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Old 2010-06-11, 13:17   #1
science_man_88
 
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Default 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
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Old 2010-06-11, 13:35   #2
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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.
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Old 2010-06-11, 13:35   #3
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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
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Old 2010-06-11, 13:38   #4
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Quote:
Originally Posted by Mini-Geek View Post
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 ?
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Old 2010-06-11, 13:46   #5
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Quote:
Originally Posted by science_man_88 View Post
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.
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Old 2010-06-11, 14:18   #6
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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.
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Old 2010-06-11, 14:52   #7
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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 ?
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Old 2010-06-11, 17:05   #8
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Quote:
Originally Posted by science_man_88 View Post
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 View Post
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?
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Old 2010-06-11, 21:09   #9
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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.
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Old 2010-06-12, 06:10   #10
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Quote:
Originally Posted by science_man_88 View Post
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
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Old 2010-06-14, 00:33   #11
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Quote:
Originally Posted by gd_barnes View Post
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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