Guest Posted July 28, 2009 Report Share Posted July 28, 2009 while studying prime numbers i found a couple interesting relationships. in the range between p(n)*(c+2) and p(n-2)*c there is at least 1 number such that it is prime and n+c is prime, where c is an even number > 0. for example, let c = 8. 2*8 = 16, 5*10 = 50 23 +8 = 31 3*8 = 18, 7*10 = 70 53 +8 = 61 5*8 = 40, 11*10 = 110 59+8 = 67 here's another one. p(n+1) <= 2*p(n) -p(n-2) Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 28, 2009 Report Share Posted July 28, 2009 while studying prime numbers i found a couple interesting relationships. in the range between p(n)*(c+2) and p(n-2)*c there is at least 1 number such that it is prime and n+c is prime, where c is an even number > 0. for example, let c = 8. 2*8 = 16, 5*10 = 50 23 +8 = 31 3*8 = 18, 7*10 = 70 53 +8 = 61 5*8 = 40, 11*10 = 110 59+8 = 67 here's another one. p(n+1) <= 2*p(n) -p(n-2) 3*8=24 Quote Link to comment Share on other sites More sharing options...
0 Guest Posted July 28, 2009 Report Share Posted July 28, 2009 well i say 3*8 = 18, for extremely small values of 8. Quote Link to comment Share on other sites More sharing options...
Question
Guest
while studying prime numbers i found a couple interesting relationships.
in the range between p(n)*(c+2) and p(n-2)*c there is at least 1 number such that it is prime and n+c is prime, where c is an even number > 0.
for example, let c = 8.
2*8 = 16, 5*10 = 50
23 +8 = 31
3*8 = 18, 7*10 = 70
53 +8 = 61
5*8 = 40, 11*10 = 110
59+8 = 67
here's another one.
p(n+1) <= 2*p(n) -p(n-2)
Link to comment
Share on other sites
2 answers to this question
Recommended Posts
Join the conversation
You can post now and register later. If you have an account, sign in now to post with your account.