The Prime Sieve is a constructive method
or algorithm for finding prime numbers. This document will analyze the method in some
detail, hopefully adding to our mathematical knowledge.
Summary of Twin Primes Conjecture Proof.
A sieve method is used to find the mathematical form of various prime numbers.
These mathematical forms split the natural #'s into various branches.
Within a given branch, an assured prime range can be established.
These assured prime ranges form an overlapping sequence by construction.
It was proved that there are an infinite number prime #'s on each branch.
The branches can be further pared to leave only the twin primes.
It is shown that the such paring cannot cut all the branches off.
Since there were an infinite # of primes on each branch, there are an infinite #
of Twin Primes.
Click here to see how the *Prime Sieve* works...
Now, we will examine the related procedure for twin primes:
Sieve-N:
Finds primes>N
The repeat distance is Prime[1]*Prime[2]*Prime[3]*...*N=2*3*5*7*11*...*N=Primorial[N]
Possible forms is LastForms*(N-1)
Includes: All primes>N and composites with prime factors>N
Excludes: All multiples of N not already excluded, (LastForms) branches
1st failure at (NextPrime[N])^{2}
2nd failure at (NextPrime[N]*NextPrime[NextPrime[N]])
Assured prime range: N < primes < (NextPrime[N]*NextPrime[NextPrime[N]])
Mathematical Form is [RepeatDistance*n_{N}+LastRepeat*c_{N}+k]
where possible c_{N} values are from 0..(N-1)
# | Mathematical Form | c_{1} values (0..0) | Old/New/Extended Primes | Bad/Failed |
1 |
1 < [n_{1}+c_{1}]
< 2^{2} = 4 (6) which simplifies to: 1 < [n_{1}] < 2^{2} = 4 (6) | 0 or undefined | 1,2,3+5,7? | 4,6 |
Next confirmed prime is 2...
We need [n_{1}] mod 2 <> 0
Try (n_{1}=2n_{2}+c_{2})
[2n_{2}+c_{2}] mod 2 <> 0
which simplifies to:
[c_{2}] mod 2 <> 0
Solution is Complement[0] of (0..1), c_{2}=1
Apply PrimeSieve-2:
Finds: primes>2
The repeat distance is: 2
Possible forms/branches is: 1*(2-1)=1
Includes: All primes>2 and composites with prime factors>2
Excludes: All multiples of 2, [2n_{2}+0]
1st failure at: 9=3*3=3^{2}
2nd failure at: 15=3*5
Assured prime range: 2 < primes < 15
Notes: At this point, all primes of this mathematical form are potentially twin, as each
is spaced 2 apart from its nearest neighbor.
# | Mathematical Form | c_{2} values (0..1) | Old/New/Extended Primes | Bad/Failed |
1 |
2 < [2n_{2}+c_{2}]
< 3^{2} = 9 (15) which simplifies to: 2 < [2n_{2}+1] < 3^{2} = 9 (15) |
Complement[0] =1 |
x, 1,(3+5+7),,(11+13),,(17?+19?),... |
9,15 |
Next confirmed prime is 3...
We need [2n_{2}+1] mod 3 <> 0
Try (n_{2}=3n_{3}+c_{3})
[6n_{3}+2c_{3}+1] mod 3 <> 0
which simplifies to:
[2c_{3}+1] mod 3 <> 0
Solution is Complement[1] of (0..2), c_{3}=0,2
Apply PrimeSieve-3:
Finds: TwinPrime pairs>3
The repeat distance is: 2*3=6
Possible forms/branches is: 1*(3-1)=2
Possible twin branches is: 1*(3-2)=1
Includes: All primes>3 and composites with prime factors>3
Excludes: All multiples of 3 not already excluded, i.e. [6n_{3}+3]
1st failure at: 25=5*5=5^{2}
2nd failure at: 35=5*7
Assured prime range: 3 < primes < 35
# | Mathematical Form | c_{3} values (0..2) | Old/New/Extended Primes | Bad/Failed |
1 |
3 < [6n_{3}+2c_{3}+1]
< 5^{2} = 25 (35) which simplifies to: 3 < [6n_{3}+1] < 5^{2} = 25 (35) & 3 < [6n_{3}+5] < 5^{2} = 25 (35) |
Complement[1] =0,2 |
1,7,13,19,,31 x, -1,5,11,17,23,29 |
25 35 |
# | Mathematical Form | c_{3} values (0..2) | Old/New/Extended Primes |
1 |
3 < [6n_{3}+2c_{3}+1]
< 5^{2} = 25 (35) which simplifies to: 3 < [6n_{3}±1] < 5^{2} = 25 (35) |
Complement[1] =0,2 |
(-1,1),(5+7),(11+13),(17+19),(23,25),(29+31),(35,37?),... |
Next confirmed prime is 5...
We need [6n_{3}±1] mod 5 <> 0
Try (n_{3}=5n_{5}+c_{5})
[30n_{5}+6c_{5}+1] mod 5 <> 0 which simplifies to: [c_{5}+1] mod 5 <> 0 Solution is Complement[4] of (0..4) c_{5}=0,1,2,3 |
[30n_{5}+6c_{5}-1] mod 5 <> 0 which simplifies to: [c_{5}-1] mod 5 <> 0 Solution is Complement[1] of (0..4) c_{5}=0,2,3,4 |
The solutions are slightly different from the regular prime sieve since I took the
starting points from ([6n_{3}+1] & [6n_{3}+5]
) to ( [6n_{3}+1]
& [6n_{3}-1] ). If you plug in the allowed values, you will see that you
get the same number sequences.
There is one more thing that makes the twin prime sieve different from the regular prime
sieve. Before, each branch had to survive the modulus operation individually.
Now, BOTH branches must obey BOTH constraints; otherwise, you would get pairing of a prime
with a composite.
Therefore, we have [30n_{5}+6c_{5}±1] mod 5 <> 0,
with the solution of:
c_{5}=Complement[1,4] of (0..4)
c_{5}=0,2,3
Thus, we have excluded the following branches:
Composites of 5 in branch [30n_{5}+6c_{5}+1]
Composites of 5 in branch [30n_{5}+6c_{5}-1]
Prime partners in branch [30n_{5}+6c_{5}+1] of a composite partner
in branch [30n_{5}+6c_{5}-1]
Prime partners in branch [30n_{5}+6c_{5}-1] of a composite partner
in branch [30n_{5}+6c_{5}+1]
Hence, only TwinPrimes remain within the defined range of the Twin Sieve.
Apply TwinSieve-5:
Finds: TwinPrime pairs>5
The repeat distance is: 2*3*5=30
Possible twin branches branches is: 1*(5-2)=3
Includes: All TwinPrimes>5 and composites with prime factors>5
Excludes: All multiples of 5 not already excluded, and prime partners of composites in the
twin branch
1st failure at: 49=7*7=7^{2}
2nd failure at: 77=7*11
Assured TwinPrime range: 5 < primes < 77
# | Mathematical Form | c_{5} values (0..4) | Old/New/Extended Primes |
1 | 5 < [30n_{5}+6c_{5}+0±1] < 7^{2} = 49 (77) |
Complement[1,4] =0,2,3 |
(-1,1),(29+31),(59+61),(89?,91),... (11+13),(41+43),(71+73),(101?+103?)... (17+19),(47,49),77 |
Next confirmed prime is 7...
We need [30n_{5}+6c_{5}±1] mod 7 <> 0 for c_{5}=0,2,3
Try (n_{5}=7n_{7}+c_{7})
[210n_{7}+30c_{7}+6c_{5}±1] mod 7 <> 0
for c_{5}=0,2,3
[210n_{7}+30c_{7}+0±1] mod 7 <> 0 which simplifies to: [2c_{7}±1] mod 7 <> 0 Solution is Complement[3,4] of (0..6) c_{7}=0,1,2,5,6 |
[210n_{7}+30c_{7}+12±1] mod 7 <> 0 which simplifies to: [2c_{7}+5±1] mod 7 <> 0 Solution is Complement[4,5] of (0..6) c_{7}=0,1,2,3,6 |
[210n_{7}+30c_{7}+18±1] mod 7 <> 0 which simplifies to: [2c_{7}+4±1] mod 7 <> 0 Solution is Complement[1,2] of (0..6) c_{7}=0,3,4,5,6 |
Apply TwinSieve-7:
Finds: TwinPrime pairs>7
The repeat distance is: 2*3*5*7=210
Possible twin branches branches is: 3*(7-2)=15
Includes: All TwinPrimes>7 and composites with prime factors>7
Excludes: All multiples of 7 not already excluded, and prime partners of composites in the
twin branch
1st failure at: 121=11*11=11^{2}
2nd failure at: 143=11*13
Assured TwinPrime range: 7 < primes < 153
# | Mathematical Form | c_{7} values (0..6) | Old/New/Extended Primes |
1 | 7 < [210n_{7}+30c_{7}+0±1] < 11^{2} = 121 (143) |
Complement[3,4] of (0..6) c_{7}=0,1,2,5,6 |
(-1,1),(209?,211?),... (29+31),(239?,241?),... (59+61),(269?,271?),... x, x, (149?,151?),... (179?,181?),... |
2 | 7 < [210n_{7}+30c_{7}+12±1] < 11^{2} = 121 (143) |
Complement[4,5] of (0..6) c_{7}=0,1,2,3,6 |
(11+13),(221?,223?),... (41+43),(251?,253?),... (71+73),(281?,283?),... (101+103),(311?,313?),... x, x, (191?+193?),... |
3 | 7 < [210n_{7}+30c_{7}+18±1] < 11^{2} = 121 (143) |
Complement[1,2] of (0..6) c_{7}=0,3,4,5,6 |
(17+19),(227?,229?),... x, x, (107+109),(217?,219?) (137+139),(247?,249?) (167?,169?),... (197?,199?),... |
Next confirmed prime is 11...
We need [210n_{7}+30c_{7}+k±1] mod 11 <> 0 for
k=0,12,18 and c_{7}=various
Try (n_{7}=11n_{11}+c_{11})
[2310n_{11}+210c_{11}+30c_{7}+k±1] mod 11
<>0
Apply TwinSieve-11:
Finds: TwinPrime pairs>11
The repeat distance is: 2*3*5*7*11=2310
Possible twin branches branches is: 15*(11-2)=135
Includes: All TwinPrimes>11 and composites with prime factors>11
Excludes: All multiples of 11 not already excluded, and prime partners of composites in
the twin branch
1st failure at: 169=13*13=13^{2}
2nd failure at: 221=13*17
Assured TwinPrime range: 11 < primes < 221
# | Mathematical Form | c_{11} values (0..10) | Old/New/Extended Primes |
1 | 11 < [2310n_{11}+210c_{11}+0±1] < 13^{2} = 169 (221) | c_{11}=Complement[1,10] of (0..10) | (-1,1),x,... |
2 | 11 < [2310n_{11}+210c_{11}+12±1] < 13^{2} = 169 (221) | c_{11}=Complement[0,9] of (0..10) | x,(221,223)... |
3 | 11 < [2310n_{11}+210c_{11}+18±1] < 13^{2} = 169 (221) | c_{11}=Complement[3,5] of (0..10) | (17+19),... |
4 | 11 < [2310n_{11}+210c_{11}+30±1] < 13^{2} = 169 (221) | c_{11}=Complement[2,4] of (0..10) | (29+31),... |
5 | 11 < [2310n_{11}+210c_{11}+42±1] < 13^{2} = 169 (221) | c_{11}=Complement[1,3] of (0..10) | (41+43),x,... |
6 | 11 < [2310n_{11}+210c_{11}+60±1] < 13^{2} = 169 (221) | c_{11}=Complement[5,7] of (0..10) | (59+61),... |
7 | 11 < [2310n_{11}+210c_{11}+72±1] < 13^{2} = 169 (221) | c_{11}=Complement[4,6] of (0..10) | (71+73),... |
8 | 11 < [2310n_{11}+210c_{11}+102±1] < 13^{2} = 169 (221) | c_{11}=Complement[7,9] of (0..10) | (101+103),... |
9 | 11 < [2310n_{11}+210c_{11}+108±1] < 13^{2} = 169 (221) | c_{11}=Complement[1,3] of (0..10) | (107+109),x,... |
10 | 11 < [2310n_{11}+210c_{11}+138±1] < 13^{2} = 169 (221) | c_{11}=Complement[4,6] of (0..10) | (137+139),... |
11 | 11 < [2310n_{11}+210c_{11}+150±1] < 13^{2} = 169 (221) | c_{11}=Complement[3,5] of (0..10) | (149+151),... |
12 | 11 < [2310n_{11}+210c_{11}+168±1] < 13^{2} = 169 (221) | c_{11}=Complement[7,9] of (0..10) | (167,169),... |
13 | 11 < [2310n_{11}+210c_{11}+180±1] < 13^{2} = 169 (221) | c_{11}=Complement[6,8] of (0..10) | (179+181),... |
14 | 11 < [2310n_{11}+210c_{11}+192±1] < 13^{2} = 169 (221) | c_{11}=Complement[5,7] of (0..10) | (191+193),... |
15 | 11 < [2310n_{11}+210c_{11}+198±1] < 13^{2} = 169 (221) | c_{11}=Complement[1,10] of (0..10) | (197+199),x,... |
Next confirmed prime is 13...
We need [2310n_{11}+210c_{11}+k±1] mod 13 <> 0 for
k=various and c_{7}=various
Try (n_{11}=13n_{13}+c_{13})
[30030n_{11}+2310c_{11}+k±1] mod 13 <>0
Apply TwinSieve-13:
Finds: TwinPrime pairs>13
The repeat distance is: 2*3*5*7*11*13=30030
Possible twin branches branches is: 135*(13-2)=1485
Includes: All TwinPrimes>13 and composites with prime factors>13
Excludes: All multiples of 13 not already excluded, and prime partners of composites in
the twin branch
1st failure at: 289=17*17=17^{2}
2nd failure at: 323=17*19
Assured TwinPrime range: 13 < primes < 323
# | Mathematical Form | c_{13} values (0..12) | Old/New/Extended Primes |
1 | 13 < [30030n_{13}+2310c_{13}+0±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (-1,1),x,... |
2 | 13 < [30030n_{13}+2310c_{13}+18±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (17+19),... |
3 | 13 < [30030n_{13}+2310c_{13}+30±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (29+31),... |
4 | 13 < [30030n_{13}+2310c_{13}+42±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (41+43),x,... |
5 | 13 < [30030n_{13}+2310c_{13}+60±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (59+61),... |
6 | 13 < [30030n_{13}+2310c_{13}+72±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (71+73),... |
7 | 13 < [30030n_{13}+2310c_{13}+102±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (101+103),... |
8 | 13 < [30030n_{13}+2310c_{13}+108±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (107+109),x,... |
9 | 13 < [30030n_{13}+2310c_{13}+138±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (137+139),... |
10 | 13 < [30030n_{13}+2310c_{13}+150±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (149+151),... |
11 | 13 < [30030n_{13}+2310c_{13}+168±1] < 17^{2} = 289 (323) | c_{13}=Complement[0] of (0..12) | x,... |
12 | 13 < [30030n_{13}+2310c_{13}+180±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (179+181),... |
13 | 13 < [30030n_{13}+2310c_{13}+192±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (191+193),... |
14 | 13 < [30030n_{13}+2310c_{13}+198±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (197+199),x,... |
15 | 13 < [30030n_{13}+2310c_{13}+222±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | x,... |
16 | 13 < [30030n_{13}+2310c_{13}+228±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (227+229),x,... |
17 | 13 < [30030n_{13}+2310c_{13}+240±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (239+241),x,... |
18 | 13 < [30030n_{13}+2310c_{13}+270±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (269+271),x,... |
19 | 13 < [30030n_{13}+2310c_{13}+282±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (281+283),x,... |
20 | 13 < [30030n_{13}+2310c_{13}+312±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (311+313),x,... |
21 | 13 < [30030n_{13}+2310c_{13}+318±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (317+319),x,... |
22 | 13 < [30030n_{13}+2310c_{13}+348±1] < 17^{2} = 289 (323) | c_{13}=Complement[] of (0..12) | (347?+349?),x,... |
23 | |||
24 | |||
25 |
Beginning Twin Primes List:
(3+5)
(5+7)
(11+13)
(17+19)
(29+31)
(41+43)
(59+61)
(71+73)
(101+103)
(107+109)
(137+139)
(149+151)
(179+181)
(191+193)
(197+199)
(227+229)
(239+241)
(269+271)
(281+283)
(311+313)
(317+319)
Now, from the earlier Prime Sieve page, we proved that every (non-all-composite)
branch has an infinite number of primes.
The Twin Prime procedure takes these SAME branches, and pares off the ones that
do not contain twin primes.
There is no limit to the TwinPrime branching procedure, as before, and there are
an infinite number of primes on each branch.
Therefore, there are an infinite # of TwinPrimes.
By related procedure, this also proves an infinite # of cousin primes, an
infinite # of sexy primes, and so
forth...
This concludes the Proof of the Twin Prime Conjecture.
JBW
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