So
potentially an unlimited number of orders with respect to both the primes and
corresponding natural numbers exist, defined as Order k primes and Order k
natural numbers respectively.

And in each
case the Order k natural numbers can be represented as a sum over the integers,
which can then be matched in corresponding fashion by a corresponding product
over the Order k primes.

So again
the Order 1 natural numbers (representing all the natural numbers) can be
matched with the Order 1 primes (representing all the primes). Thus again,

1/1

^{s}+ 1/2^{s }+ 1/3^{s }+ 1/4^{s }+ … = 1/(1 – 1/2^{s}) * 1/(1 – 1/3^{s}) * 1/(1 – 1/5^{s}) * …
Then the
Order 2 natural numbers (representing the products of prime factors which have
themselves an ordinal prime ranking) can be matched with the Order 2 primes
(i.e. as the primes with an ordinal prime ranking).

So,

1/1

^{s}+ 1/3^{s }+ 1/5^{s }+ 1/11^{s }+ … = 1/(1 – 1/3^{s}) * 1/(1 – 1/5^{s}) * 1/(1 – 1/11^{s}) * …
And then by
listing the Order 2 primes in natural number ordinal fashion, we can again pick
out those numbers with an ordinal prime ranking that then become the Order 3
primes.

And these
in turn are associated with Order 3 natural numbers (as the numbers based on
the use of these primes as factors) and so on.

However we
can easily derive in each case an alternative set of primes and natural
numbers.

So if for
example we define the Order 2 primes as those with a prime number ranking, then
the alternative set Order 2(a) can be defined as those with non prime rankings.

So the
Order 2 (a) primes are thereby 2, 7, 13, 19, 23, …

And the
corresponding Order 2(a) natural numbers (based on numbers with these primes as
factors, including as always 1) are 1, 2, 4, 7, 8, 13, 14, …

So we can
know match the sum of the Order 2(a) natural numbers with the corresponding
product of the Order 2(a) primes as follows

1/1

^{s}+ 1/2^{s }+ 1/4^{s }+ 1/7^{s }+ … = 1/(1 – 1/2^{s}) * 1/(1 – 1/7^{s}) * 1/(1 – 1/13^{s}) * …
Then by
giving the Order 2(a) primes a natural number ranking and then choosing those
with a non-prime ordinal ranking we get a new set of Order 3(a) primes, which
are associated with a corresponding set of Order 3(a) natural numbers (based on
the use of these primes as constituent factors).

So the
Order 3(a) primes are 2, 19, 29, 43, …with corresponding Order 3(a) natural
numbers 1, 2, 4, 8, 16, 19, …

Thus 1/1

^{s}+ 1/2^{s }+ 1/4^{s }+ 1/8^{s }+ … = 1/(1 – 1/2^{s}) * 1/(1 – 1/19^{s}) * 1/(1 – 1/29^{s}) * …
So in
general, just as Order k natural numbers (as sum over the integers) can be associated
with corresponding Order k primes (as products over these primes) expressions, likewise
Order k(a) natural numbers (as sum over the integers) can be associated with
corresponding Order k(a) primes (as products of these primes) expressions.

And likewise
unlimited mixing as between these two approaches is also possible.

So for
example as we have seen 3, 5, 11, 17, 31, …, are the Order 2 primes.

However we
could, when listing these primes in a natural number ordinal fashion, choose
those with a non-prime ranking. So this new “hybrid” set of primes would
thereby be

3, 17, 41,
67, 83, …

And then associated
with these primes would be a new set of “hybrid” natural numbers (based on
these primes as factors), i.e. 1, 3, 9, 17, 27, 41, 51, …

We can then
match corresponding sum over the integers and product over the primes
expressions i.e.

1/1

^{s}+ 1/3^{s }+ 1/9^{s }+ 1/17^{s }+ … = 1/(1 – 1/3^{s}) * 1/(1 – 1/17^{s}) * 1/(1 – 1/41^{s}) * …
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