1Prime number with the product inequalityLianzhongLi(Camp Hill Middle School Sichuan Camp Hill 637 700)Abstract: Positive integer n is between two adjacent prime numbers square, the well-established number of product even distribution of the formula (S) andformula (L), According to the prime number theorem, Mertens, Theor-em 3 launch ofprime numbers with the product of inequality and Corollary 1, 2. Keywords: Number theory; prime number; inequality CLC number: 015 Document code: Article ID:Theorem: prime number with the product of inequality: xppxex)11 (2)( Proband lemma.Lemma 1: If, ,，, , , consecutive primes, and | n, then 21p32pjpipjpthe n- o (mod)The number of values。 ijjipnny1)11 ()(Prove : I. The i = 1, =2 , |n 1p1p )11 ()211 (2)(1pnnnnnyiThe conclusion was established. II ,Suppose i = k, the conclusion was founded, namely:Was established. kjjkpnny1)11 ()(When i = k +1, |n，|n， |n，According to the induction hypothesis1p2pkp kjjkpnny1)11 ()( |n 1kp n o (mod) the number is .They is forof1kp1kpn1kp21, 2, 3, ., 1kpnThis a multiples. The number of multiples of the remove , 1kpn1kpnkppp,21Laccording to the induction hypothesis also remaining kjjkppn11)11 ( kjjkkjjkppn pnny1111)11 ()11 ()()11 ()11 (11 kkjjppn11)11 (kjjpn i = k +1, the conclusionWas established. 111)11 ()(kjjkpnnyBy I, II available, the conclusions are true when i is any positive integer. Lemma 1 is proved.Lemma 2：if，so 0.75 xkxpkpx11)11 ()(e)(xProve：let xkxxkln1)(1 )(ln11xxkxk = xkxpkpx11)11 ()()(ln)11 (xxpxp According to Mertens Theorem 3)ln1(ln)11 (2xOxe pxp = xkxpkpx11)11 ()()(ln)11 (xxpxp =)(ln)ln1(ln(2xxxOxe 3=)ln)(ln(ln)(2xxxOxxee is to reduce fluctuations, volatility is also reduced. xkxpxxkpx11)11 ()(limlim=)(ln)ln1(ln(2limxxxOxex e =0.75e xkxpkpx11)11 ()()2(That is 0.75e)(xLemma 2 is proved.Lemma 3 (Prime number with the product distribution of the theorem): If , ,21p，, , , for the second 32pkpip1ipconsecutive primes ，the number of prime numbers formula for2 12 iipnp(S) (n) = （） )()11 ()(1122 1nOpppiksjjkkk kkspplogOr(L), (n) = （）)()11 (1nOpnljjlpp ilogProve： n3+(83)+(248)+(4824)+)(22 1kkpp)(22 1iipp According to Lemma 1, the interval ), the number of prime 2 12, kkppnumbers can be approximately expressed as )(22 1kkpp kjjp1)11 (-Between the number of remove ,，,kpkk pp2 121p32pupLtpLjpmultiples of the few remainingnumber greater than . This is not 1kpkp14because p n caused by, but because when =, the number of-jptp32 1kpkpno in multiples, so get rid kk pp2 1 jpof the ,， multiples of the remaining number of , 21p32pupLtpLjp1kpthe number of multiples of notkp11122 1)11 ()11 (ktjjtuuiii pppppBut1122 1)11 (tuuiii ppppThis is not p is divisible by n,. This is an n by limit, - did 2 12 kkpnpkpkk pp2 1not reach thenumber of . the scope of the multiples. Multiples of the front to jp1kpproveLemma 1, remove the ,，, a multiple of, then remove21p32pjp1kp, less iskp11122 1)11 ()11 (ktjjtuuiii pppppAnd n by limit. Is actually2 12 iipnp1122 1)11 (tuuiii pppp 1122 1)11 (tuuiii pppp11122 1)11 ()11 (ktjjtuuiii pppppTherefore, the small reduction. Expression is consistent with Lemma 1,and also to avoid the backward interpretation to lead to trouble. Multiplier(so that after the primes to take、 in advance to enter, to )11 (1kp)11 (2kp)11 (kspbalance theamount of the small reduction. Therefore, the interval ) a more 2 12, kkppaccurateexpression of number of prime numbers5(1)(22 1kkpp ksjjp1)11 ( 3232 431715pp 1k At this time, just eachinterval of plus or minus 1, you can )( nmake w (S) , so that both the formula (S) in each interval with the )( noverallconsistency, but also to ensure that w (S) .)( n7Formula (L), the error of w (L) and w (k) The derivation of the same, less than lpn)(in the half. The following calculation.lpn)(According to the prime number theoremxxxln)( lpn)()(2ln2lnnpnnn pn npnlll 2 12 iipnp nppil12 12lpn lpn)()( n w(L) 21)( 21lpn)( n)( nTherefore, the mean square error of the formula (S), formula (L) O () is correct. )( nSignificance of the formula (S) and formula (L). The formula (S) is thedistrict between the number of prime numbers. Formula (L) is thecalculation looked at the overall calcula -tion of the number of prime numbers. Set formula (S), formula (L) prime number density distribution function s (x)， (x).l (S) (n)= （）22 1 111()(1)()ksikk kjjppOnp kkspplog kspp k s(x)= kppp)11 (Le