mersenneforum.org 3x*2^n-1 and 3x*2^n-1 possibly twins ?
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 2010-06-11, 13:17 #1 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26·131 Posts 3x*2^n-1 and 3x*2^n-1 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^n-1),print1(k"*"2"^"n"-1"));if(isprime(k*2^n+1),print1(","k"*"2"^"n"+1"));print("\n"))) 3*2^1-1,3*2^1+1 6*2^1-1,6*2^1+1 9*2^1-1,9*2^1+1 12*2^1-1 15*2^1-1,15*2^1+1 ,18*2^1+1 21*2^1-1,21*2^1+1 24*2^1-1 27*2^1-1 30*2^1-1,30*2^1+1 ,33*2^1+1 36*2^1-1,36*2^1+1 ,39*2^1+1 42*2^1-1 45*2^1-1 ,48*2^1+1 51*2^1-1,51*2^1+1 54*2^1-1,54*2^1+1 57*2^1-1 ,63*2^1+1 66*2^1-1 69*2^1-1,69*2^1+1 75*2^1-1,75*2^1+1 ,78*2^1+1 ,81*2^1+1 84*2^1-1 87*2^1-1 90*2^1-1,90*2^1+1 96*2^1-1,96*2^1+1 99*2^1-1,99*2^1+1 3*2^2-1,3*2^2+1 6*2^2-1 ,9*2^2+1 12*2^2-1 15*2^2-1,15*2^2+1 18*2^2-1,18*2^2+1 21*2^2-1 ,24*2^2+1 27*2^2-1,27*2^2+1 33*2^2-1 ,39*2^2+1 42*2^2-1 45*2^2-1,45*2^2+1 48*2^2-1,48*2^2+1 57*2^2-1,57*2^2+1 60*2^2-1,60*2^2+1 63*2^2-1 66*2^2-1 ,69*2^2+1 78*2^2-1,78*2^2+1 ,84*2^2+1 87*2^2-1,87*2^2+1 90*2^2-1 ,93*2^2+1 96*2^2-1 ,99*2^2+1 3*2^3-1 6*2^3-1 9*2^3-1,9*2^3+1 ,12*2^3+1 21*2^3-1 24*2^3-1,24*2^3+1 30*2^3-1,30*2^3+1 33*2^3-1 39*2^3-1,39*2^3+1 ,42*2^3+1 45*2^3-1 48*2^3-1 ,51*2^3+1 54*2^3-1,54*2^3+1 ,57*2^3+1 60*2^3-1 63*2^3-1 ,72*2^3+1 75*2^3-1,75*2^3+1 81*2^3-1 ,84*2^3+1 90*2^3-1 93*2^3-1 ,96*2^3+1 3*2^4-1 ,6*2^4+1 12*2^4-1,12*2^4+1 15*2^4-1,15*2^4+1 ,21*2^4+1 24*2^4-1 27*2^4-1,27*2^4+1 30*2^4-1 ,36*2^4+1 ,42*2^4+1 45*2^4-1 ,48*2^4+1 54*2^4-1 57*2^4-1 ,63*2^4+1 69*2^4-1 72*2^4-1,72*2^4+1 ,75*2^4+1 ,78*2^4+1 ,81*2^4+1 90*2^4-1 93*2^4-1,93*2^4+1 99*2^4-1 ,3*2^5+1 6*2^5-1,6*2^5+1 12*2^5-1 15*2^5-1 ,18*2^5+1 ,21*2^5+1 ,24*2^5+1 27*2^5-1 36*2^5-1,36*2^5+1 ,39*2^5+1 45*2^5-1 57*2^5-1 ,63*2^5+1 66*2^5-1,66*2^5+1 69*2^5-1 75*2^5-1 81*2^5-1,81*2^5+1 84*2^5-1,84*2^5+1 90*2^5-1 99*2^5-1,99*2^5+1 3*2^6-1,3*2^6+1 6*2^6-1 ,9*2^6+1 ,12*2^6+1 18*2^6-1,18*2^6+1 33*2^6-1,33*2^6+1 42*2^6-1,42*2^6+1 45*2^6-1 ,54*2^6+1 75*2^6-1,75*2^6+1 ,78*2^6+1 ,87*2^6+1 ,93*2^6+1 96*2^6-1 ,99*2^6+1 3*2^7-1 ,6*2^7+1 9*2^7-1,9*2^7+1 21*2^7-1,21*2^7+1 ,27*2^7+1 ,39*2^7+1 48*2^7-1 ,51*2^7+1 54*2^7-1 ,57*2^7+1 ,60*2^7+1 66*2^7-1 69*2^7-1 ,75*2^7+1 ,81*2^7+1 ,84*2^7+1 90*2^7-1 93*2^7-1 ,96*2^7+1 99*2^7-1 ,3*2^8+1 24*2^8-1 27*2^8-1 ,30*2^8+1 33*2^8-1 ,42*2^8+1 45*2^8-1 ,48*2^8+1 57*2^8-1,57*2^8+1 60*2^8-1,60*2^8+1 63*2^8-1 ,72*2^8+1 84*2^8-1 87*2^8-1,87*2^8+1 90*2^8-1,90*2^8+1 99*2^8-1 12*2^9-1 ,15*2^9+1 ,21*2^9+1 ,24*2^9+1 30*2^9-1,30*2^9+1 ,36*2^9+1 42*2^9-1 45*2^9-1,45*2^9+1 51*2^9-1,51*2^9+1 54*2^9-1 ,63*2^9+1 66*2^9-1 69*2^9-1 ,78*2^9+1 87*2^9-1 6*2^10-1 ,12*2^10+1 15*2^10-1,15*2^10+1 ,18*2^10+1 21*2^10-1 27*2^10-1 33*2^10-1 ,39*2^10+1 51*2^10-1 57*2^10-1,57*2^10+1 ,60*2^10+1 ,63*2^10+1 ,69*2^10+1 72*2^10-1 ,75*2^10+1 ,78*2^10+1 ,84*2^10+1 87*2^10-1 93*2^10-1,93*2^10+1 ,99*2^10+1 (10:13) gp > even when i checked for step =1 I got all 3x values so I narrowed it to that care to double check ? Last fiddled with by science_man_88 on 2010-06-11 at 13:32
 2010-06-11, 13:35 #2 kar_bon     Mar 2006 Germany 23×5×73 Posts Take a look at here. The red spots are twins! Also write those pairs you found (and use only odd k-values!) in a number and you will see, those pairs are small. And because you used even k-values, too, you listed some pairs double! The occurance of twins is much higher for very small values than at higher n.
 2010-06-11, 13:35 #3 Mini-Geek Account Deleted     "Tim Sorbera" Aug 2006 San Antonio, TX USA 17×251 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=6b-1. 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 Mini-Geek on 2010-06-11 at 13:38
2010-06-11, 13:38   #4
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by Mini-Geek 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) To put it another way, all primes except 2 and 3 are of the form p=6k+1 or p=6k-1. 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.

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 ?

2010-06-11, 13:46   #5
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

17·251 Posts

Quote:
 Originally Posted by science_man_88 should i try for a pattern for +1 and +3 or -1 and -3 ? see if it doesn't act like a sieve ?
In order for k*2^n+1,+3 to be prime, k*2^n+2 must be divisible by 6. This means that k*2^n must be 4 mod 6. This means that if k is odd, n must be even, and vice versa.
In order for k*2^n-1,-3 to be prime, k*2^n-2 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.

 2010-06-11, 14:18 #6 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26·131 Posts if I did the math correctly for k*2^n-3,+3 to be prime either k*2^n-2 must divide by 6 or k*2^n-4 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.
 2010-06-11, 14:52 #7 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26×131 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 ?
2010-06-11, 17:05   #8
Mini-Geek
Account Deleted

"Tim Sorbera"
Aug 2006
San Antonio, TX USA

17×251 Posts

Quote:
 Originally Posted by science_man_88 if I did the math correctly for k*2^n-3,+3 to be prime either k*2^n-2 must divide by 6 or k*2^n-4 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.
If k*2^n-3,+3 are primes, then they're both -1 mod 6 or they're both 1 mod 6. This means that k*2^n-2,+4 or k*2^n-4,+2 are divisible by 6. Or that k*2^n is 2 or -2 mod 6.
Quote:
 Originally Posted by science_man_88 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 ?
This makes no sense. How can 4 mod 6==0 ever be true?

 2010-06-11, 21:09 #9 science_man_88     "Forget I exist" Jul 2009 Dumbassville 26·131 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.
2010-06-12, 06:10   #10
gd_barnes

May 2007
Kansas; USA

5×2,099 Posts

Quote:
 Originally Posted by science_man_88 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.
Don't expect a response. This makes no grammatical or mathematical sense and it's not easy to tell where one sentence ends and the next one starts. Please use periods, commas, and capitalization.

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 2010-06-12 at 06:11

2010-06-14, 00:33   #11
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by gd_barnes Don't expect a response. This makes no grammatical or mathematical sense and it's not easy to tell where one sentence ends and the next one starts. Please use periods, commas, and capitalization. Is English your native language? Many people here speak other languages. If not, you might have better luck communicating in your native language.
I'll admit I missed a few words still better than most people I know in life.

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