Riemann Zeta Function: Important Number Relationships (2)

We saw yesterday the importance of the Riemann zeta function ζ(s), where s = 2 for the distribution of the prime factors of the various members of the number system.

So the probability that a number chosen at random will be composed of non-repeating prime factors = 1/ζ(2) = 6/π² = .6079 (approx). Alternatively, the probability that it will be composed of repeating prime factors (2 or more) = 1 – 1/ζ(2) = 1 – 6/π²  = .3921 (approx).
 

1/ζ(2) could also be stated as the probability that a number is square-free i.e. not divisible by the power of 2 (or higher) of any prime.

Equally it could be stated - at least in conventional (Type 1) mathematical terms - as the probability that two numbers chosen at random will not contain a common (proper) factor.

Therefore 1 – 1/ζ(2) represents the probability that a number is not square-free i.e. divisible by a prime raised to the power of 2 (or higher). Equally it represents the probability that 2 numbers chosen at random will contain a common factor. 


However, though in this context, 1/ζ(s) for s = 2 has the most general relevance for the number system, we can however give an important meaning to 1/ζ(s) where s is any other positive integer.

In this regard, it is interesting to probe the significance of the limiting case for 1/ζ(s), where s = 1.

Now clearly in this case, ζ(1) in Type 1 terms → ∞; therefore 1/ζ(s) = 0  and 1 – 1/ζ(1) = 1.  So we could interpret this as the probability that a number will be composed of 1 factor (or more), which of course is necessarily the case. Alternatively, this represents the probability that a number is divisible by the power of 1 (or higher) of some prime or alternatively the chance that a single number chosen at random will contain one factor (in common with itself) which again is necessarily the case.

Then for example when s = 3, 1 – 1/ζ(3) expresses the probability that a number chosen at random will be composed of 3 (or more) repeating prime factors. This could also be expressed as the probability that a number is divisible by a prime (raised to the power of 3 or higher) or alternatively the probability that 3 numbers chosen at random will contain a common factor.

Now ζ(3) = 1.20205693... and 1/ζ(3) = .8319 (approx).

Therefore 1 – 1/ζ(3) = .1681 (approx).

So the probability that a number chosen at random will contain 3 (or more) repeating factors is roughly 1/6. Alternatively we could express this result as the probability that a number is divisible by a prime raised to the power of 3 (or higher) or the probability that 3 numbers chosen at random will contain a common factor.

In more general terms, 1 – 1/ζ(s) expresses the probability that a number chosen at random will contain s (or more) repeating factors. Alternatively it expresses the probability that a number is divisible by a prime raised to the power of s (or higher) or the probability that s numbers chosen at random will contain a common factor.


We can use these results to obtain a more precise knowledge of the prime factor composition of the members of the number system.

For example 1 – 1/ζ(2) represents the probability that a number will contain 2 (or more) prime factors that repeat, while 1 – 1/ζ(3) represents the probability that 3 (or more) prime factors that repeat.

Therefore to obtain the probability that a number will contain at most 2 prime factors that repeat we subtract the latter from the former i.e.  1 – 1/ζ(2)  – {1 – 1/ζ(3)} = 1/ζ(3) – 1/ζ(2) = .8319 – .6079 = .224.

In more general terms we can therefore state that the probability that a number will contain at most s prime factors that repeat  = 1/ζ(s + 1) – 1/ζ(s).

To illustrate this, we will now estimate the total of those numbers (up to 100) which contain at most 3 repeating prime factors.

The probability (for the number system as a whole) is given as 1/ζ(4) – 1/ζ(3) = .9239 – 8319 = .092.

This would suggest therefore that we should find (up to 100) approximately 9 numbers that contain at most 3 repeating prime factors.

In fact the full list is given below:

8 =   2 * 2 * 2 

24 = 3 * 2 * 2 * 2

27 = 3 * 3 * 3

40 = 5 * 2 * 2 * 2

54 = 3 * 3 * 3 * 2

56 = 7 * 2 * 2 * 2

72 = 3 * 3 * 2 * 2 * 2

88 = 11 * 2 * 2 * 2

So there are 8 such numbers (with 3 or more repeating prime factors) up to 100 which is close to our prediction of 9.

When we carry on further up to 200, a further 9 numbers contain at most 3 repeating prime factors! So the early sample evidence is very much in line with the predicted result.

In fact there is a remarkable stability with respect to such results throughout the number system with variations from predicted results quite small.