Conditions: For integers k≥1, in the range 0<q (where q≥6k+1)
2 is a prime. Other than k=1, 2k is never prime.
3 is a prime. Other than k=1, 3k is never prime.
Therefore, other than 2 and 3, every other prime greater than 3 takes the form 6k±1, where k is an integer.
This is based on residue classes modulo 6. Only -1 mod 6 (also may be classified as 5 mod 6) and 1 mod 6 can be prime candidates since every other number is divisible by 2 or 3.
By Dirichlet’s Theorem, both 6k-1 and 6k+1 contain infinite primes.
No member of set {6k-1} can possibly be in set {6k+1}. They are mutually exclusive.
“Basic Prime Theory – Elimination of Composites”
If a number is 6k-1 or 6k+1, but is not (6a-1)(6n-1),(6a-1)(6b+1), or (6a+1)(6b+1), then it is prime.
(6a-1)(6b-1)=36ab-6a-6b+1=AA. This number form is always of the form 1 mod 6. So it must also always be of the form of 6k+1 for some value of k,m,n. It would be impossible for it to be 6k-1 under the modulo 6 classes already established. Some early values <100 are: 25,55,85…
(6a-1)(6b+1)=36ab+6a-6b-1=AB. This number form is always of the form -1 mod 6. So it must also always be of the form of 6k-1 for some value of k,m,n. It would be impossible for it to be 6k+1 under the modulo 6 classes already established. Some early values <100 are: 35, 65, 77, 95…
(6a+1)(6b+1)=36ab+6a+6b+1=BB. This number form is always of the form 1 mod 6. So it must also always be of the form of 6k+1 for some value of k,m,n. It would be impossible for it to be 6k-1 under the modulo 6 classes already established. Some early values <100 are: 49, 91…
So, for integers in the range 0<q, then if a number IS in the sets {6k-1}=A or {6k+1}=B, but is not in the sets {6a-1}{6b-1}={AA}, {6a-1}{6b+1}={AB}, or {6a+1}{6b+1}={BB}, it is a prime number.
BUT, the exclusion criteria for prime numbers and composites in {A} and {B} are different.
We’ve already shown there are 3 possible kinds of composites arising from 6k+1 forms which we symbolize as: AA, AB, and BB. We know that all primes >3 are in 6k+1.
So, to find primes in {A} less than some number q, we need to subtract {AB}. “{A}-{AB}”
So, to find primes in {B} less than some number q, we need to subtract {AA} and {BB}. “{B}-({AA}+{BB})”
So, to find all primes in 3<q, then {primes >3}=({A}-{AB})+({B}-({AA}+{BB}))
To find all primes in 0<q, add 2 and 3 to ({A}-{AB})+({B}-({AA}+{BB}))
“Basic Prime Theory – Twin Prime Pairs and 6k+1 Pairs”
Since every twin prime pair is separated by 2 units, then every twin prime greater that (3,5) must be of the form (6k-1,6k+1), where “k” has the same value. (Not every individual product of 6k-1 or 6k+1 is prime of course.)
(6k-1)+2=6k+1. A twin prime is always of the form (p,p+2). Since every twin prime pair <(3,5) must be (6k-1,6k+1), then every twin prime pair >3 is (6k-1=p,6k+1=p+2).
As discussed above, {A}={6k-1} and {B}={6k+1} are mutually exclusive sets.
As discussed above, {A} and {B} have different composite exclusion criteria due to their residue classes mod 6, so that {A}-{AB}={primes A} and {B}-({AA}+{BB})={primes B}.
“Basic Prime Theory – Negative Range”
Conditions: For integers ±k, in the range -q<0<q (where q≥6k+1)
-1,1 are also of the form 6k±1, when k=0
If z is in {A}, then -z is in {B} (and vice versa). So if the infinite series …-13,-7,-1,5,11,17… is in {A}, then …-17,-11,-5,1,7,13… is in {B}.
When we take all the negative numbers in {A} and negate their signs, we have the positive numbers in {B} (and vice versa).
So, in the range of -q<0<q, |A|=|B|
Since |A|=|B| and since z is in {A} and -z is in {B} (and vice versa), then any composite number in {A} in the positive range 0<q is a number which can be factored in the negative range -q>0 (and vice versa for {B}).
For example: If 25=5*5=AA in 0<q in {B}, then -25 has to be -25=–5*5=AB or -25=-5*5=BA in {A}.
For example: If 35=5*7=AB in 0<q in {A}, then -35 has to be either -35=-7*5=AA or -35=-5*7=BB in {B}.
For example: If 49=7*7=BB in 0<q in {B}, then -49 has to be -49=-7*7=AB or -49=7*-7=BA in {A}.
Negative ranges produces an additional consideration for the production of positive values: The case of a negative number times another negative number. So if AB=35=7*5, then BA=35=-5*-7. If AA=25=5*5, then BB=25=-5*-5. If BB=49=7*7, then AA=49=-7*-7. The sign relationships are maintained. (If a duplicate number is formed, it is ignored. We only care about creating a value once.)
“Basic Prime Theory – Probability I”
We’ve shown that {primes >3}=({A}-{AB})+({B}-({AA}+{BB})) for 0<q.
Since |A|=|B|, in the range of -q<0<q, for example, the odds of choosing -7 in {A} as a non-factorable number is the same as choosing 7 as a prime in {B}.
Since when we take all the negative numbers in {A} and negate their signs, we have the positive numbers in {B} (and vice versa).
So, it must be true that P|A|=P|B|, in range -q<0<q.
As shown above, positive numbers z which can be factored as {AB} in {A}, appear as negative number -z with form {AA} or {BB} in {B}, and positive numbers z which can be factored as {AA} or {BB} in {B}, appear as negative number -z with form {AB} or {BA} in {A}.
Since every value z (prime or composite) in {A} has a -z in {B} (and vice versa) the probability of a composite in the positive range 0<q of {A} is precisely equal to the probability of a number which can be factored in the negative range -q<0 of {B}. The probability of a composite in the positive range 0<q of {B} is precisely equal to the probability of a number which can be factored in the negative range -q<0 of {A}.
Since {primes >3}=({A}-{AB})+({B}-({AA}+{BB})) in 0<q, then P({A}-{AB}) in 0<q and P({B}-({AA}+{BB})) in -q<0 must be equivalent.
Since {primes >3}=({A}-{AB})+({B}-({AA}+{BB})) in 0<q, then P({B}-({AA}+{BB})) in 0<q and P({A}-({AB})) in -q<0 must be equivalent.
So, P({A}-{AB})≈P({B}-({AA}+{BB})) in 0<q
“Basic Prime Theory – Probability II”
Probability of selecting a number between 0<q and it being in A: ≈1/6
Probability of selecting a number between 0<q and it being in B: ≈1/6
Probability of selecting a number between 0<q and it being a prime in A: ≈1/6-1/6(N{36ab+6a-6b-1}/N{6k-1}) where N is the number of items in the set as the set at q.
Probability of selecting a number between 0<q and it being a prime in B: ≈1/6-1/6(N({36ab-6a-6b+1}∩{36ab+6a+6b+1})/N{6k+1}) where N is the number of items in the set as the set at q.
Probability of selecting a number between 0<q and it being a prime in A or B: ≈2(1/6)-1/6(N{36ab+6a-6b-1}/N{6k-1})-1/6(N({36ab-6a-6b+1}∩{36ab+6a+6b+1})/N{6k+1} where N is the number of items in the set at q.
