The well-known
set of primes 2, 3, 5, 7, 11, … in this context can be referred to as Order 1 primes,
with the corresponding natural numbers 1, 2, 3, 4, 5, …, in corresponding
fashion referred to as Order 1 natural numbers.

However we
can now order the cardinal primes 2, 3, 5, 7, 11, …, with respect to ordinal
natural number rankings 1, 2, 3, 4, 5, …

If we now choose
to list only those primes with a corresponding ordinal prime ranking we now
obtain a new set of cardinal primes i.e. 3, 5, 11, 17, 31, 41, …

I then
refer to this new set of cardinal primes as Order 2 primes with the
corresponding set of natural numbers (including a) based on the use of these
primes as factors as

1, 3, 5, 9,
11, 15, 17, 25, 27, 31, …

And I refer
to this new set of natural numbers as Order 2 natural numbers.

We can now
again give the Ordinal 2 cardinal primes an (Order 1) natural number ranking in
an ordinal fashion i.e. 1, 2, 3, 4, 5, …

Then if we
once again only choose those remaining primes that correspond with prime rankings
we derive yet a new set of cardinal primes i.e. 5, 11, 31, 59, 127, …

I then
refer to this latest set of primes as Order 3 primes.

And once
more we can derive the corresponding Order 3 set of natural numbers (including
1) based on the use of these primes as factors i.e. 1, 5, 11, 25, 31, 55, 59,
125, 127, …

And we can
continue on indefinitely in this fashion deriving an ever sparser set of primes
and corresponding natural numbers (based on the use of these primes as
factors).

For example
the Order 4 primes are 11, 31, 127, … and the corresponding Order 4 natural
numbers are 1, 11, 31, 121, 127, …

So we keep
interchanging in this manner as between the quantitative notion of cardinal
primes and the corresponding qualitative notion of ordinal natural number
rankings.

Indeed one
fascinating insight that derives from this approach is that the full set of natural
numbers, can thereby be expressed as Order 0 primes!

So in this
way the true independence as between the primes and natural numbers with
respect to both analytic (quantitative) and holistic (qualitative) meanings is
made readily apparent.

And what is
very interesting in this regard is that we can derive an unlimited number of
corresponding L-functions based on matching the Order k primes with the corresponding
Order k natural numbers.

So for
example, we can thereby match the Order 2 primes with the corresponding Order 2
natural numbers.

Therefore
in general terms, with convergent answers where s > 1.

1 + 1/3

^{s }+ 1/5^{s }+ 1/9^{s }+ 1/11^{s }+ … = 1/(1 – 1/3^{s}) * 1/(1 – 1/5^{s}) * 1/(1 – 1/11^{s}) * …
So for example,
where s = 4,

1 + 1/3

^{4 }+ 1/5^{4 }+ 1/9^{4 }+ 1/11^{4 }+ … = 1/(1 – 1/3^{4}) * 1/(1 – 1/5^{4}) * 1/(1 – 1/11^{4}) * …
The LHS
(for listed terms) = 1.01416…, whereas the RHS = 1.01419…

So already
the two results are very similar.

This latest
procedure - entailing Order k natural numbers with corresponding Order k primes
- is different from previous procedures where we eliminated a finite number of
terms with respect to the RHS (product over primes) with then made consequent
adjustments for all terms where the primes acted as constituent factors in the
LHS (sum over the integers) expression.

In this
case we are with each successive Order, eliminating an infinite series of
individual prime terms with respect to the RHS with then consequent adjustments
to the LHS natural numbers expression.

And the product
over primes expression is no longer adjusted for a finite number of rational
terms - but rather an infinite number - we can no longer preserve results of
the form kπ

^{s }(where k again is a rational number and s a positive even integer).
However with
respect to new Order k expressions we can again adjust for a finite number of individual
terms (with respect to the RHS prime expression).

So, as we
have seen the Order 2 function is,

1 + 1/3

^{s }+ 1/5^{s }+ 1/9^{s }+ 1/11^{s }+ … = 1/(1 – 1/3^{s}) * 1/(1 – 1/5^{s}) * 1/(1 – 1/11^{s}) * …
Therefore
if we now eliminate the individual term related to 3 as prime in the RHS, then
we must correspondingly eliminate all terms which include 3 as factor in the corresponding
LHS expression.

So 1 + 1/5

^{s }+ 1/11^{s }+ 1/17^{s }+ 1/25^{s}+ … = 1/(1 – 1/5^{s}) * 1/(1 – 1/11^{s}) * 1/(1 – 1/17^{s}) * …