Given differing primes p

_{1} and p

_{2}, where p

_{1} > p

_{2}, there is no preference for any particular residue of p

_{2} for p

_{1} mod p

_{2} over any other.

I'll note that I don't consider 0 to be a residue.

The value of the axiom can be seen from my

post on twin primes probability. Quite simply, by the axiom since primes do not prefer a particular residue modulo each other, you can determine the probability of twin primes in the following way.

If x is prime and greater than 3 the probability that x+2 is prime is given by:

prob = ((p

_{j}-2)/(p

_{j}-1))*((p

_{j-1}-2)/(p

_{j-1}-1))*...*(1/2)

where j is the number of primes up to sqrt(x+2), and p

_{j} is the jth prime, p

_{j-1} is the prime before it and so forth.

The result is easy as it is just multiplying the probability for each prime that it is NOT true that

x + 2 ≡ 0 mod p

which probability is just the result of dividing one minus the number of non-zero residues by the total number of residues together to get the total probability that a prime plus 2 is also prime.

So let's try it out. Between 5

^{2} and 7

^{2}, there are 6 primes. The probability then is given by:

prob = ((5-2)/(5-1))*((3-2)/(3-1) = (3/4)*(1/2) = 0.375

And 6*0.375 = 2.25 so you expect 2 twin primes in that interval.

The primes are 29, 31, 37, 41, 43, 47 and you'll notice, two twin primes as predicted: 29,31 and 41, 43.

However, there is an issue which shifts the probability slightly.

If you go into the actual residues it jumps out at you:

29, 31, 37, 41, 43, 47

mod 3: 2, 1, 1, 2, 1, 2

mod 5: 4, 1, 2, 1, 3, 2

Here all the residues for 5 were in evidence so the count came out right, but for random it should have been possible for ALL the residues mod 5 to be 4, but it's not because with 6 primes there isn't enough space in the interval--4*5 = 20, but 49-25=24, where only 12 are odd and only 6 are primes. So the probability is actually off! A scenario where all residues are 4 is precluded by the size of the interval for the larger prime.

That will tend to over-count because the higher residues are less likely to occur because they cannot fit. Easy explanation that jumps out at you with even a small example.

If that seems suspicious remember the prime number 5 has no idea I've clipped the interval to only go from 25 to 49. It doesn't care. But once you know there are 6 primes in that interval it's not possible for them all to be 4 mod 5, as the interval is too small.

For the smaller primes it's not an issue as if the prime is greater than interval/(prime count in interval) then that prime isn't affected and its residues can have purely random behavior. For instance, for 3 between 25 and 49, you have 24/(6) = 4, and as that is greater than 3, there is no clipping for 3.

Another example can be found using all the primes up to 100, for which I got:

prob = 0.1558 to 4 significant digits.

So that says that the probability that for a prime between 97

^{2} and 100

^{2} that adding 2 to it gives a prime is about 15.58% and there are 66 primes in that interval so there should be about 10 twin primes, and a quick count shows that there are:

(9419, 9421), (9431, 9433), (9437, 9439), (9461, 9463), (9629, 9631), (9677, 9679), (9719, 9721), (9767, 9769), (9857, 9859), (9929, 9931)

So you have this random behavior which is just about probability which follows from the prime residue axiom--since primes don't care about their residue modulo each other, you get a probabilistic behavior, which means you should be within one standard deviation about 68% of the time.

Oddly enough the equation I show above does appear in the mathematical literature on twin primes where it is part of what is called the

twin primes constant.

So there is a LOT of numerical data supporting random behavior where mathematicians appear to have mixed up random behavior with the non-random behavior of the prime distribution itself. That error is a fascinating one. But if not corrected it can be awfully convenient for continuing research where no other answer is available. Random means there is no further answer as random means--no reason.

I've explained myself why the prime distribution itself is

not random.

Going further leads to the

prime gap equation which handles prime gaps of arbitrary size.

I hypothesize that the clipping behavior for the bigger primes in the interval should actually be less significant as the gap size increases. Notice the prime gap post also hypothesizes that the max gap for primes is given by p+1 where p is the largest prime less than sqrt(x+g) where g is a positive prime gap.

But that number is significant only at lower values as later the prime gap probability dominates meaning that the gaps get pushed out further from where they are first allowed.

James Harris