| Tired of seeing ads? Click here to upgrade to Elite Membership! |
Mathematics Forum
|
| Author | Message / Information |
| cramwit Quote | Reply | This message was updated on 11/22/2007 8:44:24 PM by cramwit | Hacking Prime Intervals posted on: 11/22/2007 8:28:44 PM presume the only prime number is 2. we would know with certainty that 1/2 of all numbers would be prime. If we found that only 3 & 2 were prime we would know that there would be 1/2*2/3=1/3 of all numbers subsequent to 3 would be prime. in fact this is true up until the next prime squared. [correct me if wrong] the expected primes over that interval is ((pj-1)/pj))! ie. 1/2*2/3*4/5*6/7....(pj-1)/pj so from pj+1 to (p(j+1))^2 [inclusive,exclusive] there is an expected ((pj-1)/pj))! primes on that interval & i think that might be provable within a rounding error where either a prime was at either ending of the interval or open ends of (pj-1) just missing a prime on either end of the interval. so on the interval for 3-8 inclusive [skipping 3^2=9] of which 3,5,7 are prime, so 3/6 or 0.5 of an expected 0.5 are prime for 4-24 inclusive of which 5,7,11,13,17,19,23 are prime so 7/21 = .3333_ are prime out of an expected 0.3333_ for 8-120 we have 27/113 = 0.238938 of an expected 0.22857 So with increasing accuracy [excepting the initial perfect proportions] the proportion of primes over a thusly specified interval approaches the expected, and is [provably?] fixed within the rounding specifications. I think that might be provable. Are there any glaring errors in my logic? Errors about provability? |
|
cramwit
Quote | Reply | |
Hacking Prime Intervals
replied on: 11/22/2007 9:12:09 PM forgot to include the case of 6-48 with 12 primes, interval length 43 12/43= 0.279069 of an expected 0.26666_ i think i screwed up the interval from 8-120, with an extra prime so it should be 26 primes on interval of 113; 26/113= 0.230088 of an expected 0.22857 [which is actually closer] making a bit of a hash of the particulars, need to get some more clear headed time on it. i think the essential proposition may hold in any event. |
|
cramwit
Quote | Reply | |
Hacking Prime Intervals
replied on: 11/23/2007 2:00:42 PM say we are examining a given indeterminate interval for the 1/7th elimination of potential primes. if the interval was exactly a factor of 7 in length we know there would be exactly 1/7th of them eliminated from primality. no matter how one shifts the sevens when one slips off the near end another comes back on the far end to replace it. so we simply have to consider the non-zero modulo cases. . . . . . 7000000 (zero modulo) if we add non-factors of 7 . . 0000007000000 we can have a maximum of 6 more non-seven factors before we add a factor of seven this is the maximum undershoot of the 1/7th proportion conversely we can add another factor of 7 only followed by non-factors of 7 . . . . . 70000007 . . . . . 700000070 etc the worst case is where a factor of seven is the next addition, that is the worst case overshooting the 1/7th proportion as we add more non-factors it goes closer back to the 1/7th proportion. so essentially it is x number of factors of 7 +- 6/7 of a 'pj' interval or approximately +- one interval or a margin of error of within one pj interval since the interval in question is p(j+1)^2-pj we know that p(j+1) must be at least 2 greater than pj so the minimal length of the interval is (pj+2)^2-pj = pj^2 + 4pj +4 - pj = pj^2 + 3pj + 4 the number of factors of 7 [intervals] is (pj^2 + 3pj +4 +-pj)/pj or from (pj^2 + 2pj + 4)/pj to (pj^2 + 4pj + 4)/pj as pj grows large the variance of 2pj is ever diminished by the quadratic we have to consider the p(j-1) prime intervals of course their modulo variance is +-p(j-1) over an interval of pj^2+(2,4)pj+4 so it is even proportionally smaller & each smaller primes potential error is ever diminishing i think the errors are additive, ie. they have no bearing on one another so the very worst cases are where all the modulos include an extra prime factor or where all the modulos maximally exclude a prime factor so max error is +-(pj+p(j-1)+p(j-2)...)/(pj+2)^2) or sum(pj)/pj^2 something like that it sounds less absolutely certain than i had hoped/thought. anyway, only [well less than] 1/2 numbers are prime so even a sum(pi) at worst is pj^2/4 since the actual interval is p(j+1)^2 - pj that drops lower but safely we can say the error can not exceed 25% of the expected value statistically the probability of it even approaching 25% becomes virtually impossible, but in theory it is a potential so the absolute limit lends provability. so i think we can say with certainty that from pj+1 to p(j+1)^2-1 [inclusive] the proportion of primes will be (pj-1)!/pj! with an absolute error limit of less than 25% from the expected (pj-1)!/pj! |
|
LinkBot
|
Gamers Wanted is looking for people to write game reviews and post news, |
|
|
| Tired of seeing ads? Click here to upgrade to Elite Membership! |
ChatArea.com Help & News Forums | Terms of Use | Contact ChatArea.com | Advertising
Powered By ChatArea.com - Get your free Society today! © Copyright 2003 Wewp!