In those entries I suggested that a certain constant
relationship (based on π) would hold with respect to the number system as a
whole regarding the distribution of those numbers with non-repeating prime factors
and repeating prime factors - or structures as I referred to them - respectively.

So based on some initial empirical investigation at various intervals
of the number system, I concluded then that the average frequency of those
numbers with non-repeating prime factors would be π/(π + 2) = .611 (approx)
with the average frequency of the remaining numbers with repeating prime
factors 2/(π + 2) = .389 (approx).

However when reading about - what are referred to in the
literature as - square-free numbers, I realised that this in fact represented an
alternative way of stating my position.

Thus the average frequency of square-free numbers (i.e. numbers
that are not divisible by prime factors raised to the power of 2 or higher)
would by definition represent those numbers with non-repeating prime factors.

And it is well-known that the average frequency of such numbers
(or alternatively the probability of obtaining a square-free number) = 6/π

^{2}, which of course is the corresponding value of the Riemann zeta function for 1/ζ(s) where s = 2.
Now 1/ζ(2) = 6/π

^{2 }= .608 (approx) which is very close to π/(π + 2) = .611 (approx).
In fact, in reaching my earlier erroneous conclusion,
I had initially considered 6/π

^{2 }the most likely estimate. However as the sample values that I took at various intervals (to estimate the frequency of numbers with non repeating prime factors) repeatedly averaged out very close to .611, I then changed to an alternative π estimate.
So now in retrospect, it is apparent that the consistency
of my estimates represented in fact sample bias over the number intervals that
I chose.

In any case it is quite easy to establish the correct result
(without resort to empirical data).

The probability that a number is divisible by the square of a
prime p = 1/p

^{2}.
So for example in the simplest case where p = 2, we would
expect 1 in 4 numbers to be divisible by 2

^{2}.
Therefore the probability that a number is square-free with
respect to a given prime (i.e. not divisible by the square of a particular prime)
= 1 – 1/p

^{2}.
However as well as considering divisibility by the square of
2, we equally have to consider all the other primes i.e. 3, 5, 7, ….

So therefore the probability that a number is generally square-free
with respect to any prime = ∏(1 – 1/p

And this result can be easily generalised as the probability that a number cannot be divided by a prime (raised to the power of n) = 1 – 1/p

^{2}) = 1/ζ(2) = 6/π^{2}.And as we have seen, this equally represents the probability that a number is composed of non-repeating prime factors, or alternatively, the average frequency of numbers with non-repeating prime factors.And this result can be easily generalised as the probability that a number cannot be divided by a prime (raised to the power of n) = 1 – 1/p

^{n}. And then when we allow for divisibility by all the primes we get ∏(1 – 1/p^{n}) = 1/ζ(n).
Then returning to the case for s = 2, this then entails that the probability that a number is not
square-free i.e. is composed of at least one repeating prime factor = 1 – 6/π

^{2}.
However there is yet another way of stating this important relationship.

The probability that a number is square-free, in fact
represents the probability that any two randomly chosen numbers will contain no
common factors.

And once again this probability is given as 1/ζ(2) = 6/π

^{2}.
Let us consider for example the two numbers 6 and 8.
These contain 2 as a common prime factor. Alternatively we could express this
situation as the fact that the product of 6 and 8 (= 48) is necessarily divisible
by the square of 2.

Therefore if two numbers chosen at random contain common prime factors
then the product of those numbers cannot be square-free.

However, clearly if the two numbers do not have common
prime factors, then the product of these numbers is square-free.

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