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Old 2018-12-14, 21:10   #3
Nov 2016

ACC16 Posts

Originally Posted by sweety439 View Post
We use the extension of the thread, e.g. for R4 k=25, we need n such that 5*2^n-1 and (5*2^n+1)/3 (for even n) (5*2^n-1)/3 and 5*2^n+1 (for odd n) are both primes.
Thus, for R4: (we should find n such that all of these formulas take prime values)

k  formula(s)
1  {(1*2^n-1)/3,1*2^n+1} (for even n) or {1*2^n-1,(1*2^n+1)/3} (for odd n)
2  {2*4^n-1}
3  {3*4^n-1}
4  {2*2^n-1,(2*2^n+1)/3} (for even n) or {(2*2^n-1)/3,2*2^n+1} (for odd n)
5  {5*4^n-1}
6  {6*4^n-1}
7  {(7*4^n-1)/3}
8  {8*4^n-1}
9  {3*2^n-1,3*2^n+1}
10 {(10*4^n-1)/3}
11 {11*4^n-1}
12 {12*4^n-1}
13 {(13*4^n-1)/3}
14 {14*4^n-1}
15 {15*4^n-1}
16 {(4*2^n-1)/3,4*2^n+1} (for even n) or {4*2^n-1,(4*2^n+1)/3} (for odd n)
17 {17*4^n-1}
18 {18*4^n-1}
19 {(19*4^n-1)/3}
20 {20*4^n-1}
21 {21*4^n-1}
22 {(22*4^n-1)/3}
23 {23*4^n-1}
24 {24*4^n-1}
25 {5*2^n-1,(5*2^n+1)/3} (for even n) or {(5*2^n-1)/3,5*2^n+1} (for odd n)

Last fiddled with by sweety439 on 2018-12-14 at 21:20
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