Differing Orders of Primes and Natural Numbers (2)

In yesterday's blog entry I dealt with the notion of - what I refer to as - the different orders of primes and natural numbers respectively.

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) * …