20100611, 13:17  #1 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
3x*2^n1 and 3x*2^n1 possibly twins ?
I've been doing some code in pari and the above seems to work for all twins within the range I was checking if I figure out a few things i may be able to extend it later has the above ever been shown ?
Code:
(10:01) gp > for(n=1,10,forstep(k=3,100,[3],if(isprime(k*2^n1),print1(k"*"2"^"n"1"));if(isprime(k*2^n+1),print1(","k"*"2"^"n"+1"));print("\n"))) 3*2^11,3*2^1+1 6*2^11,6*2^1+1 9*2^11,9*2^1+1 12*2^11 15*2^11,15*2^1+1 ,18*2^1+1 21*2^11,21*2^1+1 24*2^11 27*2^11 30*2^11,30*2^1+1 ,33*2^1+1 36*2^11,36*2^1+1 ,39*2^1+1 42*2^11 45*2^11 ,48*2^1+1 51*2^11,51*2^1+1 54*2^11,54*2^1+1 57*2^11 ,63*2^1+1 66*2^11 69*2^11,69*2^1+1 75*2^11,75*2^1+1 ,78*2^1+1 ,81*2^1+1 84*2^11 87*2^11 90*2^11,90*2^1+1 96*2^11,96*2^1+1 99*2^11,99*2^1+1 3*2^21,3*2^2+1 6*2^21 ,9*2^2+1 12*2^21 15*2^21,15*2^2+1 18*2^21,18*2^2+1 21*2^21 ,24*2^2+1 27*2^21,27*2^2+1 33*2^21 ,39*2^2+1 42*2^21 45*2^21,45*2^2+1 48*2^21,48*2^2+1 57*2^21,57*2^2+1 60*2^21,60*2^2+1 63*2^21 66*2^21 ,69*2^2+1 78*2^21,78*2^2+1 ,84*2^2+1 87*2^21,87*2^2+1 90*2^21 ,93*2^2+1 96*2^21 ,99*2^2+1 3*2^31 6*2^31 9*2^31,9*2^3+1 ,12*2^3+1 21*2^31 24*2^31,24*2^3+1 30*2^31,30*2^3+1 33*2^31 39*2^31,39*2^3+1 ,42*2^3+1 45*2^31 48*2^31 ,51*2^3+1 54*2^31,54*2^3+1 ,57*2^3+1 60*2^31 63*2^31 ,72*2^3+1 75*2^31,75*2^3+1 81*2^31 ,84*2^3+1 90*2^31 93*2^31 ,96*2^3+1 3*2^41 ,6*2^4+1 12*2^41,12*2^4+1 15*2^41,15*2^4+1 ,21*2^4+1 24*2^41 27*2^41,27*2^4+1 30*2^41 ,36*2^4+1 ,42*2^4+1 45*2^41 ,48*2^4+1 54*2^41 57*2^41 ,63*2^4+1 69*2^41 72*2^41,72*2^4+1 ,75*2^4+1 ,78*2^4+1 ,81*2^4+1 90*2^41 93*2^41,93*2^4+1 99*2^41 ,3*2^5+1 6*2^51,6*2^5+1 12*2^51 15*2^51 ,18*2^5+1 ,21*2^5+1 ,24*2^5+1 27*2^51 36*2^51,36*2^5+1 ,39*2^5+1 45*2^51 57*2^51 ,63*2^5+1 66*2^51,66*2^5+1 69*2^51 75*2^51 81*2^51,81*2^5+1 84*2^51,84*2^5+1 90*2^51 99*2^51,99*2^5+1 3*2^61,3*2^6+1 6*2^61 ,9*2^6+1 ,12*2^6+1 18*2^61,18*2^6+1 33*2^61,33*2^6+1 42*2^61,42*2^6+1 45*2^61 ,54*2^6+1 75*2^61,75*2^6+1 ,78*2^6+1 ,87*2^6+1 ,93*2^6+1 96*2^61 ,99*2^6+1 3*2^71 ,6*2^7+1 9*2^71,9*2^7+1 21*2^71,21*2^7+1 ,27*2^7+1 ,39*2^7+1 48*2^71 ,51*2^7+1 54*2^71 ,57*2^7+1 ,60*2^7+1 66*2^71 69*2^71 ,75*2^7+1 ,81*2^7+1 ,84*2^7+1 90*2^71 93*2^71 ,96*2^7+1 99*2^71 ,3*2^8+1 24*2^81 27*2^81 ,30*2^8+1 33*2^81 ,42*2^8+1 45*2^81 ,48*2^8+1 57*2^81,57*2^8+1 60*2^81,60*2^8+1 63*2^81 ,72*2^8+1 84*2^81 87*2^81,87*2^8+1 90*2^81,90*2^8+1 99*2^81 12*2^91 ,15*2^9+1 ,21*2^9+1 ,24*2^9+1 30*2^91,30*2^9+1 ,36*2^9+1 42*2^91 45*2^91,45*2^9+1 51*2^91,51*2^9+1 54*2^91 ,63*2^9+1 66*2^91 69*2^91 ,78*2^9+1 87*2^91 6*2^101 ,12*2^10+1 15*2^101,15*2^10+1 ,18*2^10+1 21*2^101 27*2^101 33*2^101 ,39*2^10+1 51*2^101 57*2^101,57*2^10+1 ,60*2^10+1 ,63*2^10+1 ,69*2^10+1 72*2^101 ,75*2^10+1 ,78*2^10+1 ,84*2^10+1 87*2^101 93*2^101,93*2^10+1 ,99*2^10+1 (10:13) gp > Last fiddled with by science_man_88 on 20100611 at 13:32 
20100611, 13:35  #2 
Mar 2006
Germany
2·5·293 Posts 
Take a look at here.
The red spots are twins! Also write those pairs you found (and use only odd kvalues!) in a number and you will see, those pairs are small. And because you used even kvalues, too, you listed some pairs double! The occurance of twins is much higher for very small values than at higher n. 
20100611, 13:35  #3 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
10257_{8} Posts 
It is a known fact that all primes except 2 and 3 are congruent to ±1 mod 6. (trivially proven by the prime factors when a number is any other value mod 6, e.g. any number 4 mod 6 is divisible by 2, and number 3 mod 6 is divisible by 3, etc.) To put it another way, all primes except 2 and 3 are of the form p=6b+1 or p=6b1.
So in order for k*2^n±1 to be twin primes, k must be divisible by 3 (so that k*2^n becomes a multiple of 6). To put it another way, 3x*2^n±1 can be twin primes. Last fiddled with by MiniGeek on 20100611 at 13:38 
20100611, 13:38  #4  
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
Quote:
thanks mini and kar_bon should i try for a pattern for +1 and +3 or 1 and 3 ? see if it doesn't act like a sieve ? 

20100611, 13:46  #5  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
4,271 Posts 
Quote:
In order for k*2^n1,3 to be prime, k*2^n2 must be divisible by 6. This means that k*2^n must be 2 mod 6. This means that if k is odd, n must be odd, and if k is even, n must be even. 

20100611, 14:18  #6 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
if I did the math correctly for k*2^n3,+3 to be prime either k*2^n2 must divide by 6 or k*2^n4 and k*2^n+2 must divide by 6. if the 2 situation works then k,n is either both odd or both even. if the other situation is true then k*2^n is 4 mod 6.

20100611, 14:52  #7 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 
if(4 mod 6==0) then k*2^n is 2 mod 6
if(2 mod 6==0) then k*2^n is 4 mod 6 if(0 mod 6==0) then k*2^n is 0 mod 6 is this right ? 
20100611, 17:05  #8  
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
4271_{10} Posts 
Quote:
This makes no sense. How can 4 mod 6==0 ever be true? 

20100611, 21:09  #9 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
I must of messed up on the first part the mod thing is for every 4th up above it when mod 6 give k*2^n must be 2 mod 6
use this same way to figure the rest. 
20100612, 06:10  #10  
May 2007
Kansas; USA
10591_{10} Posts 
Quote:
Is English your native language? Many people here speak other languages. If not, you might have better luck communicating in your native language. Last fiddled with by gd_barnes on 20100612 at 06:11 

20100614, 00:33  #11  
"Forget I exist"
Jul 2009
Dumbassville
20300_{8} Posts 
Quote:


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