My Math Forum Prime Numbers and the relationship between n and P_n

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 April 22nd, 2019, 01:36 PM #1 Member   Joined: Feb 2013 From: London, England, UK Posts: 38 Thanks: 1 Prime Numbers and the relationship between n and P_n We often hear it said that either there is no relation between the natural, or counting numbers, $n$, and their counterparts the primes, $P_{n}$, or that if there is, it is so recondite as to be completely obscure. This is something of a sacred cow, but I don't believe it to be true. I'm not claiming the following is an exact relationship by any means, but it is a start, and a step in the right direction; it holds pretty well for the first 100,000 primes with remarkable accuracy: 2*$n$*$P_{n}$ $\approx$ $P_{n^2}$ Examples: $n$----------$P_{n}$-------2*$n$*$P_{n}$-------$P_{n^2}$ 1-----------2------------4------------2 2-----------3------------12-----------7 3-----------5------------30-----------23 4-----------7------------56-----------53 5-----------11-----------110----------97 6-----------13-----------156----------151 7-----------17-----------238----------227 8-----------19-----------304----------311 9-----------23-----------414----------419 10----------29-----------580----------541 11----------31-----------682----------661 12----------37-----------888----------827 13----------41-----------1066---------1009 14----------43-----------1204---------1193 15----------47-----------1410---------1427 16----------53-----------1696---------1619 17----------59-----------2006---------1879 ... ... ... 77----------389----------59906--------58579 78----------397----------61932--------60289 79----------401----------63358--------62081 80----------409----------65440--------63809 ... ... ...
 April 22nd, 2019, 07:53 PM #2 Senior Member   Joined: Mar 2019 From: iran Posts: 150 Thanks: 9 2nP(n) = 2n*nln(n) = 2n^2ln(n) = n^2 ln(n^2) = P(n^2) Thanks from idontknow

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