## Friday, December 9, 2016

### Prime Number Magic (1)

We now will look at the frequency of occurrence (or probability) of the specific prime factors with respect to the overall number system. These are governed by a small set of remarkably simple rules.

Firstly to calculate the frequency with which a particular prime factor will occur, one simply obtains the reciprocal of the prime (in cardinal terms)   –  1.

So this could be expressed as 1/(P–  1).

For example, for 2 the answer = 1. This entails that on average we would expect to find 2 occurring once in the factor composition of a number. Now of course, we would not find 2 in every factor, but this would be balanced out with multiple occurrences of 2 in other factors!

Put another way, if we counted through a sequence of 100 sets of factors, expressing the unique prime composition of each number (taken from any section of the number system), we would expect to find the factor 2 occurring 100 times.

Then for 3,  1/(P–  1) = 1/2. So again in a random sequence of 100 numbers, we would expect to find the factor 3 occurring 50 times.

And just one more example to illustrate! For the prime factor 31,  1/(P–  1) = 1/30. Therefore, likewise in any random sequence  of 100 numbers, we would expect the prime, 31 to occur on average about 3.3 times.

One might next consider the frequency of occurrence of distinct prime factors.

So again though the first prime factor 2 occurs on average 100 times in every sequence of 100 numbers, many of these numbers involve multiple occurrences of 2. So in terms of distinct prime factors, these multiple cases of 2 count (like single occurrences) as 1.

Thus the calculation of the distinct factor occurrences of 2, tells us the fraction of numbers in which 2 (either in single or multiple terms) occurs.

What is surprising here is that we now move from the cardinal to the ordinal ranking of the prime with respect to the simple formula involved (which holds for all prime factors)!

Here, we use the number representing the ordinal ranking of the prime and divide it by the same number + 1.

This can be expressed therefore as Po/(Po + 1).

As 2 is the first prime its ordinal ranking = 1. Therefore the proportion of distinct occurrences of 2 = 1/2.

So in every sequence of 100 numbers, we would expect, on average, 50 distinct occurrences of 2 (as a factor). Alternatively this means that 50% of numbers would not contain 2 (in single or multiple form) as a factor.

As 3 is the second prime, its ordinal ranking = 2. Therefore Po/(Po  +  1) = 2/3. So of the 50 expected occurrences in the prime factor composition of a sequence of 100 number, distinct occurrences of 3 would represent 1/3 of the total. Alternative 2/3 of numbers would not contain 3 (in single or multiple form) as a factor.

Finally, to again illustrate, 31 is the 11th prime. Therefore its ordinal ranking = 11. Then using the formula, this implies that 11/12 of all occurrences involve 31 as a distinct prime factor (i.e. about 3 in 100). This equally implies that relatively few multiple occurrences of 31 thereby take place.

The next task is calculation of the frequency with which a prime occurs just once in the factor composition of a number!

This is based directly on the previous result as its square.

So the proportion of a single occurrences (in relation to total occurrences) of a prime factor = {Po/(Po + 1)}2.

So for 2,  {Po/(Po + 1)}=  1/4. This means that of the 100 occurrences of 2 (as a prime factor) in a sequence of 100 numbers, that 25 on average would represent numbers where 2 occurs just once.

In the case of 3, {Po/(Po + 1)}= 4/9. This entails that of the 50 occurrences of 3 in a sequence of 100 numbers, about 22 on average would represent numbers where 3 occurs just once.

In the case of 31, {Po/(Po + 1)}= 121/144.  This entails that in a larger sequence - say of 1000 - where 31 on average, would occur 33 times, that 28 (approx). would represent single occurrences of the factor!

We will now look at the probability of obtaining 1, 2, 3, 4, or more occurrences of the same prime (in the factor composition of a number).

This is related especially closely to the Zeta 2 function
i.e. ζ2(s2) = 1 + s2+ s2+ s2+ s2+….

So s2  represents the common ratio which is calculated directly from the previous result as
1 – {Po/(Po + 1)}2.

Then to obtain the probability of each of the cases where 1, 2, 3, , 4,....n occurrences of the same prime factor occur we obtain,

ζ2(s2) = (1 + s2+ s2+ s2+ s2+….) * Po/(Po + 1)}2.

So as we have seen, 1 * {Po/(Po + 1)}represents the probability that a specific prime will occur just once (in the prime composition of a number).

s21 * Po/(Po + 1)}then represents the probability that a specific prime will occur 2 times, s2Po/(Po + 1)}2 the probability that a specific prime will occur 3 times, and so on.

So we will now calculate the probability that 2 will occur exactly 2 times (in the factor composition of a number).

Here s21  1 – {Po/(Po + 1)}2  = 1 – 1/4 = 3/4.

Then s21 * Po/(Po + 1)}= 3/4 * 1/4 = 3/16 = .1875. So in a sequence of 100 numbers, 19 (approx) entail a prime factor composition where 2 occurs twice.

The probability that 2 will then occur exactly 3 times as factors in the same number = (3/4)2 * 1/4  = 9/64 = .140625. So 14 numbers on average, entail a prime factor sequence, where 2 occurs exactly 3 times.

Again to illustrate let us now calculate the probability that 3 will occur exactly 4 times (in the factor composition of a number).

Here the common ratio, s21 = 1 – (2/3)2 = 5/9.

and  (5/9)3 * 4/9 = .0762 (approx) is the probability that 3 will occur exactly 4 times (relative to all occurrences of 3).

Therefore in a sequence of 100 (where 3 as a factor occurs on average 50 times) about 4 would relate to factors where 3 occurs 4 times. So this would imply on average just one such occurrence (with 3 occurring here 4 times).