Extended Riemann Connections (1)

As we have already seen, the frequency of - what I term - the Type 2 Class - where 1 or more prime factors of a number occur at most 2 times (with therefore all other factors occurring just once) is given as 1/{ζ(3) – 1/ζ(2)} = .224 (approx) i.e. 22.4%.

Once again as I illustrated in the last entry for the numbers between 8000000000000001 - 16,

8000000000000001 = 4447 * 1423 * 409 * 163 * 43 * 7 * 7 * 3 * 3 ,
8000000000000004 = 37884167 * 138563 * 127 * 3 * 2 * 2 ,
8000000000000006 = 601775236949 * 23 * 17 * 17 * 2,
8000000000000010 = 88888888888889 * 5 * 3 * 3 * 2  and
8000000000000012 = 105263157894737 * 19 * 2 * 2,

belong to this class (i.e. Class 2).

In the first case, we have two factors that occur at most twice; in each of the other cases just one factor that occurs at most twice!

And - though with much less probability - it is possible that 2 or more, 3 or more, 4 or more and in general terms, n or more prime factors could occur at most twice. 

Therefore a more refined estimation should allow one to calculate the exact frequency of occurrence (with respect to the number system as a whole) of all these sub-classes of Class 2.

So once again, 1/ζ(3) – 1/ζ(2) = .224 (approx) represents the frequency of occurrence for all cases where 1 or more prime factors occur at most 2 times.

To calculate the more restricted frequency of occurrence for all cases where now 2 or more prime factors occur at most 2 times we obtain  {ζ(3) – 1/ζ(2)}²/2 = .02508... i.e. .025 (approx).

So we would expect on average just 2.5% of all numbers to belong to this subclass (of Class 2)

Then to calculate the frequency of occurrence now for all cases where 3 or more prime factors occur at most 2 times we obtain {ζ(3) – 1/ζ(2)}³/4 = .0028 (approx).

Therefore, we would expect just .28% (i.e. slightly less than 3 numbers in a 1000) to belong to this class.

Then the frequency of occurrence where 4 or more prime factors occur at most 2 times = {ζ(3) – 1/ζ(2)}⁴/8   =  .0003 (approx) i.e. just 3 in 10,000 numbers.

In more general terms to calculate the frequency of occurrence, when n or more factors occur at most 2 times we obtain {ζ(3) – 1/ζ(2)}ⁿ/2^{n – 1}


Alternatively, we could start from the first result i.e. the frequency of occurrence where 1 or more prime factors occur at most 2 times and continually multiply by a constant factor of {ζ(3) – 1/ζ(2)}/2 to find the more restricted - and progressively less frequent - occurrence of all the other subclasses.

Armed with the above information we are then in a position to find the frequency of occurrence where exactly 1, exactly 2, exactly 3, and - in more general terms - exactly n prime factors occur at most twice. And these now comprise further subclasses of the subclasses already defined!

So for example the frequency of occurrence of all those numbers where 1 factor occurs at most exactly twice represents a further subclass of the - already defined - subclass, where 1 or more factors occurs at most twice!

Thus to find the frequency of occurrence of this new subclass we simply subtract the frequency of occurrence where 2 or more primes occur at most twice from the corresponding frequency of occurrence where 1 or more primes occur at most twice,

i.e. 1/ζ(3) – 1/ζ(2) – {ζ(3) – 1/ζ(2)}²/2 = .224 – .025 = .199 (approx) = 19.9%.

Therefore we can conclude that with respect to the number system as a whole, that very close to 1 in 5 members will belong to that subclass of numbers, where 1 prime factor occurs at most twice (with all other factors occurring just once).

Then, to find the frequency of occurrence of all those numbers, where 2 factors occur at most exactly twice, we subtract the frequency of occurrence, where 3 or more primes occur at most twice from the corresponding frequency of occurrence, where 2 or more occur at most twice,

i.e. {ζ(3) – 1/ζ(2)}²/2 – {ζ(3) – 1/ζ(2)}³/4 = .025 – .0028 = .022 (approx) = 2.2%.

In more general terms to calculate the frequency of occurrence, where n factors occur at most exactly twice, we subtract the frequency of occurrence, where (n + 1) or more primes occur at most twice from the corresponding frequency of occurrence where n or more occur at most twice,

i.e.  {ζ(3) – 1/ζ(2)}ⁿ/2^{n – 1} – {ζ(3) – 1/ζ(2)}^{(n + 1)}/2ⁿ


It should of course be clear that the a geometric type progression connect the various terms where 1, 2, 3, ... n primes occur exactly twice. So here again, the 1st term, corresponding to the frequency of occurrence where exactly 1 prime factor occurs at most twice =   

1/ζ(3) – 1/ζ(2) – {ζ(3) – 1/ζ(2)}²/2 = .224 – .025 = .199 (approx)

The corresponding terms, corresponding to the frequency of occurrence of exactly 1, 2, 3, ... n prime factors occurring at most twice require the successive multiplication of .199 by a common ratio =
{1/ζ(3) – 1/ζ(2)}/2 = .112 (approx).

So, alternatively, for example the estimated frequency of occurrence of numbers (with respect to the number system as a whole) where exactly 2 prime factors occur at most twice = .199 * .112 = .022 (approx).

Then the estimated frequency of occurrence of numbers where exactly 3 prime factors occur at most twice = .022 * .112 = .0025 (approx).

And we can continue on in this manner to generate the corresponding frequency of occurrence where exactly 4, 5,....n prime factors occur at most twice.