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,

Once again the Zeta 2 function can be given as,

ΞΆ

_{2}(s

_{2}) = 1 + s

_{2}

^{1 }+ s

_{2}

^{2 }+ s

_{2}

^{3 }+ s

_{2}

^{4 }+….

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, s

_{2}= 1 – {P

_{o}/(P

_{o}+ 1)}

^{2}.

Then the probability that a specific prime factor will occur 2 or more times in a number is given as s

_{2}

^{1}, 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 s

_{2}

^{2 }(the 3rd term in the series). Then the probability that a specific prime factor will occur 4 or more times is given as s

_{2}

^{3}. And in general terms, the probability that a prime will occur n or more times is given as s

_{2}

^{n - 1}.

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

_{2}.

This is given as (2P

_{o }+ 1)/(P

_{o }+ 1)

^{2}, where P

_{o }= the ordinal ranking of the prime.

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

Therefore s

_{2 }= (2 + 1)/2

^{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 s

_{2}

^{3 }= (3/4)

^{2 }entails that 9/16 of all occurrences of 2 as a prime factor belong to numbers where 2 occurs 3 or more times and s

_{2}

^{3 }= (3/4)

^{3 }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, s

_{2}= {(2 * 5) + 1}/6

^{2 }= 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.

s

_{2}

^{1 }= 11/36 (= .3055...) then represents the probability that 11 will belong to numbers entailing 2 or more occurrences of that factor.

s

_{2}

^{2}= (11/36)

^{2 }(= ..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 on 10 that this factor will belong to numbers where this factor occurs 3 or more times!

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 – s

_{2}

^{1}= 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) = s

_{2}

^{1}– s

_{2}

^{2}=

11/36 – (11/36)

^{2 }= .3055 – .0939 = .2116 (approx).

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