In like manner therefore we can use this extended sequence to provide increasingly good approximations for the value of 1/ζ(s) where s is a positive integer, which as we saw yesterday plays an important role with respect to the distribution of the primes.

So once again in general terms 1/ζ(s) represents the probability that a number, chosen at random will not contain s or more repeating prime factors.

Likewise as we have seen, it expresses the probability that the number will not be divisible by a prime factor raised to the power of s or alternatively the probability that s numbers chosen at random will not contain a common (proper) factor.

Now the famed Fibonacci sequence relates to the equation,

x

^{2 }– x – 1 = 0 with phi (the golden ratio = 1.618... the positive real valued solution for x to this equation).
Alternatively, the ratio can be estimated by starting with the digits 0, 1 and then successively adding the last digit (in a 2-digit sequence) to the previous digit to thereby obtain

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, .....

Then the ratio (i.e. phi) can be approximated to increasing degrees of accuracy by dividing the last number in the sequence by the previous number.

Therefore in this case we get 144/89 = 1.617977... which is already a very good approximation to the true value 1.618033....

Now in general terms - what I refer to as - the extended Fibonacci sequence relates to the equation

x

^{s }– x^{ s – 1}^{}– ...... – x – 1 = 0.
Now in every case for s ≥ 1, a positive real-valued solution for x will arise.

Though we could algebraically obtain a solution where s = 3 and 4, the simplest approach (which can be extended to larger values of s) is by an extension of the method we used to obtain the terms in the Fibonacci (2-digit) case.

^{3 }– x

^{2 }– x – 1 = 0) we start with the 3 digits

0, 0, 1, and now add - starting with the last in the sequence - the 3 previous digits.

In this way we obtain the following sequence,

0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149,....

The estimate for x (as the positive real valued solution for x) here = 149/81 = 1.839506... which is very close to the true estimate 1.839286....).

Now the basic contention here is that x – 1 (where x is the solution to the general equation,

x

^{s }– x^{ s – 1}^{}– ...... – x – 1 = 0) offers an excellent approximation for the corresponding value of
1/ζ(s).

For example, when s = 3, as the relevant solution to the general equation for x = 1.839 (approx), then x – 1 = .839.

And the corresponding value of 1/ζ(s) for s = 3 is .832 (approx). So the Fibonacci type estimate is already very close and steadily improves in accuracy for larger values of s.

Once again as 1/ζ(s) represents the probability that a number, chosen at random will not contain s (or more) repeating prime factors, then 1
– 1/ζ(s) therefore represents the probability that the number will contain s (or more) repeating factors.

Now

^{}.the ratio of the probability that a number will contain s (or more) repeating factors to the probability that it will not contain these factors = {1 – 1/ζ(s)}/1/ζ(s) = ζ(s) – 1.
And ∑{ζ(s) – 1} = 1, where s = 2, 3, 4, 5,...... .

Thus the ratio of the probabilities that a number will contain a certain number of (proper) factors (or more) to the probability that it will not contain those factors = 1 (when summed for s ≥ 2).

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