1/1

^{s}+ (1 + k

_{1})/2

^{s}+ (1 + k

_{2})/3

^{s}+ (1 + k

_{1})/4

^{s}+ (1 + k

_{3})/5

^{s}+ {(1 + k

_{1})(1 + k

_{2})}/6

^{s}+ …

= 1/{1 – (1 + k

_{1})/(2

^{s}+ k

_{1})} * 1/{1 – (1 + k

_{2})/(3

^{s}+ k

_{2})} * 1/{1 – (1 + k

_{3})/(5

^{s}+ k

_{3})} * …,

where k

_{1},

_{ }k

_{2}, k

_{3}, …

_{ }are rational numbers which can be either positive or negative.

In that entry, I demonstrated the - surely - interesting fact that when s is 2, 4, 6, …

and k

_{1}= k

_{2 }= k

_{3, }… = 1, that the value of the two expressions (sum over the integers and product over the primes respectively) is a rational number.

In this case there is a clear relationship as between the values of the standard zeta function for s and 2s respectively.

So - again when s is a positive even integer - the value of the zeta expression (for both sum over the integers and product over the primes expression) is t

_{1}π

^{s}(where t

_{1}is a rational number).

Likewise the value of the expression when the dimension is 2s is t

_{2}π

^{2s}(where t

_{2}is a rational number).

Let the ratio t

_{2}/t

_{1}= t

_{3.}

Then the value of the extended zeta function where k

_{1}= k

_{2 }= k

_{3, }… = 1, is t

_{3}/t

_{1 }i.e. t

_{2}/(t

_{1})

^{2}.

So when s = 2, ζ(2) = π

^{2}/6 and when s = 4, ζ(4) = π

^{4}/90.

So t

_{1 }= 6 and t

_{2 }= 90.

Therefore t

_{2}/(t

_{1})

^{2 }= 90/36 = 5/2

And this is the value of the extended expression where k

_{1}= k

_{2 }= k

_{3, }… = 1.

Another interesting - if trivial - case arises when k

_{1}= k

_{2 }= k

_{3, }… = – 1.

This leads to the elimination on both sides of all terms entailing k so that we are left with the identity 1 = 1.

This is useful to remind us that 1 precedes all subsequent relationships entailing prime factors. So 1 is not obtained from these relationships but rather serves as a necessary precondition for their use.

Thus internally the very notion of a prime entails unit components which then have both a quantitative identity as (independent of each other) and a qualitative identity as (interdependent with other) respectively.

So far in the use of the extended general formula we have shown how to generate expressions where the numerator > 1.

However by picking the value of k

_{1}, k

_{2},

_{ }k

_{3, }…, appropriately we can likewise generate expressions where the denominator - rather than the numerator term - increases.

For example when k

_{1 }= – 1/2, k

_{2 }= – 2/3, k

_{3 }= – 4/5, …, we have

1/1

^{s}+ (1 – 1/2)/2

^{s}+ (1 – 2/3)/3

^{s}+ (1 – 1/2)/4

^{s}+ (1 – 4/5)/5

^{s}+ {(1 – 1/2)(1 – 2/3)}/6

^{s}+ …

= 1/{1 – (1 – 1/2)/(2

^{s}– 1/2)} * 1/{1 – (1 – 2/3)/(3

^{s}– 2/3)} * 1/{1 – (1 – 4/5)/(5

^{s}– 4/5)} * …,

i.e.

1/1

^{s }+ 1/(2.2

^{s}) + 1/(3.3

^{s}) + 1/(2.4

^{s}) + 1/(5.5

^{s}) + 1/(6.6

^{s}) + …

= 1/1 – {(1/2)/(2

^{s}– 1/2)} * 1/1 – {(1/3)/(3

^{s}– 2/3)} * 1/1 – {(1/5)/(5

^{s}– 4/5)} * …,

So for example, when s = 2

1/1

^{2 }+ 1/(2.2

^{2}) + 1/(3.3

^{2}) + 1/(2.4

^{2}) + 1/(5.5

^{2}) + 1/(6.6

^{2}) + …

= 1/{1 – [(1/2)/(2

^{2}– 1/2)]} * 1/{1 – [(1/3)/(3

^{2}– 2/3)]} * 1/{1 – [(1/5)/(5

^{2}– 4/5]} * …,

= 1/(1 – 1/7) * 1/(1 – 1/25) * 1/(1 – 1/121) * …= 7/6 * 25/24 * 121/120 * …

And likewise k

_{1}, k

_{2},

_{ }k

_{3, }…, can be chosen so that both numerator and denominator are multiples of values that occur in the standard case.

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