We have already defined Zeta 2, the finite equation (complementary with Zeta 1) as

s^1 + s^2 + s^3 + s^4 +.......s^n = 0

Then dividing by the trivial solution i.e. s = 0 we obtain

1 + s^1 + s^2 + ....... s^(n - 1) = 0.

For any given value of n, we obtain the (n - 1) non-trivial roots of 1.

For example in the simple case where n = 2, we thereby obtain the equation

1 + s = 0. So s = - 1 represents the non-trivial 2nd root of 1. As + 1 will always be a root, regardless of the value of n, we refer to it as the trivial root!

Now, if n is a prime number, the non-trivial roots of 1 will be unique in nature (and the basis for all other natural number roots).

So in this complementary sense, we can understand the prime numbers as comprising unique circles of interdependence (with respect to their ordinal natural number roots).

Therefore for example with respect to the 5 roots of 1, we have a unique circle of interdependent roots (as points on the unit circle in the complex plane) with the 4 non-trivial roots arising as the solution to the equation,

1 + s^1 + s^2 + s^3 + s^4 = 0.

Now the series involved i.e.

1 + s^1 + s^2 + s^3 + .... + s^(n - 1) can be shown to have a fascinating relationship to the primes when we take the value of s = 1.

For example, once more when n = 5, we derive the series

y = 1 + s^1 + s^2 + s^3 + s^4

Now setting s = 1, we obtain y = 1 + 1 + 1 + 1 + 1 = 5 (which is the same prime number).

Now if we the differentiate this series (with respect to s) we obtain

dy/ds = 1 + 2s + 3(s^2) + 4(s^3);

Now setting s = 1, we obtain 1 + 2 + 3 + 4 = 10 (which is divisible by 5)

Then by deriving the second derivative we obtain the new expression

2 + 6s + 12(s^2)

So setting s = 1 we obtain 2 + 6 + 12 = 20 (again divisible by 5)

Differentiating once more we obtain

6 + 24s and setting s = 1 the sum = 30 (again divisible by 5).

Basically if n is a prime number when we set s = 1 in our series (= y) and repeatedly differentiate the expression with respect to s till we obtain a linear expression in s, the resulting sum will always be divisible by n.

If n is not a prime number then this conclusion will not apply.

For example when n = 6,

y = 1 + s^1 + s^2 + s^3 + s^4 + s^5

setting s = 1 we obtain y = 1 + 1 + 1 + 1 + 1 + 1 = 6.

Then differentiating (with respect to s) we obtain

dy/ds = 1 + 2s + 3(s^2) + 4(s^3) + 5(s^4).

And setting s = 1 we obtain 1 + 2 + 3 + 4 + 5 = 15

However 15 is not divisible by 6.

If 2 is a factor of n, when we obtain the 1st differential of our original expression for y (with respect to s) and set s = 1, the resulting sum will not be divisible by n.

Then if 3 is a factor of n and we obtain the second differential of the original expression for y (with respect to s) and set s = 1, the resulting sum will not be divisible by n.

For example, 3 is also a prime factor of 6 and the second differential of our expression (where n = 6) is

2 + 6s + 12(s^2) + 20(s^3).

So setting s = 1, we obtain 2 + 6 + 12 + 20 = 40 (which is not divisible by 6).

Thus in general terms, if k is a factor of n then we obtain the (k - 1)th differential of the original expression (with respect to y, and set s = 1, the resulting sum will not be divisible by k.

So therefore it is only when n is prime that the sums with respect to all differentials of the original expression for y (until the linear form is obtained) will all be divisible by n.

There are some other highly interesting aspects to the ultimate sum for the expression (when it is reduced to linear format).

For example when n = 5 the ultimate linear form of the expression = 6 + 24s.

So setting s = 1 the resulting sum = 6 + 24 = 3! + 4!

This is a pattern that universally holds, so the ultimate sum

= (n - 2)! + (n - 1)!

Alternatively it can be expressed as n * (n - 2)!

Also when we look at the factors of the sum (where again n = 5) we can see that it is 30 = 2 * 3 * 5 (which is the product of all the prime factors up to and including 5).

Again this universally holds. So if n is prime, the ultimate sum will be an expression involving the product of all the prime factors from 2 to n inclusive (and only these primes) with many of these factors perhaps recurring!

Also the ultimate sum also will always be divisible by the sum of all the natural numbers (up to and including n).

Where n = 5, for example the sum of the natural numbers = 1 + 2 + 3 + 4 + 5 = 15.

And of course 30 (the ultimate result) is divisible by 15!

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