\frac{p^{2}_{j}+4p_{j}+3}{p_{i}}\right\rfloor$ and $i,j\in\mathbb{N}$. Then,

1. (1)

the set of all composite numbers in $[p_{j}+1,2p_{j}]$ is given by

$\mathbb{C}([p_{j}+1,2p_{j}])$* $=$ $\bigcup^{j}_{i=1}$ $p_{i}\{Q_{i,j}+1,Q_{i,j}+2,\ldots,Q_{i,j}^{{}^{\prime}}\}$*

$=$* $\bigcup^{j-1}_{i=1}$ $p_{i}\{Q_{i,j}+1,Q_{i,j}+2,\ldots,Q_{i,j}^{{}^{\prime}}\}$ $\bigcup$ $\{2p_{j}\}$;* 2. (2)

the set of all composite numbers in $[p_{j}+1,p^{2}_{j}]$ is given by

$\mathbb{C}([p_{j}+1,p^{2}_{j}])$* $=$ $\bigcup^{j}_{i=1}$ $p_{i}\{Q_{i,j}+1,Q_{i,j}+2,\ldots,Q_{i,j}^{{}^{\prime\prime}}\}$;* 3. (3)

the set of all composite numbers in $[p_{j}+1,p^{2}_{j}+4p_{j}+3]$ is given by

$\mathbb{C}([p_{j}+1,p^{2}_{j}+4p_{j}+3])$* $=$ $\bigcup^{j}_{i=1}$ $p_{i}\{Q_{i,j}+1,Q_{i,j}+2,\ldots,Q_{i,j}^{{}^{\prime\prime\prime}}\}$;* 4. (4)

the number of primes in $[p_{j}+1,2p_{j}]$ is given by

$\pi([p_{j}+1,2p_{j}])$* $=\pi(2p_{j})-j$;* 5. (5)

the number of primes in $[p_{j}+1,p^{2}_{j}]$ is given by

$\pi([p_{j}+1,p^{2}_{j}])$* $=\pi(p^{2}_{j})-j$ and* 6. (6)

the number of primes in $[p_{j}+1,p^{2}_{j}+4p_{j}+3]$ is given by

$\pi([p_{j}+1,p^{2}_{j}+4p_{j}+3])$* $=\pi(p^{2}_{j}+4p_{j}+3)-j$.*

Proof.

Given, $1\leq i<j$, $Q_{i,j}=\left\lfloor\frac{p_{j}}{p_{i}}\right\rfloor$ and $Q_{i,j}^{{}^{\prime}}=\left\lfloor\frac{2p_{j}}{p_{i}}\right\rfloor$. Then,

1. (1)

the set of all composite numbers in $[p_{j}+1,2p_{j}]$ is

$\mathbb{C}([p_{j}+1,2p_{j}])$ = $\bigcup^{j}_{i=1}\{$multiples of $p_{i}$ in $[p_{j}+1,2p_{j}]\}$.

= $\bigcup^{j}_{i=1}$ $([p_{i}]:p_{j}+1,2p_{j})$.

= $\{2p_{j}\}$ $\bigcup$ $(\bigcup^{j-1}_{i=1}$ $\{p_{i}(Q_{i,j}+1),p_{i}(Q_{i,j}+2),\ldots,p_{i}Q_{i,j}^{{}^{\prime}}\})$

using Lemma 2.1 (5) and also $2p_{j}$ is the only composite

number with $p_{j}$ as a divisor in $[p_{j}+1,2p_{j}]$.

Proof of $(2)$ and $(3)$ are similar to $(1)$.

Results $(4)$-$(6)$ follow from $\pi([p_{j}+1,k])$ = $\pi([1,k])$ - $\pi([1,p_{j}])$ and $\pi([1,p_{j}])$ = $j$ = $|\{p_{1},p_{2},...,p_{j}\}|$ for any $k\geq p_{j}+1$.

Hence, the lemma is true. ∎

Prime numbers in $[p_{j}+1,2p_{j}],$  $j\geq 2$.

Here, for $j\geq 2$, we obtain all prime numbers contained in the interval $[p_{j}+1,2p_{j}]$ by removing its composite numbers.

Theorem 3.1**.**

Let $1\leq i<j$, $i,j\in\mathbb{N}$, $Q_{i,j}=\left\lfloor\frac{p_{j}}{p_{i}}\right\rfloor$ and $Q_{i,j}^{{}^{\prime}}=\left\lfloor\frac{2p_{j}}{p_{i}}\right\rfloor$. Then, the set $[p_{j}+1,2p_{j}]$ $\setminus$ $\mathbb{C}([p_{j}+1,2p_{j}])$

1. (1)

= $[p_{j}+2,2p_{j}-1]$ $\setminus$ $\mathbb{C}([p_{j}+2,2p_{j}-1])$. 2. (2)

is non-empty. 3. (3)

contains prime number(s) as its element(s). 4. (4)

= $\phi([p_{j}+1,2p_{j}])$ $=\phi([p_{j}+2,2p_{j}-1])$

= the set of all prime numbers contained in $[p_{j}+1,2p_{j}]$. 5. (5)

= $[p_{j}+1,2p_{j}]$ $\setminus$ $(\bigcup^{j}_{i=1}$ $p_{i}\{Q_{i,j}+1,Q_{i,j}+2,\ldots,Q_{i,j}^{{}^{\prime}}\})$. 6. (6)

= $[p_{j}+2,2p_{j}-1]$ $\setminus$ $(\bigcup^{j-1}_{i=1}$ $p_{i}\{Q_{i,j}+1,Q_{i,j}+2,\ldots,Q_{i,j}^{{}^{\prime}}\})$.

Proof.

For $j\geq 2$, the set of all prime numbers contained in $[p_{j}+1,2p_{j}]$ is obtained by removing all composite numbers contained in $[p_{j}+1,2p_{j}]$.

For $j\geq 1$, by Bertrand’s postulate, $[p_{j}+1,2p_{j}]$ contains at least one prime and for $j\geq 2$, $p_{j}+1$ is composite. Hence, results $(1)$, $(2)$ and $(3)$ are true.

By the definition of $\phi(S)$, $(4)$ is true.

For $j\geq 2$, using $(1)$, $[p_{j}+2,2p_{j}]$ contains at least one prime. And any prime number contained in $[p_{j}+2,2p_{j}]$ is greater than $p_{j}$ and less than $2p_{j}$. But for $j\geq 2$, any composite number in $[p_{j}+2,2p_{j}]$ contains $p_{1}$, $p_{2}$, $\ldots$ or $p_{j}$ as a factor. Hence, after removing all multiples of $p_{1}$, $p_{2}$, $\ldots$, $p_{j}$ from $[p_{j}+2,2p_{j}]$, the resultant set contains only prime(s). This implies,

$[p_{j}+1,2p_{j}]$ $\setminus$ $\mathbb{C}([p_{j}+1,2p_{j}])$

$=[p_{j}+2,2p_{j}]$ $\setminus$ $\mathbb{C}([p_{j}+2,2p_{j}])$ = $\phi([p_{j}+2,2p_{j}])$

= $[p_{j}+2,2p_{j}-1]$ $\setminus$ ($\bigcup^{j-1}_{i=1}$ $p_{i}\{Q_{i,j}+1,Q_{i,j}+2,\ldots,Q_{i,j}^{{}^{\prime}}\})$

using Lemma 3.1 where $Q_{i,j}=\left\lfloor\frac{p_{j}}{p_{i}}\right\rfloor$ and $Q_{i,j}^{{}^{\prime}}=\left\lfloor\frac{2p_{j}}{p_{i}}\right\rfloor$ and $1\leq i<j$. And thereby $(5)$ and $(6)$ are true. ∎

In the above result, we could see that $\phi([p_{j}+2,2p_{j}])$, the set of all prime numbers in the interval $[p_{j}+1,2p_{j}]$, is a non-empty set for every $j\geq 2$, $j\in\mathbb{N}$. Consider the following example to calculate different primes using Theorem 3.1 with the notation of $MIpP_{j}$ $=$ maximum value in $IpP_{j}$, $MpP_{j}$ $=$ maximum value in $pP_{j}$ and $\pi(IpP_{j})=$ number of prime numbers in $IpP_{j}$ $=\pi(pP_{j})=|pP_{j}|$, $j=$ $1,2,\ldots$.

Example 1**.**

$p_{1}=2$, $IpP_{1}=\{2\}$ $=pP_{1}$, $MIpP_{1}=2=MpP_{1}$.

$\Rightarrow$* $IpP_{2}=[MIpP_{1}+1,2MpP_{1}]=[3,4]$, $MIpP_{2}=4$,*

$pP_{2}=\phi(IpP_{2})=\phi([3,4])=\{3\}$, $p_{2}=3$, $MpP_{2}=3$.

$\Rightarrow$* $IpP_{3}=[MIpP_{2}+1,2MpP_{2}]=[5,6]$, $MIpP_{3}=6$,*

$pP_{3}=\phi(IpP_{3})=\phi([5,6])$* = $\{5\}$, $p_{3}=5$, $MpP_{3}=5$.*

$\Rightarrow$* $IpP_{4}=[MIpP_{3}+1,2MpP_{3}]$ = $[7,10]$, $MIpP_{4}=10$,*

$pP_{4}=\phi(IpP_{4})=\{7\}$, $p_{4}=7$, $MpP_{4}=7$.

$\Rightarrow$* $IpP_{5}=[MIpP_{4}+1,2MpP_{4}]=[11,14]$, $MIpP_{5}=14$,*

$pP_{5}=$* $\phi(IpP_{5})=\phi([11,14])$ = $\{11,13\}$,*

$MpP_{5}=13$, $p_{5}=11$, $p_{6}=13$.

$\Rightarrow$* $IpP_{6}=[MIpP_{5}+1,2MpP_{5}]=[15,26]$, $MIpP_{6}=26$,*

$pP_{6}=\phi(IpP_{6})=\phi([15,26])$* = $\{17,19,23\}$, $MpP_{6}=23$,*

$p_{7}=17$, $p_{8}=19$, $p_{9}=23$.

$\Rightarrow$* $IpP_{7}=[MIpP_{6}+1,2MpP_{6}]=[27,46]$, $MIpP_{7}=46$,*

$pP_{7}=\phi(IpP_{7})$* $=\phi([27,46])$ $=\{29,31,37,41,43\}$,*

$p_{10}=29$, $p_{11}=31$, $p_{12}=37$, $p_{13}=41$, $p_{14}=43$, $MpP_{7}=43$.

$\Rightarrow$* $IpP_{8}=[MIpP_{7}+1,2MpP_{7}]=[47,86]$, $MIpP_{8}=86$,*

$pP_{8}=\phi(IpP_{8})=\phi([47,86])$* $=\{47,53,59,61,67,71,73,79,83\}$,*

$p_{15}=47$, $p_{16}=53$, $p_{17}=59$, $p_{18}=61$, $p_{19}=67$,

$p_{20}=71$, $p_{21}=73$, $p_{22}=79$, $p_{23}=83$, $MpP_{8}=83$.

$\Rightarrow$* $IpP_{9}=[MIpP_{8}+1,2MpP_{8}]=[87,166]$, $MIpP_{9}=166$,*

$pP_{9}=\phi(IpP_{9})=\phi([87,166])$* $=\{89,97,101,103,107,109,$*

$113,127,131,137,139,149,151,157,163\}$, $MpP_{9}=163$,

$p_{24}=89$, $p_{25}=97$, $p_{26}=101$, $p_{27}=103$, $p_{28}=107$,

$p_{29}=109$, $p_{30}=113$, $p_{31}=127$, $p_{32}=131$, $p_{33}=137$,

$p_{34}=139$, $p_{35}=149$, $p_{36}=151$, $p_{37}=157$, $p_{38}=163$.

$\Rightarrow$* $IpP_{10}=[MIpP_{9}+1,2MpP_{9}]=[167,326]$, $MIpP_{10}=326$,*

$pP_{10}$* $=\phi(IpP_{10})$ $=\phi([167,326])=\{167=p_{39},173,179,181$,*

$191,193,197,199,211,223,227,229,233,239,241,251,257,263$,

$269,271,277,281,283,293,307,311,313,317=p_{66}\}$, $\dots$.

With the notation, $pP_{1}=\{2\}$, $pP_{2}=\{3\}$, $pP_{3}=\{5\}$, $pP_{4}=\{7\}$, $pP_{5}=$ $\{11,13\}$, $pP_{6}=\{17,19,23\}$, $\dots$, $IpP_{i+1}=[MIpP_{i}+1,$ $2MpP_{i}]$ and $pP_{i+1}=\phi(IpP_{i+1})$ where $MIpP_{i}=$ max $IpP_{i}=$ the maximum value in $IpP_{i}$ and $MpP_{i}=$ $max~{}pP_{i}$, $i=2,3,\ldots$, we obtain the following result. Hereafter, we call $pP_{i}$ as the $i^{th}$ pocket of prime(s), $i\in\mathbb{N}$.

Theorem 3.2**.**

Let $pP_{1}=\{2\}$, $IpP_{1}=\{2\}$, $IpP_{2}=[3,4]$, $pP_{2}=\{3\}$, $IpP_{3}=[5,6]$, $pP_{3}=\{5\}$, $IpP_{4}=[7,10]$, $pP_{4}=\{7\}$, $IpP_{5}=[11,14]$, $pP_{5}=\{11,13\}$, $\dots$ where $IpP_{i+1}=[MIpP_{i}+1,$ $2MpP_{i}]$, $pP_{i+1}=\phi(IpP_{i+1})$, $MIpP_{i}=$ max $IpP_{i}=$ the maximum value in $IpP_{i}$ and $MpP_{i}=$ max $pP_{i}=$ the maximum value in $pP_{i}$, $i=2,3,\ldots$. Then,

1. (1)

The set of all prime numbers,

$\mathbb{P}$* = $\phi(\mathbb{N})=$ $pP_{1}\cup pP_{2}\cup pP_{3}\cup\ldots$ $=$ $\bigcup^{\infty}_{j=1}$ $pP_{j}$.* 2. (2)

the set of all pockets of primes,

$\{pP_{1},pP_{2},\ldots,pP_{i},\ldots\}$* partitions the set of all primes. $\Box$*

Observation 1**.**

1. (1)

To obtain the set of all primes $\mathbb{P}$, one can consider different set of pockets of primes. 2. (2)

A set of pockets of primes whose union is the set of all primes need not be a partition of $\mathbb{P}$.

We have generated prime numbers from successive intervals $IpP_{j+1}$; in each successive interval we obtain all its primes and thereby the corresponding pockets of primes $pP_{j+1}=\phi([MIpP_{j}+1,2MpP_{j}])$ for $j=1,2,\ldots$, starting with $pP_{1}=\{2\}$, $IpP_{1}=\{2\}$, $IpP_{j+1}=$ $[MIpP_{i}+1,$ $2MpP_{i}]$, $MIpP_{j}=$ max $IpP_{j}$ and $MpP_{j}=$ max $pP_{j}$. Now, the question is ‘Is it possible to find out any better method to generate primes?’ The following method based on $IpP_{j+1}$ $=[MIpP_{j}+1,$ $MpP_{j}^{2}]$ and $pP_{j+1}$ $=\phi(IpP_{j+1})$ $=\phi([MIpP_{j}+2,$ $MpP_{j}^{2}-2])$ where $MIpP_{j}$ $=$ maximum value in $IpP_{j}$ and $MpP_{j}$ $=$ maximum value in $pP_{j}$ with $p_{1}=2$, $pP_{1}=\{2\}$, $IpP_{2}=\{3\}=pP_{2}$ and $p_{2}=3$ $=MIpP_{2}=$ $MpP_{2}$ for $j=2,3,\ldots$ is a better method. To begin with, let us calculate prime numbers in $[p_{j}+1,p_{j}^{2}],$  $j\geq 2$.

Prime numbers in $[p_{j}+1,p_{j}^{2}],$  $j\geq 2$.

Here, we obtain prime numbers by considering successive intervals $IpP_{j+1}$ as $[p_{j}+1,p_{j}^{2}]$  instead of $[p_{j}+1,2p_{j}]$ in the previous method for $j\geq 2$ with $p_{1}=2$, $p_{2}=3$, $\ldots$. Interval $[p_{j}+1,p_{j}^{2}]$, in general, contains more primes than in $[p_{j}+1,2p_{j}]$ for $j\geq 2$. Similar to Theorem 3.1, we get Theorems 3.3 as follows.

Theorem 3.3**.**

For $j\geq 2$, the set $[p_{j}+1,p_{j}^{2}]$ $\setminus$ $\mathbb{C}([p_{j}+1,$ $p_{j}^{2}])$

1. (1)

is non-empty; 2. (2)

contains prime number(s) as its element(s); 3. (3)

$=\phi([p_{j}+1,p_{j}^{2}])=$* $\phi([p_{j}+2,p_{j}^{2}-2])$;* 4. (4)

$=[p_{j}+2,p_{j}^{2}-2]$* $\setminus$ $(\bigcup^{j}_{i=1}$ $p_{i}\{Q_{i,j}+1,Q_{i,j}+2,\ldots,Q_{i,j}^{{}^{\prime\prime}}\})$*

where $Q_{i,j}=\left\lfloor\frac{p_{j}}{p_{i}}\right\rfloor$ and $Q_{i,j}^{{}^{\prime\prime}}=\left\lfloor\frac{p_{j}^{2}-2}{p_{i}}\right\rfloor$, $1\leq i\leq j$.

Proof.

By Bertrand’s postulate, $[p_{j}+1,2p_{j}]$ contains at least one prime, $j\in\mathbb{N}$. For $j\in\mathbb{N}$, $[p_{j}+1,p_{j}^{2}]$ $\supseteq$ $[p_{j}+1,2p_{j}]$ and so $[p_{j}+1,p_{j}^{2}]$ contains at least one prime since $p_{j}\geq 2$. Also, for $j\geq 2$, $p_{j}$ is odd prime and $p_{j}^{2}$ an odd composite number. And hence the set of all prime numbers contained in $[p_{j}+1,p_{j}^{2}]$ is same as the set of all prime numbers contained in $[p_{j}+2,p_{j}^{2}-2]$ and is obtained by removing all composite numbers contained in $[p_{j}+2,p_{j}^{2}-2]$. Hence $(1)$, $(2)$ and $(3)$ are true.

For $j\geq 2$, using $(1)$, $[p_{j}+2,p_{j}^{2}-2]$ contains at least one prime. And any prime number contained in $[p_{j}+1,p_{j}^{2}]$ is greater than $p_{j}$ and less than $p_{j}^{2}$ for $j\geq 2$. Also, for $k,l\in\mathbb{N}$, prime numbers $p_{j+k},p_{j+l}>p_{j}$ and $p_{j+k}p_{j+l}>p_{j}^{2}$. This implies, $p_{1}$, $p_{2}$, $\ldots$, $p_{j-1}$ or $p_{j}$ is a divisor of every composite number contained in $[p_{j}+1,p_{j}^{2}]$ and after removal of all composite numbers that are multiples of $p_{1}$, $p_{2}$, $\ldots$ or $p_{j}$ from $[p_{j}+1,p_{j}^{2}]$, the resultant set contains only prime(s). This implies,

$[p_{j}+1,p_{j}^{2}]$ $\setminus$ $\mathbb{C}([p_{j}+1,p_{j}^{2}])$ = $\phi([p_{j}+1,p_{j}^{2}])$

= $[p_{j}+2,p_{j}^{2}-2]$ $\setminus$ $\mathbb{C}([p_{j}+2,p_{j}^{2}-2])$ since $p_{j}+1$ is even for $j\geq 2$.

= $[p_{j}+2,p_{j}^{2}-2]$ $\setminus$ ($\bigcup^{j}_{i=1}$ $p_{i}\{Q_{i,j}+1,Q_{i,j}+2,\ldots,Q_{i,j}^{{}^{\prime\prime}}\})$ follows from Lemma 3.1 where $Q_{i,j}=\left\lfloor\frac{p_{j}}{p_{i}}\right\rfloor$ and $Q_{i,j}^{{}^{\prime\prime}}=\left\lfloor\frac{p_{j}^{2}-2}{p_{i}}\right\rfloor$, $1\leq i\leq j$.

Hence, result $(4)$ is true. ∎

In the above result, $\phi([p_{j}+1,p_{j}^{2}])$, the set of all prime number(s) in the interval $[p_{j}+1,p_{j}^{2}]$, is a non-empty set for every $j$ using Bertrand’s postulate, $j\in\mathbb{N}$. Now, let us calculate different pockets of primes $pP_{j+1}$ $=\phi([MIpP_{j}+2,MpP_{j}^{2}-2])$ with $p_{1}=2$, $IpP_{1}=\{2\}$ $=pP_{1}$, $\pi(pP_{1})=|pP_{1}|=1$, $MIpP_{1}=2=MpP_{1}$, $IpP_{2}=$ $[MIpP_{1}$ $+1,MpP_{1}^{2}]=[3,4]$, $pP_{2}=\phi(IpP_{2})=\{3\}$, $\pi(pP_{2})=1$, $MIpP_{2}=4$, $MpP_{2}=3$ $=p_{\pi(pP_{1})+\pi(pP_{2})}=p_{2}$, $IpP_{j+1}$ $=[MIpP_{j}+1$, $MpP_{j}^{2}]$, $pP_{j+1}$ $=\phi([MIpP_{j}+2,MpP_{j}^{2}-2])$, $MpP_{j+1}=$ max  $pP_{j+1}$ $=p_{\pi(pP_{1})+\pi(pP_{2})+\ldots+\pi(pP_{j})+\pi(pP_{j+1})}$ and $\pi(pP_{j+1})=|pP_{j+1}|$, $j\geq 2$ and $j\in\mathbb{N}$. Here, we obtain $pP_{j}$ from $IpP_{j}$ using Theorem 3.3.(4).

Example 2**.**

Let $p_{1}=2$, $IpP_{1}=\{2\}=pP_{1}$, $\pi(pP_{1})=1=|pP_{1}|$, $MIpP_{1}$ = 2 = $MpP_{1}$.

$\Rightarrow$* $IpP_{2}=[MIpP_{1}+1,MpP_{1}^{2}]=[3,4]$, $pP_{2}=\phi(IpP_{2})=\{3\}$,*

$\pi(pP_{2})=|pP_{2}|=1$, $MpP_{2}=3=p_{\pi(pP_{1})+\pi(pP_{2})}=p_{2}$, $MIpP_{2}=4$.

$\Rightarrow$* $IpP_{3}=[MIpP_{2}+1,MpP_{2}^{2}]=[5,9]$, $MIpP_{3}=9$,*

$pP_{3}=\phi(IpP_{3})=\{5,7\}$, $\pi(pP_{3})=|pP_{3}|=2$,

$MpP_{3}=7=p_{\pi(pP_{1})+\pi(pP_{2})+\pi(pP_{3})}=p_{4}$.

$\Rightarrow$* $IpP_{4}=[MIpP_{3}+1,MpP_{3}^{2}]$ $=[10,49]$, $MIpP_{4}=49$,*

$pP_{4}=\phi(IpP_{4})$* $=\phi([10,49])=\phi([11,47])$*

$=\{11,13,17,19,23,29,31,37,41,43,47\}$, $\pi(pP_{4})=11$,

$p_{\pi(pP_{1})+\pi(pP_{2})+\pi(pP_{3})+1}=p_{5}=11$, $p_{6}=13$, $p_{7}=17$, $p_{8}=19$,

$p_{9}=23$, $p_{10}=29$, $p_{11}=31$, $p_{12}=37$, $p_{13}=41$, $p_{14}=43$,

$p_{15}=47=MpP_{4}=p_{\pi(pP_{1})+\pi(pP_{2})+\pi(pP_{3})+\pi(pP_{4})}$.

$\Rightarrow$* $IpP_{5}=[MIpP_{4}+1,MpP_{4}^{2}]$ $=[50,2209]$, $MIpP_{5}=2209$,*

$pP_{5}=\phi(IpP_{5})=\phi([MIpP_{4}+1,MpP_{4}^{2}])$* $=\phi([51,2209])$*

$=\phi([53,2207])$* = $\{53,59,61,67,71,73,79,83,89,97$,*

$101,103,107,109,113,127,131,137,139,149,151,157,163,167,173$,

$179,181,191,193,197,199,211,223,227,229,233,239,241,251,257$,

$263,269,271,277,281,283,293,307,311,313,317,331,337,347,349$,

$353,359,367,373,379,383,389,397,401,409,419,421,431,433,439$,

$443,449,457,461,463,467,479,487,491,499,503,509,521,523,541$,

$547,557,563,569,571,577,587,593,599,601,607,613,617,619,631$,

$641,643,647,653,659,661,673,677,683,691,701,709,719,727,733$,

$739,743,751,757,761,769,773,787,797,809,811,821,823,827,829$,

$839,853,857,859,863,877,881,883,887,907,911,919,929,937$,

$941,947,953,967,971,977,983,991,997,1009,1013,1019,1021$,

$1031,1033,1039,1049,1051,1061,1063,1069,1087,1091,1093,1097$,

$1103,1109,1117,1123,1129,1151,1153,1163,1171,1181,1187,1193$,

$1201,1213,1217,1223,1229,1231,1237,1249,1259,1277,1279,1283$,

$1289,1291,1297,1301,1303,1307,1319,1321,1327,1361,1367,1373$,

$1381,1399,1409,1423,1427,1429,1433,1439,1447,1451,1453,1459$,

$1471,1481,1483,1487,1489,1493,1499,1511,1523,1531,1543,1549$,

$1553,1559,1567,1571,1579,1583,1597,1601,1607,1609,1613,1619$,

$1621,1627,1637,1657,1663,1667,1669,1693,1697,1699,1709,1721$,

$1723,1733,1741,1747,1753,1759,1777,1783,1787,1789,1801,1811$,

$1823,1831,1847,1861,1867,1871,1873,1877,1879,1889,1901,1907$,

$1913,1931,1933,1949,1951,1973,1979,1987,1993,1997,1999,2003$,

$2011,2017,2027,2029,2039,2053,2063,2069,2081,2083,2087,2089$,

$2099,2111,2113,2129,2131,2137,2141,2143,2153,2161,2179,2203$,

$2207\}$, $\pi(pP_{5})=314$,

$p_{15+1}=p_{16}=53$, $p_{17}=59$, $\ldots$, $p_{15+\pi(pP_{5})}=p_{329}=MpP_{5}=2207$.

$\Rightarrow$* $IpP_{6}=[MIpP_{5}+1,MpP_{5}^{2}]$ $=[2210,4870849])$, $MIpP_{6}=4870849$,*

$pP_{6}=\phi([MIpP_{5}+2,MpP_{5}^{2}-2])$* $=\phi([2211,4870847])$*

$=\{2213=p_{330},2221=p_{331},\ldots,4870843=p_{340059}\}$,

$\pi(pP_{6})=339730$, $MpP_{6}=4870843=p_{329+npP_{6}}=p_{340059}$, $\dots$.

Continuing the above process of obtaining prime numbers from successive pockets of primes, one can generate consecutive prime numbers up to any given limit. The algorithm used here is $p_{1}=2$, $IpP_{1}$ $=\{2\}=pP_{1}$, $\pi(pP_{1})=|pP_{1}|=1$, $MIpP_{1}=2=MpP_{1}$, $p_{2}=3$, $IpP_{2}$ $=\{3\}=$ $pP_{2}$, $MIpP_{2}=3=MpP_{2}$, $\pi(pP_{2})=1$, then $IpP_{i+1}=[MIpP_{i}+1,MpP_{i}^{2}]$, $MpP_{i}=$ max   $pP_{i}$ $=$ the maximum value of $pP_{i}=$ the biggest prime in $pP_{i}=$ the biggest prime in $IpP_{i}$, $pP_{i+1}=$ the ${(i+1)}^{th}$ pocket of primes $=\phi([MIpP_{i}+2,MpP_{i}^{2}-2])$, $\pi(pP_{i+1})=|pP_{i+1}|$, $i=2,3,\ldots$.

Remark 2**.**

The above algorithm works since $MpP_{i}^{2}$ is composite and odd for every $i\geq 2$ whereas $MpP_{1}^{2}=4$ is even. Thus, corresponding to the above algorithm, we get Theorem 3.4.

Theorem 3.4**.**

Let $p_{1}=2$, $IpP_{1}$ $=\{2\}=pP_{1}$, $p_{2}=3$, $IpP_{2}$ $=\{3\}=$ $pP_{2}$, $p_{2}=3=MIpP_{2}=MpP_{2}$, $IpP_{i+1}=[MIpP_{i}+1,MpP_{i}^{2}]$ and $pP_{i+1}=\phi([MIpP_{i}+2,MpP_{i}^{2}-2])$ where $MIpP_{i}=max~{}IpP_{i}$ and $MpP_{i}=max~{}pP_{i}$, $i=2,3,\ldots$. Then, the pockets of primes are

1. (1)

$pP_{1}=\phi(\{2\})=\{2\}$, $pP_{2}$ $=\phi(\{3\})=\{3\}$, $pP_{3}=\{5,7\}$,

$pP_{4}=\{11=p_{5},13,17,19,23,29,31,37,41,43,47=p_{15}\}$;

$pP_{5}=\{53=p_{16},59=p_{17},61,\ldots,2207=p_{329}=MpP_{5}\}$,

$pP_{6}=\{2213=p_{330},2221=p_{331},\dots,4870843=p_{340059}\}$,

.   .   .  . 2. (2)

The set of all prime numbers,

$\mathbb{P}=$* $\phi(\mathbb{N})$ $=\bigcup_{j=1}^{\infty}$ $pP_{j}$*

$=pP_{1}$* $\bigcup$ $pP_{2}$ $\bigcup$ $(\bigcup_{j=2}^{\infty}$ $\phi([MIpP_{j}+2,MpP_{j}^{2}-2]))$*

$=pP_{1}$* $\bigcup$ $pP_{2}$*

$\bigcup$* $(\lim_{n\rightarrow\infty}$ $(\bigcup_{j=2}^{n}$ $\phi([MIpP_{j}+2,MpP_{j}^{2}-2])))$.* 3. (3)

The set of all pockets of primes, $\{pP_{1},pP_{2},\ldots\}$ partitions the set of all prime numbers, $\mathbb{P}$.

Proof.

Here, for $i\neq j$ and $i,j\in\mathbb{N}$, $pP_{i}\cap pP_{j}=\emptyset$ and each pocket of prime(s) is non-empty by Bertrand Postulate. Also, the set of all intervals $IpP_{j}$ partition $\mathbb{N}\setminus\{1\}$ and each interval $IpP_{j}$ covers pocket of prime(s) $pP_{j}$, $j=1,2,\ldots$. Then, the theorem is true from the following.

$\mathbb{P}=$ $\phi(\mathbb{N})$ $=\bigcup_{j=1}^{\infty}$ $pP_{j}$ $=\lim_{n\rightarrow\infty}(\bigcup_{j=1}^{n}$ $pP_{j}$)

$=pP_{1}$ $\bigcup$ $pP_{2}$ $\bigcup$ $(\lim_{n\rightarrow\infty}$ $(\bigcup_{j=2}^{n}$ $\phi([MpP_{j}+2,MpP_{j}^{2}-2])))$. ∎

Remark 3**.**

In the two methods that we have discussed to generate prime numbers using successive pockets of prime(s) $pP_{1}$, $pP_{2}$, $\ldots$ with $IpP_{1}$ $=\{2\}$ $=pP_{1}$, $IpP_{2}=$ $\{3\}$ $=pP_{2}$, $MIpP_{j}=max~{}IpP_{j}$ and $MpP_{j}$ $=max~{}pP_{j}$, in the first method we have $IpP_{j+1}=$ $[MIpP_{j}+1,$ $2MpP_{j}]$ and $pP_{j+1}=\phi(IpP_{j+1})=\phi([MIpP_{j}$ $+1,$ $2MpP_{j}])$ for $j=2,3,\ldots$ and in the second method $IpP_{j+1}=$ $[MIpP_{j}+1,MpP_{j}^{2}]$ and $pP_{j+1}=\phi(IpP_{j+1})=$ $\phi([MIpP_{j}+2,$ $MpP_{j}^{2}-2])$ for $j=2,3,\ldots$. Let $\pi(pP_{j})=|pP_{j}|$, $j=1,2,\ldots$. It is easy to observe the following.

1. (1)

The second method is a better one in terms of number of primes generated in successive pockets of primes. The number of elements in the successive pockets of primes in the two methods are as follows.

Method-1:* $\pi(pP_{1})=1$, $\pi(pP_{2})=1$, $\pi(pP_{3})=1$, $\pi(pP_{4})=1$, $\pi(pP_{5})=2$, $\pi(pP_{6})=3$, $\pi(pP_{7})=5$, $\pi(pP_{8})=9$, $\pi(pP_{9})=15$, $\pi(pP_{10})=28$, $\ldots$.*

Method-2:* $\pi(pP_{1})=1$, $\pi(pP_{2})=1$, $\pi(pP_{3})=2$, $\pi(pP_{4})=11$, $\pi(pP_{5})=314$, $\pi(pP_{6})=339730$, $\ldots$.*

See Table $1$ for more details. 2. (2)

To generate prime numbers, if we consider any bigger interval of the form $[MIpP_{j}+1,$ $MpP_{j}^{2}+k]$ in the previous methods, then the method may fail since $[MIpP_{j}+1,MpP_{j}^{2}+k]$ may contain composite numbers of the form $MpP_{j}^{2}+m$ for some $k$ and $m$ such that $1\leq m\leq k$ and each of its prime divisor is $>MpP_{j}$. 3. (3)

In the previous remark, let us ask the question, whether there exists such a value of $k$ for which the algorithm works, in general, to generate prime numbers? If it exists, is it possible to find out such value(s) of $k$ ? And what is the maximum general value of $k$ ? Yes, such values of $k$ exist. See the following. 4. (4)

For $p_{1}=2$, $p_{2}=3$, $pP_{1}=\{2,3\}$, $IpP_{1}=[2,3]$ and $j\geq 2$, $p_{j}$ is an odd prime and the next prime $p_{j+1}\geq p_{j}+2$ and so $p_{j+1}p_{j+1}$ $\geq$ ${(p_{j}+2)}^{2}$ $=p_{j}^{2}+4p_{j}+4$ $>p_{j}^{2}+4p_{j}+3$. This implies, $k$ $=4p_{j}+3$ is a possible value of $k$, in general, for which the following algorithm works.

$p_{1}=2$, $p_{2}=3$, $IpP_{1}=[2,3]$, $pP_{1}=\{2,3\}$,

$\pi(pP_{1})=2$, $MIpP_{1}=3=MpP_{1}=p_{2}$,

$k=4MpP_{j}+3$,

$IpP_{j+1}=[MIpP_{j}+1,MpP_{j}^{2}+k]$,

$pP_{j+1}=\phi([MIpP_{j}+1,MpP_{j}^{2}+k-1])$,

$\pi(pP_{j+1})=$* $|pP_{j+1}|$,*

$MIpP_{j+1}=$* $max~{}IpP_{j+1}$ and*

$MpP_{j+1}=$* $max~{}pP_{j+1}$ $=p_{\pi(pP_{1})+\pi(pP_{2})+\ldots+\pi(pP_{j+1})}$.*

Here, $MpP_{j}^{2}+4MpP_{j}+3$ is even for $j=1,2,\ldots$. 5. (5)

Continuing the above process of generating primes from successive pockets of primes, one can generate consecutive prime numbers up to any given limit. Thus, corresponding to the third method, we get the following important result.

Remark 4**.**

In the above algorithm if we take $k=4MpP_{j}+4$ instead of $4MpP_{j}+3$, then the algorithm fails when $MpP_{j}$ and its successive prime are twin primes (two primes of difference two). Thus, $k=4MpP_{j}+3$ is a possible, in general, maximum value of $k$ used in the third method to generate successive primes and correspondingly we get the most important result, Theorem 3.5.

Theorem 3.5**.**

Let $j\in\mathbb{N}$ and $k\in\mathbb{N}_{0}$. Then,

1. (1)

$k=4MpP_{j}+3$* is a possible, in general, maximum value of $k$ for which the following algorithm works to generate successive prime numbers from successive pocket of primes and doesn’t fail.*

$p_{1}=2$, $p_{2}$ $=3$, $IpP_{1}$ $=[2,3]$, $pP_{1}=\{2,3\}$,

$\pi(pP_{1})=|pP_{1}|=2$, $MpP_{1}=$ $3=MIpP_{1}=p_{\pi(pP_{1})}=p_{2}$,

$IpP_{j+1}=[MIpP_{j}+1,MpP_{j}^{2}+k]$, $MIpP_{j}=$ $max$ $IpP_{j}$,

$pP_{j+1}=\phi([MIpP_{j}+1,MpP_{j}^{2}+k-1])$, $\pi(pP_{j+1})$ $=|pP_{j+1}|$ and $MpP_{j+1}$ $=max~{}pP_{j+1}=$ $p_{\pi(pP_{1})+\pi(pP_{2})+\ldots+\pi(pP_{j+1})}$ for $j=1,2,\ldots$. 2. (2)

Let $IpP_{1}=[2,3]$, $pP_{1}=\{2,3\}$, $p_{1}=2$, $p_{2}=3$, $\pi(pP_{1})=2$ and $MIpP_{1}=3=MpP_{1}=p_{\pi(pP_{1})}=p_{2}$. Then, the set of all primes,

$\mathbb{P}$* $=$ $\bigcup_{j=1}^{\infty}$ $pP_{j}$ $=$ $\lim_{n\rightarrow\infty}(\bigcup_{j=1}^{n}$ $pP_{j})$*

$=pP_{1}$* $\bigcup(\lim_{n\rightarrow\infty}(\bigcup_{j=1}^{n}\phi([MIpP_{j}+1,MpP_{j}^{2}+4MpP_{j}+2])))$*

where $IpP_{j+1}=[MIpP_{j}+1,MpP_{j}^{2}+4MpP_{j}+3]$, $MIpP_{j}$ $=$ $max~{}IpP_{j}$, $pP_{j+1}=\phi(IpP_{j+1})=\phi([MIpP_{j}+1,MpP_{j}^{2}+4MpP_{j}$ $+2])$, $\pi(pP_{j+1})$ $=|pP_{j+1}|$ and $MpP_{j+1}=$ $p_{\pi(pP_{1})+\pi(pP_{2})+\ldots+\pi(pP_{j+1})}$ = $max~{}pP_{j+1}$, $j=$ $1,2,\ldots$. $\Box$

Definition 2**.**

Let $pP_{1}=\{2,3\}$, $p_{1}=2$, $p_{2}=3$, $\pi(pP_{1})=2$, $IpP_{1}=[2,3]$, $MIpP_{1}$ $=3$ $=MpP_{1}$, $IpP_{j+1}=[MIpP_{j}+1,$ $MpP_{j}^{2}$ $+4MpP_{j}+3]$, $MIpP_{j+1}=$ $max~{}IpP_{j+1}$, $pP_{j+1}=\phi([MIpP_{j}+1,MpP_{j}^{2}$ $+4MpP_{j}+2])$, $\pi(pP_{j+1})=$ $|pP_{j+1}|$ and $MpP_{j+1}=$ $p_{\pi(pP_{1})+\pi(pP_{2})+\ldots+\pi(pP_{j+1})}=$ $max$ $pP_{j+1}$, $j=1,2,\ldots$. Then, $pP_{1}$, $pP_{2}$, $\ldots$ are called the standard pockets of primes.

Theorem 3.6**.**

For $j\geq 2$, the set $[p_{j}+1,p_{j}^{2}+4p_{j}+3]$ $\setminus$ $\mathbb{C}([p_{j}+1,p_{j}^{2}+$ $4p_{j}+3])$

1. (1)

is non-empty; 2. (2)

contains prime number(s) as its element(s); 3. (3)

$=\phi([p_{j}+1,p_{j}^{2}+4p_{j}+3])=$* $\phi([p_{j}+2,p_{j}^{2}+4p_{j}+2])$;* 4. (4)

$=[p_{j}+2,p_{j}^{2}+4p_{j}+2]$* $\setminus$ $(\bigcup^{j}_{i=1}$ $p_{i}\{Q_{i,j}+1,Q_{i,j}+2,\ldots,Q_{i,j}^{{}^{\prime\prime\prime}}\})$*

where $Q_{i,j}=\left\lfloor\frac{p_{j}}{p_{i}}\right\rfloor$ and $Q_{i,j}^{{}^{\prime\prime\prime}}=\left\lfloor\frac{p_{j}^{2}+4p_{j}+2}{p_{i}}\right\rfloor$, $1\leq i\leq j$.

Proof.

Proof is similar to Theorem 3.3. ∎

Now, let us calculate the standard pockets of primes using Theorem 3.6.

Example 3**.**

$p_{1}=2$, $p_{2}=3$, $IpP_{1}=[2,3]$, $pP_{1}=\{2,3\}$, $MIpP_{1}=3$ $=MpP_{1}$, $\pi(pP_{1})=2$;

$\Rightarrow IpP_{2}=[MIpP_{1}+1,MpP_{1}^{2}+4MpP_{1}+3]=[4,24]$, $MIpP_{2}=24$,

$pP_{2}=\phi([MIpP_{1}+2,MpP_{1}^{2}+4MpP_{1}+2])$* $=\phi([5,23])$ $=\{5,7$,*

$11,13,17,19,23\}$, $\pi(pP_{2})=7$, $p_{3}=5$, $p_{4}=7$, $p_{5}=11$, $p_{6}=13$,

$p_{7}=17$, $p_{8}=19$, $p_{9}=23$, $MpP_{2}=23=p_{\pi(pP_{1})+\pi(pP_{2})}=p_{9}$;

$\Rightarrow$* $IpP_{3}=[MIpP_{2}+1,MpP_{2}^{2}+4MpP_{2}+3]=[25,624]$,*

$MIpP_{3}=$* $624$, $pP_{3}=\phi([MIpP_{2}+2,MpP_{2}^{2}+4MpP_{2}+2])$*

$=\phi([26,623])=\{29,31,37,41,43,47,53,59,61,67,71,73,79,83$,

$89,97,101,103,107,109,113,127,131,137,139,149,151,157,163$,

$167,173,179,181,191,193,197,199,211,223,227,229,233,239,241$,

$251,257,263,269,271,277,281,283,293,307,311,313,317,331,337$,

$347,349,353,359,367,373,379,383,389,397,401,409,419,421,431$,

$433,439,443,449,457,461,463,467,479,487,491,499,503,509,521$,

$523,541,547,557,563,569,571,577,587,593,599,601,607,613,617$,

$619\}$,    $\pi(pP_{3})=105$,

$MpP_{3}=619=p_{9+\pi(pP_{3})}=p_{114}$, $p_{10}=29$, $p_{11}=31$, $\ldots$, $p_{114}=619$;

$\Rightarrow$* $IpP_{4}=[MIpP_{3}+1,MpP_{3}^{2}+4MpP_{3}+3]=\phi([625,385640])$,*

$MIpP_{4}=$* $385640$, $pP_{4}=\phi([MIpP_{3}+2,MpP_{3}^{2}+4MpP_{3}+2])$*

$=\phi([626,385639])=\{631,641,\ldots,385639\}$,

$\pi(pP_{4})=32622$, $MpP_{4}=$ $385639=p_{114+\pi(pP_{4})}=p_{32736}$,

$p_{115}=631$, $p_{116}=641$, $\ldots$, $385639=MpP_{4}=p_{32736}$;

$\Rightarrow$* $IpP_{5}=[MIpP_{4}+1,MpP_{4}^{2}+4MpP_{4}+3]$*

$=\phi([385641,148718980880])$, $MIpP_{5}=$ $148718980880$,

$pP_{5}=\phi([MIpP_{4}+2,MpP_{4}^{2}+4MpP_{4}+2])$**

$=\phi([385642,$* $148718980879])$ $=\{385657=p_{32737}$,*

$385661=p_{32738},\ldots,MpP_{5}$* = biggest prime $\leq 148718980879\}$,*

$\pi(pP_{5})=|pP_{5}|$, $MpP_{5}=$ $p_{32736+\pi(pP_{5})}$;    $\ldots$.

On Twin Prime Conjecture One great open problem in number theory for many years is the twin prime conjecture, which states that there are infinitely many primes $p$ such that $p+2$ is also prime [21]. Here, we restate the twin prime conjecture in terms of the standard pockets of primes that may help to settle the conjecture in an easier way.

Conjecture 1**.**

[Twin Prime Conjecture in terms of pockets of primes] Each pocket of primes of the standard pockets of primes, starting from the second, contains at least one pair of twin primes.

Remark 5**.**

1. (1)

If we prove the above conjecture, then the Twin prime conjecture is also proved since the number of pockets of primes in the standard pockets of prime is infinite. 2. (2)

Finding number of twin primes in each pocket of primes in the standard pockets of primes and obtaining any general formula seems to be an interesting open problem.

4. Algorithm to generate primes

To generate prime numbers to any given limit, the third method is the fastest and the best among the three methods that we have discussed. Here, we present an algorithm corresponding to the third method only that is based on Theorems 3.5 and 3.6. It is easy to check whether a given natural number is prime or not by comparing with the list of primes already generated provided the number is within the list.

Generating primes with $[MIpP_{k}+1,MpP_{k}(MpP_{k}+4)+3],$ $k\geq 1$.

Algorithm:

$p_{1}=2$, $p_{2}=3$, $IpP_{1}=[2,3]$, $pP_{1}=\{2,3\}$,

$\pi(pP_{1})=2$, $MIpP_{1}=3=MpP_{1}$,

$j$ $=$ $2$,    ($j=$ Number of primes already generated)

$x=8$,   ($x=$ Number of pockets of primes that are to be considered)

Do $10$ $k=1$ to $x$,

$IpP_{k+1}=[MIpP_{k}+1,MpP_{k}(MpP_{k}+4)+3]$,

$IP_{k+1}=[MIpP_{k}+2,MpP_{k}(MpP_{k}+4)+2]$,

$C(IP_{k+1})=\{\}$,

Do $20$ $i=1$ to $j$,

$Q_{i,k}=\left\lfloor\frac{MpP_{k}}{p_{i}}\right\rfloor$,

$Q_{i,k}^{{}^{\prime}}=\left\lfloor\frac{MpP_{k}(MpP_{k}+4)+2}{p_{i}}\right\rfloor$,

$C_{i}(IP_{k+1})=\{p_{i}(Q_{i,k}+1),p_{i}(Q_{i,k}+2),\ldots,p_{i}Q_{i,k}^{{}^{\prime}}\}$,

$20$ $C(IP_{k+1})$ $=$ $C(IP_{k+1})$ $\bigcup$ $C_{i}(IP_{k+1})$

$pP_{k+1}$ $=$ $IP_{k+1}$ $\setminus$ $C(IP_{k+1})$,  ($C(IP_{k})=$ set of all composites in $IP_{k}$.)

$\pi(pP_{k+1})$ $=$ $|pP_{k+1}|$,

$MIpP_{k+1}=$ $max~{}IpP_{k+1}$,

$MpP_{k+1}=$ $max~{}pP_{k+1}$,

$tpP_{k+1}$ $=$ $pP_{k+1}$,   (Here, $tpP_{k+1}$ represents temporary $pP_{k+1}$.)

$y$ $=$ $j$,

$j$ $=$ $j+\pi(pP_{k+1})$,

$p_{j}=$ $MpP_{k+1}$,

print $(k+1$, ′th  pocket  of  primes  with  No.  of  primes  = ’, $\pi(pP_{k+1})$,

’ starting  with ’, $y+1$, ′th  prime  to ’, $j$, ′th  prime.  They  are’)

print $(p_{y+l},~{}l=1$ to $\pi(pP_{k+1}))$,

$10$ Print $(^{\prime}MpP_{k+1}=^{\prime},p_{j},^{\prime}=^{\prime},j,$ ′th  prime.’)

End

In the three methods that we have discussed so far, we could see that the values of $\pi(pP_{k})$, in each method, play an important roll and thereby we define the following.

Definition 3**.**

The sequence $\pi(pP_{1})$, $\pi(pP_{2})$, $\ldots$ is called the sequence of order of pockets of primes, $pP_{1}$, $pP_{2}$, $\ldots$.

Table $1$ shows different sequences of order of pockets of primes in the three methods up to $k$ $=$ $8$ in the first method, $k=6$ in the second and $k=4$ in the third.

Remark 6**.**

A new study is needed on the behaviour of sequences of order of pockets of primes in the three methods.

Remark 7**.**

While generating prime numbers using the above method, one need not start with $IpP_{1}$ $=[2,3]$, $pP_{1}=\{2,3\}$, $IpP_{2}$ $=$ $[4,24]$, $\ldots$. If $p_{1}$, $p_{2}$, $\ldots$, $p_{k}$ are already known primes, then by taking $IpP_{1}=[p_{1},p_{k}]$ and $pP_{1}=\{p_{1},p_{2},\ldots,p_{k}\}$ in the above method, one can generate primes by considering successive pockets of primes, $pP_{j+1}$, $j=1,2,\ldots$ and $k\geq 3$.

5. Riemann zeta function and prime number generation

In this section, we relate prime number generation with Riemann zeta function. Riemann zeta function $\zeta(z)$ [9] is given by

[TABLE]

the series is convergent for $Re(z)>1$.

Proposition 5.1**.**

[TABLE]

and $n_{1}>n_{2}>\ldots$, then $p_{k+1}=n_{1}$, $p_{k+2}=n_{2}$, $\ldots$, $p_{k+\pi(pP_{k+1})}=n_{\pi(pP_{k+1})}$ and $p_{k+1}$, $p_{k+2}$, $\ldots$, $p_{k+\pi(pP_{k+1})}\in pP_{k+1}$ $=\phi([p_{k}+1,p_{k}(p_{k}+4)+3])$ where $\pi(pP_{k+1})=|pP_{k+1}|$ and $Re(z)>1$, $k\in\mathbb{N}$.

Proof.

For $Re(z)>1$, we have

[TABLE]

[TABLE]

• $\frac{1}{2^{z}}(1+\frac{1}{2^{z}}+\frac{1}{3^{z}}+\frac{1}{4^{z}}+~{}.~{}.~{}.~{})$.

[TABLE]

$=1+$ sum of the terms of $\zeta(z)$ without multiples of $\frac{1}{2^{z}}$.

Similarly, we get,

[TABLE]

= 1+ sum of the terms of $\zeta(z)$ without multiples of $\frac{1}{2^{z}}$ or $\frac{1}{3^{z}}$.

Continuing the above process and using Theorem 3.6, we get the result. ∎

Remark 8**.**

Number of primes in a pocket of primes plays an important role in the process of generating primes from successive pockets of primes and also measure the power/speed of the algorithm that generates primes. Legendre [11] obtained a formula to calculate $\pi(x)$, the number of primes $\leq x$, $x\in\mathbb{N}$. Based on this, we obtain, the number of primes in $[p_{k}+1,p_{k}(p_{k}+4)+3]$ as

$\pi([p_{k}+1,p_{k}(p_{k}+4)+3])=\pi(p_{k}(p_{k}+4)+3)-\pi(p_{k})$**

$=\pi(p_{k}(p_{k}+4)+3)-k$, $k\in\mathbb{N}$.

Acknowledgement. I express my sincere thanks to the Central University of Kerala, Kasaragod, Kerala, India - 671 316 and St. Jude’s College (SJC), Thoothoor, Kanyakumari District, Tamil Nadu, India - 629 176 for providing facilities to do this research work. I also express my sincere thanks to Prof. S. Krishnan (late) and Dr. V. Mohan, Department of Mathematics, Thiyagarayar College of Engineering, Madurai, Tamil Nadu, India - 625 015, Prof. M. I. Jinnah, Mr. Albert Joseph and Prof. L. John, University of Kerala, Trivandrum, Kerala, India - 695 581; Rev. Fr. J. G. Jesudas (late), Correspondent, Rev. Fr. C. M. George (late), Principal, Dr. K. Vareethaiah and Dr. S. Amirthaiyan, St. Jude’s College, Thoothoor, India and Dr. Oscar Fredy, Royal Liverpool University Hospital, Liverpool, U.K. for their support and encouragement to do research.