tag:blogger.com,1999:blog-1523608782014773920.post6000573489995911689..comments2024-02-14T19:41:03.209-08:00Comments on Prime Numbers: From Euclid to the infinitude of 6n+1 and 6n-1 primes.Anonymoushttp://www.blogger.com/profile/07751661372310417574noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-1523608782014773920.post-40620805830115679482024-02-14T19:41:03.209-08:002024-02-14T19:41:03.209-08:00I think you are so close! Grab your own 2nd Table ...I think you are so close! Grab your own 2nd Table of https://barkerhugh.blogspot.com/2011/11/semi-prime-distribution-on-n-factor.html and do the sums - then apply your thoughts again! Tell me about the results! This instruction should leed you to a blind peer review - to my approach - which is so similar to yours! I´d love to see the "slightly stronger conjecture than Goldbach" to be true ;) Find the Minimum - and let´s compare results ;Ddanielmhttps://www.blogger.com/profile/09100037061911507130noreply@blogger.comtag:blogger.com,1999:blog-1523608782014773920.post-26450413469661094992017-02-15T13:28:14.620-08:002017-02-15T13:28:14.620-08:00herhangi bir a sayısında toplam 6n+1 asallarının t...herhangi bir a sayısında toplam 6n+1 asallarının toplam miktarı <br />toplam <br />P(6n+1)=a*(1*2*6*12*...*(p(max6n+1)-1))/(2*3*7*13*....*p(max6n+1))<br /><br />herhangi bir a sayısında toplam 6n-1 asallarının toplam miktarı <br />toplam <br />P(6n-1)=a*(1*2*4*10*...*(p(max6n-1)-1))/(2*3*5*11*....*p(max6n-1))<br />a=(Pmax(6n+1))^2+c<br />1<=c<(Pmax(6n+1))*2+3doktor0906https://www.blogger.com/profile/09377314350760393016noreply@blogger.comtag:blogger.com,1999:blog-1523608782014773920.post-58830410106562727302017-02-15T12:19:24.729-08:002017-02-15T12:19:24.729-08:00This comment has been removed by the author.doktor0906https://www.blogger.com/profile/09377314350760393016noreply@blogger.comtag:blogger.com,1999:blog-1523608782014773920.post-71838873993647735302017-02-15T12:06:49.831-08:002017-02-15T12:06:49.831-08:00This comment has been removed by the author.doktor0906https://www.blogger.com/profile/09377314350760393016noreply@blogger.comtag:blogger.com,1999:blog-1523608782014773920.post-11275318424493602482017-02-14T21:06:24.798-08:002017-02-14T21:06:24.798-08:001 * 2 * 4 * 6 * .... (P-1)
2 * 3 * 5 * 7 * .........1 * 2 * 4 * 6 * .... (P-1) <br />2 * 3 * 5 * 7 * ...... p<br /><br />yukardaki bu formül her hangi bir n sayısını çarptığında n sayıda kaç tane asal sayı olacağını sana ortalama olarak verir yani n sayısındaki asal sayı(P) <br />p=n*(1*2*4*6*10*12*...pmax-1)/(2*3*5*7*11*13*...pmax) olur.<br /><br />pmax<=karekökndoktor0906https://www.blogger.com/profile/09377314350760393016noreply@blogger.comtag:blogger.com,1999:blog-1523608782014773920.post-36026047136591799472017-02-14T21:00:39.377-08:002017-02-14T21:00:39.377-08:00This comment has been removed by the author.doktor0906https://www.blogger.com/profile/09377314350760393016noreply@blogger.com