Extended Riemann Connections (3)

We have looked at the frequency of numbers where one or more prime factors can occur at most 2 and 3 times respectively, showing how all the more refined sub-categories that can then arise, can be calculated.

We will now look briefly in more detail at the case where 1 or more factors can occur at most 4 times before suggesting a general formula with which to calculate all possible cases.

Now once more, the relative frequency of numbers where 1 or more prime factors can occur at most 4 times  = 1/ζ(5) – 1/ζ(4) = .040 (approx) i.e. 4.0%.

However, again we can make more refined predictions for the cases where 2 or more, 3 or more or in general terms, n or more primes can occur at most 4 times.

Once again the key to this is the common ratio. So once we have found the common ratio we successively multiply the frequency of the case where 1 or more primes can occur at most 4 times by the common ratio, to obtain all the other results.

Now the common ratio in this case is given as {1/ζ(5) – 1/ζ(4)}/8 = .005 (approx).

Notice that a definite pattern is emerging?

We have already seen how with respect to the (internal) frequency of prime factors associated with each of the major classes, that such a definite pattern emerged.

So for those factors associated with numbers where 1 or more primes occur at most 2 times (Class 2), the overall frequency → 1/2 of  Class 1 (where each prime factor occurs just once in a number).

Then, for those factors associated with numbers where 1 or more primes occur at most 3 times (Class 3) the overall frequency → 1/4 of Class 1.

And continuing for those factors associated with numbers where 1 or more primes occur at most 4 times (Class 4) the overall frequency → 1/8 of Class 1.

So the fractions involved are successive terms in the simple geometric series 1/2, 1/4, 1/8, ....
the sum of which = 1.

Thus in this case we simply divide the starting key estimate (for where 1 or more primes can occur at most n times) by the appropriate fraction (as successive terms in this simple geometric series) to calculate the common ratio, which is then applied to all the more restricted sub-classes that we have mentioned.

So in general terms, where we start by considering the relative frequency of numbers where 1 or more primes occur at most n times, the common ratio is given as

{1/ζ(n + 1) – 1/ζ(n)}/2^{n – 1}

Therefore in this case to calculate the frequency where 2 or more primes can occur at most 4 times we obtain {1/ζ(5) – 1/ζ(4)} * {1/ζ(5) – 1/ζ(4)}/8 = .0002 (approx).

So this means that on average only 2 in 10,000 numbers will be comprised of a prime factor combination, where 2 or more factors can occur at most 4 times!

And then to obtain the frequency associated with the numbers where 3 or more primes can occur at most 4 times we multiply in turn .0002 by the common ratio (i.e. .005).


Now once again, armed with these results, we can then calculate the frequency for the occurrence of all exact combinations (made up of 4 prime factors).

Therefore the frequency of occurrence for numbers where exactly 1 prime factor occurs at most 4 times = frequency where 1 or more prime factors occur at most 4 times –  frequency where 2 or more prime factors occur at most 4 times

= .04 – .0002 = .0398.

And we can continue on in this fashion to calculate the corresponding frequencies associated with exactly 2, exactly 3,.....exactly n primes occurring at most 4 times.


Now the case where at least 1 prime occurs 4 times in the factor composition of a number, leaves open the possibility of varying combinations of other primes where 2 or 3 occurrences may be involved.

For example continuing in the same stretch of numbers, from which I have been illustrating (8000000000000001 +), the following number occurs,

8000000000000144 = 12758387 * 3221 * 23 * 23 * 23 * 2 * 2 * 2 * 2  

Now this represents an example of a number where exactly 1 prime factor occurs at most 4 times, with exactly 1 other prime factor occurring at most 3 times (with the remaining factors occurring but once).

So the issue now is to calculate the relative frequency (with respect to the number system as a whole) of this subset of numbers where exactly 1 prime occurs at most 4 times and exactly 1 other prime occurs 3 times, (with all others occurring once).

Therefore the first requirement is to calculate the relative frequency of numbers where exactly 1 prime factor occurs at most 4 times.

And - as we have seen above the result here = .0398 (approx).

The next step is then to combine this with the relative frequency of numbers where exactly 1 other prime occurs at most 3 times, which as illustrated in yesterday's entry = .0899.

Therefore the relative frequency of that subset of numbers where exactly 1 prime factor occurs at most 4 times and exactly 1 other factor occurs 3 times (with all other factors occurring once),

 = .0398 *  0899 = .0036 (approx).

Therefore on average we would expect somewhere between 3 and 4 numbers in every 1000 to belong to this subset of numbers.


We therefore now have the means to calculate for every possible subset of numbers (based on the varying combinations of prime factors) the relative frequency (probability) with which it is associated with respect to the number system as a whole.

And in one way this is quite remarkable and requiring nothing more than the Riemann zeta function (for the positive integers)  .