The continuity of prime numbers can lead to even continuity(The ultimate)

we take , we an If a group of continuous prime numbers 3,5,7,11,... P , we can get a group of continuous even numbers 6,8,10,12,..., 2n . Then if an adjacent prime number q is followed, the original group of even numbers 6,8,10,12,..., 2n will be finitely extended to 2 (n + 1) or more adjacent even numbers. My purpose is to prove the continuity of prime numbers will to even long 2 (n + 1) can be


∴ (a): {p-3≥2,p-3≠2}→(p-3＞2)→(p-3≥4)
The results of Bertrand Chebyshev theorem are as follows: (2×3)＞(prime number p1)＞3. ∴ (p＞2×3＞3)→(p＞p1＞3)→(P and 3 are adjacent) contradiction. Negation hypothesis. ∴ (3+2)=p, ∴ 3+p=8. ∴ 6→8 The proof process uses the known prime number, and the sieve method is not used in the derivation process, Prime number theorem and Bertrand Chebyshev theorem. By quoting the prime number theorem, we get a finite number of continuous prime numbers, and then generate even numbers according to the requirements of this manuscript. Use the extreme rule again ψ, Force even numbers to continue. least one prime P, which conforms to n < p < 2n−2. Another slightly weaker argument is: for all integers n greater than 1, there is at least one prime P, which conforms to n < p < 2n.

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In this paper, the even number generation rules are as follows ① Only odd primes are allowed as elements. ② Only two prime numbers can be added.(any combination of two odd primes). ③ Two prime numbers can be used repeatedly: (3 + 3), or (3 + 5), or (P + P). ④ meet the previous provisions, and all prime combination to the maximum.
⑤ Take only one of(p a +p b)and(p b+p a ) .

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[References cited 2] The theorem of infinite number of primes: the n bit after each prime can always find another prime. For example, 3 is followed by 5 and 13 is followed by 17; There must be an adjacent prime pi after the prime P.

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Here we only discuss the following cases: prime number sequence and even number sequence  in today's words: the prime number in the prime number sequence discussed in this paper is adjacent and continuous, and the first number is 3. An even number in an even number sequence is contiguous and the first number is 6.

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Remember: all the primes I'll talk about below refer to odd primes (excluding 2). It is not logically proved that any even number greater than 4 satisfies 1  .
Take the minimum prime number 3 from the front of the prime sequence, According to the rule, 3 can only get {3+3=6} Nonexistence: 5+p=6 【Because 1 in 5+1=6 is not defined as a prime number. If 1 is defined as a prime number, this paper will come to the same conclusion】 Note: 5{3}.
Prime number 3, limit is used according to 4  , cannot be: 6→8.
If you want to: 6→8, you must add an adjacent prime number 5. Take the prime number that is greater than p1 and adjacent to p1 as p0,

Starting from (Analysis Ⅱ)
The principle of mathematical complete induction: it is correct in the front, until  What man knows: in n A α When we reach the last p1 , all of them conform to: (lower)≤ (upper).
(2) Take: p0 -p1=2 With the same logic, all the results of the same (T station) are obtained: p0-p1＞2a1 If pc and p1 in n C are not in the same (T station).
In (6), a1=1.. Go straight to the next (T station). I'm not sure how big a1 is. The natural number a1 is to be determined.
(T station)in n C and n A ,