连续素数3元组猜想

Applied Mathematics and Sciences: An International Journal (MathSJ) Vol.10, No.3, September 2023A POSITIVE INTEGER 𝑵 SUCH THAT𝒑𝒏 + 𝒑𝒏+𝟑 ~ 𝒑𝒏+𝟏 + 𝒑𝒏+𝟐FOR ALL 𝒏 ≥ 𝑵Pham Minh Duc, Luong Van TamVNU University of Science, 334, Nguyen Trai, Thanh Xuan Trung Ward, Thanh Xuan District, Hanoi, VietnamABSTRACTAccording to Bertrand's postulate,we have 𝑝𝑛 + 𝑝𝑛 ≥ 𝑝𝑛+1. Is it true that for all 𝑛 > 1 then 𝑝𝑛−1 + 𝑝𝑛 ≥𝑝𝑛+1? Then 𝑝𝑛 + 𝑝𝑛+3 > 𝑝𝑛+1 + 𝑝𝑛+2where 𝑛 ≥ 𝑁, 𝑁 is a large enough value?KEYWORDSBertrand’s Postulate, Rosser's theorem, L’Hospital Rule, prime number.1. INTRODUCTIONIn 1845, Bertrand conjectured what became known as Bertrand’s postulate: twice any prime strictly exceeds the next prime. Tchebichef presented his proof of Bertrand’s postulate in 1850 and published it in 1852. It is now sometimes called the Bertrand-Chebyshev theorem. Surprisingly, a stronger statement seems not to be well known, but is elementary to prove: The sum of any two consecutive primes strictly exceeds the next prime, except for the only equality 2 + 3 = 5. After I conjectured and proved this statement independently, a very helpful referee pointed out that Ishikawa published this result in 1934 (with a different proof). This observation is a special case of a much more general result, Theorem 1, that is also elementary to prove (given the prime number theorem), and perhaps not previously noticed: If𝑝𝑛 denotes the 𝑛th prime, 𝑛 = 1, 2, 3, … with 𝑝1 = 2, 𝑝2 = 3, 𝑝3 = 5, …, and then there exists a positive integer 𝑁 such that 𝑝𝑛 + 𝑝𝑛+3 ~ 𝑝𝑛+1 + 𝑝𝑛+2for ll 𝑛 ≥ 𝑁. We give the nonnegative integers, then there exists a large enough positive integer 𝑁 such that, for all 𝑛 ≥ 𝑁, 𝑝𝑛 + 𝑝𝑛+3 ~ 𝑝𝑛+1 + 𝑝𝑛+m网页链接 Rosser's theorem for all 𝑛 ≥ 6, we have𝑛(ln𝑛 + lnln𝑛 − 1) < 𝑝𝑛 < 𝑛(ln𝑛 + lnln𝑛)(𝑛 + 1)[ln(𝑛 + 1) + lnln(𝑛 + 1) − 1] < 𝑝𝑛+1 < (𝑛 + 1)[ln(𝑛 + 1) + lnln(𝑛 + 1)](𝑛 + 2)[

上边复制粘贴的是Ai找到的Nagura定理，可惜我不会英语看不懂