For example
the (infinite) series of triangular numbers i.e. 1, 3, 6, 10, 15, …, provides
one such ordered sequence (where we are always enabled to provide the next term
in the sequence i.e. n(n + 1)/2, with n = 1, 2, 3, …

So if we
now apply this ordering to the primes by choosing the 1

^{st}, 3^{rd}, 6^{th}, 10^{th}15^{th}… numbers, then we have 2, 5, 13, 29, 47, …
So the
corresponding sum over the integers expression is thereby now based (including
1) on using these primes as sole factors.

Thus

1/1

^{s}+ 1/2^{s}+1/4^{s}+ 1/5^{s}+ 1/8^{s}+ 1/10^{ s}… = 1/(1 – 1/2^{s}) * 1/(1 – 1/5^{s}) * 1/(1 – 1/13^{s}) * …
In fact the
prototype Riemann zeta function itself corresponds on an ordering that is
equivalent to the denominators of the same zeta function (for s = 1).

So if we
now take an ordering based on the denominators of the function for s = 2, it
will correspond to the sum of squares 1, 4, 9, 16, 25,…

Thus choosing
our primes based on these ordinal rankings we have 2, 7, 23, 53, 97,…

Then the sum
over integers expression matching the product over primes expression (using this
selection) is based solely (including 1) on numbers based on the factors of such
primes.

Thus,

1/1

^{s}+ 1/2^{s}+1/4^{s}+ 1/7^{s}+ 1/8^{s}+ 1/14^{ s}… = 1/(1 – 1/2^{s}) * 1/(1 – 1/7^{s}) * 1/(1 – 1/23^{s}) * …
Of course
as always we can then choose to omit any specific prime (or combination of
primes) from the RHS expression with corresponding adjustments (based on omission
of natural numbers where these are factors) on the LHS.

Therefore
in choosing to omit 2 from the RHS (product over primes), we thereby choose to
omit all numbers with 2 as a common factor in the LHS (sum over the integers) expression.

So,

1/1

^{s}+ 1/7^{s}+ 1/23^{s}+ 1/49^{s}+ 1/53^{s}+ … = 1/(1 – 1/7^{s}) * 1/(1 – 1/23^{s}) * 1/(1 – 1/53^{s}) * …
When in the
above s = 4 the sum of listed terms for LHS expression = 1.000420367, while the
corresponding sum of terms for RHS expression = 1.000420368 (which already
shows an extremely close matching).

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