## Sunday, December 11, 2016

### Prime Number Magic (2)

Even closer links can be demonstrated as between the distribution of the specific primes and the Zeta 2 function.

Once again the Zeta 2 function can be given as,

ζ2(s2) = 1 + s2+ s2+ s2+ s2+….

Now, if we concern ourselves with the relative probability that a specific prime occurs 1 or more, 2 or more, 3 or more, 4 or more times, and in general terms n or more times, then the answer is directly given by the Zeta 2 function.

So clearly where a specific prime factor e.g. 2 occurs 1 or more times in a number, this represents 100% of all possible occurrences of that factor. Thus the probability (i.e. that the occurrence of 2 belongs to all of those numbers where it occurs 1 or more times) = 1.

So the answer here is given by the 1st term in the series.

As we have seen in the previous entry, s2 = 1 – {
Po/(Po + 1)}2.

Then the probability that a specific prime factor will occur 2 or more times in a number is given as s21, which is the second term in the Zeta 2 series.

The probability that a specific prime factor will occur 3 or more times in a number is given as
s2(the 3rd term in the series). Then the probability that a specific prime factor will occur 4 or more times is given as s23. And in general terms, the probability that a prime will occur n or more times is given as s2n - 1.

In fact it is possible to provide - again in general terms - an expression for s2.

This is given as (2P+ 1)/(P+ 1)2, where Po   = the ordinal ranking of the prime.

For example 2 is the 1st prime, its ordinal ranking = 1.

Therefore s= (2 + 1)/2 = 3/4.

This entails 3/4 of all occurrences of 2 as a prime factor belong to numbers where 2 occurs 2 or more times.

And then of course s2= (3/4) entails that 9/16 of all occurrences of 2 as a prime factor belong to numbers where 2 occurs 3 or more times and s2= (3/4)entails in turn that 27/64 of all occurrences of 2  as a prime factor belong to numbers where 2 occurs 4 or more times.

And of course we can continue on in this fashion to obtain the probability of all possible occurrences of 2 as a prime factor.

And this approach is perfectly general, working in every case.

For example, 11 is the 5th prime.

Therefore in this instance, s2 = {(2 * 5) + 1}/6= 11/36.

So 1 - the trivial case - again represents the probability that all possible occurrences of 11 as a prime factor relate to numbers where 11 occurs 1 or more times.

s2= 11/36 (= .3055...) then represents the probability that 11 will belong to numbers entailing 2 or more occurrences of that factor.

s22  = (11/36)(= ..0939...) then represents the probability that 11 will belong to numbers entailing 3 or more occurrences of that factor.

As we can see the probability of subsequent occurrences - other than in single form - rapidly declines as we move towards larger prime factors. In the case of 11 for example, there is less than 1 chance in 10 that this factor will belong to numbers where this factor occurs 3 or more times (i.e. in numbers that contain 11 as a factor)!

In fact we can use this information  - directly related to the Zeta 2 function - to make the calculations shown in the last blog entry.

For example to find the probability that 11 will belong to numbers , where that factor occurs exactly once, we simply subtract the probability of 2 or more from the corresponding probability of 1 or more occurrences (i.e. the 2nd term from the 1st) = 1 –  s21 = 1 – 11/36 = 25/36 = .694...

And to find the probability that 11 will belong to numbers, where that factor occurs exactly twice, we simply subtract in turn the  probability of 3 or more from the corresponding probability of 2 or more occurrences (i.e. the 3rd term from the 2nd) = s21 –  s22  =
11/36 – (11/36)2  = .3055 – .0939 =  .2116 (approx).