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Bcz we can find out any consecutive prime number by just subtraction of two numbers of series i.e Pn*, and Ipn. Ipn series Number of series Commonn Diffrance Prime number in range Remarks I2 1 2 2 TO 4 Series in AP I3 1 6 3 TO 9 Series in AP I5 2 30 5 TO 25 Series in AP, and Got prime by subtarction 30*N-I5 where N is natural no I7 8 210 7 TO 49 Series in AP, and Got prime by subtarction 210*N-I7 where N is natural no I11 48 2310 11 TO 121 Series in AP, and Got prime by subtarction 2310*N-I11 where N is natural no I13 480 30030 13 TO 169 Series in AP, and Got prime by subtarction 30030*N-I13 where N is natural no I17 5760 510510 17 TO 289 Series in AP, and Got prime by subtarction 510510*N-I17 where N is natural no I19 92160 9699690 19 TO 361 Series in AP, and Got prime by subtarction 9699690*N-I19 where N is natural no I23 1658880 223092870 23 TO 529 Series in AP, and Got prime by subtarction 223092870*N-I23 where N is natural no I29 36495360 6469693230 29 TO 841 Series in AP, and Got prime by subtarction 6469693230*N-I29 where N is natural no I31 1021870080 200560490130 31 TO 961 Series in AP, and Got prime by subtarction 200560490130*N-I31 where N is natural no I37 30656102400 7420738134810 37 TO 1369 Series in AP, and Got prime by subtarction 7420738134810*N-I37 where N is natural no I41 1103619686400 304250263527210 41 TO 1681 Series in AP, and Got prime by subtarction 304250263527210*N-I41 where N is natural no . . . . . . . . . . . . . . . Ipn (p1-1)*(p2-1)*(p3-1)*(p4-1)*………….(pn-1)=k Pn To Pn^2 Series in AP, and Got prime by subtarction {NPn*±IPn }

Prime ={NPn*±IPn } i.e 30N±I5, Prime Number range 5<=P<25, I5 have 2 AP series with C.D=6, and first element = 5 & 7 5,11,17,23….infinite, C.D=6, T1=5 7,13,19,25…..infinite, C.D=6, T1=7 in example below you can see with help of I5 AP series we can find all Prime Number range 5<=P<25 P5* I5 I5 Prime=(P5*-I5) Prime= (I5-P5*) 30 25 35 5 5 30 23 37 7 7 30 19 41 11 11 30 17 43 13 13 30 13 47 17 17 30 11 49 19 19 30 7 53 23 23 60 55 65 5 5 60 53 67 7 7 60 49 71 11 11 60 47 73 13 13 60 43 77 17 17 60 41 79 19 19 60 37 83 23 23

Thanks for reply plz find support of my post Ipn series--- It is well define set of posative number. Range of continues Prime ----It contains each and every prime number in fixed range. No of Ipn sub series...This Ipn series have subseries it is also well define. Common difference......As subseries are in A.P so common difference is given % of Ipn element w.r.t. + Ve integer........its %of Ipn series wrt posative integer.i.e how many Ipn are in set of all +ve integer start from some fix number Ipn series Range of continues Prime No of Ipn sub series Common difference % of Ipn element w.r.t. + Ve integer I2 2 ≤ P < 4 1 2 100% I3 3 ≤ P < 9 1 2 50% I5 5 ≤ P < 25 2 6 33.33% I7 7 ≤ P < 49 8 30 22% I11 11 ≤ P < 121 48 210 20% I13 13 ≤ P < 169 480 2310 19% I17 17 ≤ P < 289 5760 30030 18% I19 19 ≤ P < 361 92160 510510 17% . . . . IpN................ Tends to 0% means near abt all are primes Ex-I3 have one AP Serie and C.d = 2 3, 5, 7, 9, 11, 13------∞ i) Contain all rpimes and specific odd positive integer from 3 to infinite ii) Continious prime 3≤P<9 ii) I3 Contain only 50.00% of positive integer Ex-I5 have two AP Serie and C.d = 6 5 11 17 23 29 35 41 ------∞ 7 13 19 25 31 37 43 -----∞ i) Contain all prime and specific odd positive integer from 5 to infinite ii) Continuous prime 5≤P<25 ii) I5 Contains only 33.33% of positive integer br satish kumar singh

Ipn series--- It is well define set of posative number. Range of continues Prime ----It contains each and every prime number in fixed range. No of Ipn sub series...This Ipn series have subseries it is also well define. Common difference......As subseries are in A.P so common difference is given % of Ipn element w.r.t. + Ve integer........it %of Ipn series wrt posative integer.i.e how many Ipn are in set of all +ve integer Ex-I3( it is Ipn series) have one AP Serie and C.d = 2 3, 5, 7, 9, 11, 13------∞ i) Contain all primes and specific odd positive integer from 3 to infinite ii) Continious prime 3≤P<9 ii) I3 Contain only 50.00% of positive integer Ex-I5( it is Ipn series) have two AP Serie and C.d = 6 5 11 17 23 29 35 41 ------∞ (sub series 1) 7 13 19 25 31 37 43 -----∞ (sub series 2) i) Contain all prime and specific odd positive integer from 5 to infinite ii) Continuous prime 5≤P<25 ii) I5 Contains only 33.33% of positive integer like this we can go for any series and any prime number. br// satish Me to intruded same funda for prime number find out with mod 3#,5#,7#,11# 3#=6 gives all prime 3>=p<9 5#=30 gives all all prime 5>=p<=25 7#=210 gives all prime 7>=p<=49 11#=2310 gives all prime 11>=p<=121 . . . . Pn#=...gives all prime Pn#>=p<=(Pn)^2 so we can find out any prime number in any range.Only thing is that i have required more advance compute which is comfortable with large numbers. Br satish kumar singh