On the singular series in the Jiang prime k-tuple theorem

doi:10.15406/paij.2018.02.00134

eISSN: 2576-4543

Physics & Astronomy International Journal

Review Article Volume 2 Issue 6

On the singular series in the Jiang prime k-tuple theorem

Chun Xuan Jiang

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China Aerospace Science and Technology Corporation, China

Correspondence: Chun Xuan Jiang, China Aerospace Science and Technology Corporation, P. O. Box 142-206, Beijing 100854, P. R. China

Received: September 18, 2018 | Published: November 16, 2018

Citation: Jiang CX. On the singular series in the Jiang prime k-tuple theorem. Phys Astron Int J. 2018;2(6):514-517. DOI: 10.15406/paij.2018.02.00134

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Abstract

Using Jiang function we prove Jiang prime k-tuple theorem. We find true singular series. Using the examples we prove the Hardy-Littlewood prime k-tuple conjecture with wrong singular series. Jiang prime k-tuple theorem will replace the Hardy-Littlewood prime k-tuple conjecture.

Theorems

Jiang prime k-tuple theorem with true singular series.^1,2

We define the prime k-tuple equation

$p, p + n_{i},$ $p, p + n_{i},$ (1)

where $2 | n_{i}, i = 1, \dots k - 1.$ $2| n_{i}, i =1, \dots k - 1.$

we have Jiang function^1,2

$J_{2} (a)) = \prod_{P} (P - 1 - χ (P)),$ $J_{2} (a))= \prod_{P} (P - 1 - χ (P)),$ (2)

where $c o = \prod_{P} P, χ (P)$ $c o = \prod_{P} P, χ (P)$ is the number of solutions of congruence

$\prod_{i = 1}^{k - 1} (q + n_{i}) \equiv 0 (mod P), q = 1, \dots, p - 1$ $\prod_{i =1}^{k - 1} (q + n_{i}) \equiv 0 (mod P), q =1, \dots, p - 1$ (3)

which is true.

If $χ (P) < P - 1$ $χ (P)< P - 1$ then $J_{2} (ω) \neq 0$ $J_{2} (ω) \neq 0$ . There exist infinitely many primes $P$ $P$ such that each of $P + n_{i}$ $P + n_{i}$ is prime. If $χ (P) = P - 1$ $χ (P)= P - 1$ then $J_{2} (ω) 0 =$ $J_{2} (ω)=0$ . There exist finitely many primes $P$ $P$ such that each of $P + n_{i}$ $P + n_{i}$ is prime. $J_{2} (ω)$ $J_{2} (ω)$ is a subset of Euler function $\emptyset (c o)$ $\emptyset (c o)$ .²

If $J_{2} (ω) \neq 0$ $J_{2} (ω) \neq 0$ , then we have the best asymptotic formula of the number of prime $P$ $P$ .^1,2

$π_{k} (N,2) = | {P \leq N : P + n_{i} = p r i m e} | - \frac{J_{2} (c o) c o^{k - 1} N}{\emptyset^{k} (c o) {log}^{k} N} = C (k) \frac{N}{{log}^{k} N}$ $π_{k} (N,2)=|{P \leq N : P + n_{i} = p r i m e}| - \frac{J_{2} (c o) c o^{k - 1} N}{\emptyset^{k} (c o) \log^{k} N} = C (k) \frac{N}{\log^{k} N}$ (4)

$\emptyset (c o) = \prod P (P - 1)$ $\emptyset (c o)= \prod_{P} (P - 1)$

$C (k) = \prod_{P} (1 - \frac{1 + χ (P)}{P}) 1 (- \frac{1}{P})^{- k}$ $C (k)= \prod_{P} (1 - \frac{1 + χ (P)}{P})(1 - \frac{1}{P})^{- k}$ (5)

is Jiang true singular series.

Example 1

Let $k = 2, P, P + 2$ $k =2, P, P + 2$ , twin primes theorem.

From (3) we have

$χ (2 0,) = χ (P) 1 = i f P > 2$ $χ (2)=0, χ (P)=1 i f P >2$ (6)

Substituting (6) into (2) we have

$J_{2} (a)) = \prod_{P \geq 3} (P - 2) \neq 0$ $J_{2} (a))= \prod_{P \geq 3} (P - 2) \neq 0$ (7)

There exist infinitely many primes $P$ $P$ such that $P + 2$ $P + 2$ is prime. Substituting (7) into (4) we have the best asymptotic formula

$π_{k} (N,2) = | {P \leq N : P + 2 = p r i m e} | - 2 \prod_{P \geq 3} (1 - \frac{1}{{(P - 1)}^{2}}) \frac{N}{{log}^{2} N}$ $π_{k} (N,2)=|{P \leq N : P + 2= p r i m e}| - 2 \prod_{P \geq 3} (1 - \frac{1}{{(P - 1)}^{2}}) \frac{N}{\log^{2} N}$ (8)

Example 2

Let $k = 3, P, P + 2, P + 4.$ $k =3, P, P + 2, P + 4.$

From (3) we have

$χ (2 0,) = χ (32) =$ $χ (2)=0, χ (3)=2$ (9)

From (2) we have

$J_{2} (ω) 0 =$ $J_{2} (ω)=0$ (10)

It has only a solution $P = 3, P + 25, = P + 47 =$ $P =3, P + 2=5, P + 4=7$ . One of $P, P + 2, P + 4$ $P, P + 2, P + 4$ is always divisible by 3. Example 2 is not admissible.

Example 3

Let $k = 4, P, P + n$ $k =4, P, P + n$ , where $n = 2,6,8.$ $n =2,6,8.$

From (3) we have

$χ (2 0,) = χ (3 1,) = χ (P) 3 = i f P > 3$ $χ (2)=0, χ (3)=1, χ (P)=3 i f P >3$ . (11)

Substituting (11) into (2) we have

$J_{2} (ω) = \prod_{P \geq 5} (P - 4) \neq 0$ $J_{2} (ω)= \underset{P \geq 5}{\prod_{}} (P - 4) \neq 0$ , (12)

There exist infinitely many primes $P$ $P$ such that each of $P + n$ $P + n$ is prime. Example 3 is admissible.

Substituting (12) into (4) we have the best asymptotic formula

$π_{4} (N,2) = | {P \leq N : P + n = p r i m e} | - \frac{27}{3} \prod_{P \geq 5} \frac{P^{3} (P - 4) N}{{(P - 1)}^{4} {log}^{4} N}$ $π_{4} (N,2)=|{P \leq N : P + n = p r i m e}| - \frac{27}{3} \underset{P \geq 5}{\prod_{}} \frac{P^{3} (P - 4) N}{{(P - 1)}^{4} \log^{4} N}$ (13)

Example 4

Let $k = 5, P, P + n$ $k =5, P, P + n$ , where $n = 2,6,8,12.$ $n =2,6,8,12.$

From (3) we have

$χ (2 0,) = χ (3 1,) = χ (5 3,) = χ (P) 4 = i f P > 5$ $χ (2)=0, χ (3)=1, χ (5)=3, χ (P)=4 i f P >5$ (14)

Substituting (14) into (2) we have

$J_{2} (ω) = \prod_{P \geq 7} (P - 5) \neq 0$ $J_{2} (ω)= \underset{P \geq 7}{\prod_{}} (P - 5) \neq 0$ (15)

There exist infinitely many primes $P$ $P$ such that each of $P + n$ $P + n$ is prime. Example 4 is admissible. Substituting (15) into (4) we have the best asymptotic formula

$π_{5} (N,2) = | {P \leq N : P + n = p r i m e} | - \frac{15^{4}}{2^{11}} \prod_{P \geq 7} \frac{(P - 5) P^{4} N}{{(P - 1)}^{5} {log}^{5} N}$ $π_{5} (N,2)=|{P \leq N : P + n = p r i m e}| - \frac{15^{4}}{2^{11}} \underset{P \geq 7}{\prod_{}} \frac{(P - 5) P^{4} N}{{(P - 1)}^{5} \log^{5} N}$ (16)

Example 5

Let $k = 6, P, P + n$ $k =6, P, P + n$ , where $n = 2,6,8,12,14.$ $n =2,6,8,12,14.$

From (3) and (2) we have

$χ (2 0,) = χ (3 1,) = χ (5 4,) = J_{2} (50) =$ $χ (2)=0, χ (3)=1, χ (5)=4, J_{2} (5)=0$ (17)

It has only $a$ $a$ solution $P = 5, P + 27, = P + 611, = P + 813, = P + 1217, = P + 1419 =$ $P =5, P + 2=7, P + 6=11, P + 8=13, P + 12=17, P + 14=19$ . One of $P + n$ is always divisible by 5. Example 5 is not admissible.

The Hardy-Littlewood prime k-tuple conjecture with wrong singular series.³⁻¹⁷

This conjecture is generally believed to be true, but has not been proved.¹⁸

We define the prime k-tuple equation

$P, P + n_{i}$ (18)

where $2| n_{i}, i =1, \dots, k - 1.$

In 1923 Hardy et al.³ conjectured the asymptotic formula

$π_{k} (N,2)=|{P \leq N : P + n_{i} = p r i m e}| - H (k) \frac{N}{\log^{k} N}$ , (19)

where

$H (k)= \prod_{P} (1 - \frac{V (P)}{P}) (\begin{array}{l} 1 - \underline{1} \\ P \end{array})$ (20)

Is Hardy-Littlewood wrong singula series,

$v (P)$ is the number of solutions of congruence

$\prod_{i =1}^{k - 1} (q + n_{i}) \equiv 0(mod P)$ , $q =1, \dots, P$ . (21)

which is wrong.

From (21) we have $v (P)< P$ and $H (k) \neq 0$ . For any prime k-tuple equation there exist infinitely many primes $P$ such that each of $P + n_{i}$ is prime, which is false.

Conjecture 1

Let $k =2, P, P + 2$ , twin primes theorem

From (21) we have

$v (P)=1$ (22)

Substituting (22) into (20) we have

$H (2)= \prod_{P} \frac{P}{P - 1}$ (23)

Substituting (23) into (19) we have the asymptotic formula

$π_{2} (N,2)=|{P \leq N : P + 2= p r i m e}| - \underset{p}{\prod_{}} \frac{P N}{P - 1 \log^{2} N}$ (24)

which is wrong see example l. They do not get twin primes formula (8).

Conjecture 2

Let $k =3, P, P + 2, P + 4.$

From (21) we have

$v (2)=1, v (P)=2 i f P >2$ (25)

Substituting (25) into (20) we have

$H (3)=4 \underset{P \geq 3}{\prod_{}} \frac{P^{2} (P - 2)}{{(P - 1)}^{3}}$ (26)

Substituting (26) into (19) we have asymptotic formula

$π_{3} (N,2)=|{P \leq N : P + 2= p r i m e, P + 4= p r i m}| - 4 \underset{P \geq 3}{\prod_{}} \frac{P^{2} (P - 2) N}{{(P - 1)}^{3} \log^{3} N}$ (27)

which is wrong see example 2.

Conjecture 3

Let $k =4, P, P + n$ , where $n =2,6,8.$

From (21) we have

$v (2)=1, v (3)=2, v (P)=3 i f P >3$ (28)

Substituting (28) into (20) we have

$H (4)= \frac{27}{2} \prod_{P >3} \frac{P^{3} (P - 3)}{{(P - 1)}^{4}}$ (29)

Substituting (29) into (19) we have asymptotic formula

$π_{4} (N,2)=|{P \leq N : P + n = p r i m e}| - \frac{27}{2} \prod_{P >3} \frac{P^{3} (P - 3) N}{{(P - 1)}^{4} \log^{4} N}$ (30)

Which is wrong see example 3.

Conjecture 4

Let $k =5, P, P + n$ , where $n =2,6,8,12$

From (21) we have

$v (2)=1, v (3)=2, v (5)=3, v (P)=4 if P >5$ (31)

Substituting (31) into (20) we have

$H (5)= \frac{15^{4}}{4^{5}} \prod_{P >5} \frac{P^{4} (P - 4)}{{(P - 1)}^{5}}$ (32)

Substituting (32) into (19) we have asymptotic formula

$π_{5} (N,2)=|{P \leq N : P + n = p r i m e}| - \frac{15^{4}}{4^{5}} \prod_{P >5} \frac{P^{4} (P - 4) N}{{(P - 1)}^{5} \log^{5} N}$ (33)

Which is wrong see example 4.

Conjecture 5

Let $k =6, P, P + n$ , where $n =2,6,8,12,14.$

From (21) we have

$v (2)=1, v (3)=2, v (5)=4, v (P)=5 if P >5$ (34)

Substituting (34) into (20) we have

$H (6)= \frac{15^{5}}{2^{13}} \prod_{P >5} \frac{(P - 5) P^{5}}{{(P - 1)}^{6}}$ (35)

Substituting (35) into (19) we have asymptotic formula

$π_{6} (N,2)=|{P \leq N : P + n = p r i m e}| - \frac{15^{5}}{2^{13}} \prod_{P >5} \frac{(P - 5) P^{5} N}{{(P - 1)}^{6} \log^{6} N}$ (36)

which is wrong see example 5.

Conclusion

The Jiang prime k-tuple theorem has true singular series.The Hardy-Littlewood prime -tuple conjecture has wrong singular series. The tool of additive prime number theory is basically the Hardy-Littlewood wrong prime k-tuple conjecture which are wrong.³⁻¹⁷ Using Jiang true singula series we prove almost all prime theorems. Jiang prime k-tuple theorem will replace Hardy-Littlewood prime k-tuple Conjecture. There cannot be really modern prime theory without Jiang function.