The 'The Sieve of Eratosthenes - Part 2

Previous post: The Sieve of Eratosthenes - Backing Up

To continue examining the data sets formed by the 'Sieve of Eratosthenes', let's look at some definitions for the sieve sets and remainder sets. Refer to the previous post for an explanation of the process and for examples of the sets, The Sieve of Eratosthenes - Backing Up.

Reviewing the 'Sieve of Eratosthenes'

Let's define the process a little better and define the resultant sets of numbers from each iteration of the process.

1. The starting point is the entire set of all natural numbers.

2. On the first iteration we sieve out all the numbers that are divisible by the prime number 2.

This produces the first sieve set soe2, and the first remainder set soe2_rem.

The soe2 set contains all numbers that have a smallest common prime factor of 2. 

The soe2_rem set contains all numbers not having any factor of the prime number 2. 

The soe2_rem set only has numbers with factors of prime numbers 3 and greater. 

3. On the second iteration we are only working with the soe2_rem set.

From the soe2_rem set we sieve out all numbers that are divisible by the prime number 3.

This produces the second sieve set soe3, and the second remainder set soe3_rem.

The soe3 set contains all numbers that have a smallest common prime factor of 3. 

The soe3_rem set contains all numbers not having any factors of the prime numbers 2, 3.  

The soe3_rem set only has numbers with factors of prime numbers 5 and greater. 

4. On the third iteration we are only working with the soe3_rem set.

From the soe3_rem set we sieve out all numbers that are divisible by the prime number 5.

This produces the third sieve set soe5, and the third remainder set soe5_rem.

The soe5 set contains all numbers that have a smallest common prime factor of 5. 

The soe5_rem set contains all numbers not having any factors of the prime numbers 2, 3, 5.  

The soe5_rem set only has numbers with factors of prime numbers 7 and greater. 

5. On the fourth iteration we are only working with the soe5_rem set.

From the soe5_rem set we sieve out all numbers that are divisible by the prime number 7.

This produces the fourth sieve set soe7, and the fourth remainder set soe7_rem.

The soe7 set contains all numbers that have a smallest common prime factor of 7. 

The soe7_rem set contains all numbers not having any factors of the prime numbers 2, 3, 5, 7.  

The soe7_rem set only has numbers with factors of prime numbers 11 and greater. 

We continue this iterative process for each prime number.

Sieve of Eratosthenes set definitions:

For each P (prime),

Each soe set can be defined as: 

The subset of all natural numbers that have a smallest common prime factor of P.

Each soe_rem set can be defined as: 

The subset of all natural numbers that only contains the number 1 and the natural numbers with prime factors greater than the prime number P.

Factorization Model for the 'Sieve of Eratosthenes' sets and remainder sets:

To get to our conclusion that the infinite series of all natural numbers has an equality with an infinite product where each term of the infinite product is a prime geometric series we need to look at factorization and the model for representing any natural number (positive integer) as an infinite product. For this we can refer to Wiki, Canonical representation of a positive integer.

Any positive integer can be uniquely represented as an infinite product taken over all the positive prime numbers, as $${\displaystyle n=2^{n_1}{3}^{n_2}{5}^{n_3}{7}^{n_4}\cdots =\prod _{i=1}^{\mathrm{\infty }}{p}_i^{n_i},}$$ where a finite number of the n_i are positive integers, and the others are zero.

Imagine if you can, going to a web page where you could create any natural number. The web page would have a form with the prime numbers and a drop down for each exponent value. The page would have infinite width to accommodate all the prime numbers and the dropdowns for each exponent selection would include every infinite positive integer and also 0. In fact let's create an abbreviated version of what that would look like, except instead of a dropdown for the choice of an exponent for each prime number, let's just show the selections themselves. This is of course an abbreviated version for demonstration purposes.

You have to imagine a column for every prime number and a row for continuing every integer exponent for each prime number.

20	30	50	70	110	130	170	190	230	290
21	31	51	71	111	131	171	191	231	291
22	32	52	72	112	132	172	192	232	292
23	33	53	73	113	133	173	193	233	293
24	34	54	74	114	134	174	194	234	294
25	35	55	75	115	135	175	195	235	295
26	36	56	76	116	136	176	196	236	296
27	37	57	77	117	137	177	197	237	297
28	38	58	78	118	138	178	198	238	298
29	39	59	79	119	139	179	199	239	299

With this you can actually select the prime number with its exponent value. You must select a prime number with its exponent value for each and every prime number, (make one selection from each column), then push a button to create the number and you have your number.

Factorization of the soe2 and soe2_rem sets:

Now let's change it up a bit and see what would happen if I restricted your choice of exponents for the prime number 2.

If I restricted the choice of exponents for the prime number 2, so that you can only select 2⁰, what numbers could you create?

20	30	50	70	110	130	170	190	230	290
	31	51	71	111	131	171	191	231	291
	32	52	72	112	132	172	192	232	292
	33	53	73	113	133	173	193	233	293
	34	54	74	114	134	174	194	234	294
	35	55	75	115	135	175	195	235	295
	36	56	76	116	136	176	196	236	296
	37	57	77	117	137	177	197	237	297
	38	58	78	118	138	178	198	238	298
	39	59	79	119	139	179	199	239	299

The only numbers you could create are the odds, the soe2_rem set.
The subset of all natural numbers that only contains numbers with prime factors greater than the prime number 2, with the exception that the number 1 is always in the remainder set.
( 1, 3, 5, 7, 9, 11,... )

If I then changed that up and restricted the choice of exponents for the prime number 2, so that you can only select integer powers of 2 and you could never select 2⁰, what numbers could you create?

	30	50	70	110	130	170	190	230	290
21	31	51	71	111	131	171	191	231	291
22	32	52	72	112	132	172	192	232	292
23	33	53	73	113	133	173	193	233	293
24	34	54	74	114	134	174	194	234	294
25	35	55	75	115	135	175	195	235	295
26	36	56	76	116	136	176	196	236	296
27	37	57	77	117	137	177	197	237	297
28	38	58	78	118	138	178	198	238	298
29	39	59	79	119	139	179	199	239	299

Now you could not create any odd numbers, the only numbers you could create are the evens, the soe2 set.
The subset of all natural numbers that have a smallest common prime factor of 2.
( 2, 4, 6, 8, 10, 12,... )

But what is this telling us?

It is telling us this, the evens are composed exactly of each power of 2 times the odds.

$${\displaystyle 2^1\times (1,3,5,7,9,11,...)}$$ $${\displaystyle 2^2\times (1,3,5,7,9,11,...)}$$ $${\displaystyle 2^3\times (1,3,5,7,9,11,...)}$$ $${\displaystyle ⋮}$$ $${\displaystyle 2^n\times (1,3,5,7,9,11,...)}$$

This directly implies that the evens (the soe2 set) is equal to the 2 geometric series times the odds (the soe2_rem set). We should literally be able to write the infinite series of all evens as equal to the product of the 2 geometric series times the infinite series of the odds and this should preserve every term of the infinite series of evens:

$$2+4+6+8+10+12+14+16+18+20+\cdots =\left(\begin{array}{c}\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\\ \end{array}\right)\times (1+3+5+7+9+11+13+15+17+19+\cdots )$$

This allows us to write the following equation:
$$\begin{array}{}\text{(1)}& x=1+2+3+4+5+6+7+8+9+10+\cdots \end{array}$$

We can reorder the infinite series on the right hand side, to be the soe2 infinite subseries (the evens) + the soe2_rem infinite subseries (the odds).
$$x=(2+4+6+8+10+\cdots )+(1+3+5+7+9+\cdots )$$

Since we know from our factorization model that the soe2 infinite subseries (the evens) is just the prime number 2 geometric series times the soe2_rem infinite subseries (the odds), we can write the right hand side of our equation as:
$$\begin{array}{}\text{(2)}& x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\\ \end{array}\right)\times (1+3+5+7+9+11+13+15+17+19+\cdots )\end{array}$$

Notice we have added the term 2⁰, to the first term of the product on the right hand side. This preserves all of the terms that are the odds from the infinite series expression on the right hand side of equation $\text{(1)}$.

And all of the terms that are the integer powers of 2, in the first term of the product on the right hand side produces all of the evens from the infinite series expression on the right hand side of equation $\text{(1)}$.

If we substitute for the variable x in equation $\text{(2)}$ with the value of x from equation $\text{(1)}$, we get the following:

$$\begin{array}{}\text{(3)}& 1+2+3+4+5+6+7+8+9+10+11+\cdots =\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\\ \end{array}\right)\times (1+3+5+7+9+11+13+15+17+19+\cdots )\end{array}$$

Since the second term of the product on the right hand side of equation $\text{(3)}$ is the odds (the soe2_rem set as an infinite series), we can examine the factorization of that expression.

The soe2_rem set (odds) factorization model.
The subset of all natural numbers that only contains numbers with prime factors greater than the prime number 2, with the exception that the number 1 is always in the remainder set.

Factorization of the soe2_rem set (the odds) into the soe3 and soe3_rem sets:

Once again you have to imagine a column for every prime number greater than 2 and a row for continuing every integer exponent for each prime number. 

The below table can create all of the natural numbers representing the soe2_rem set (odds).

30	50	70	110	130	170	190	230	290
31	51	71	111	131	171	191	231	291
32	52	72	112	132	172	192	232	292
33	53	73	113	133	173	193	233	293
34	54	74	114	134	174	194	234	294
35	55	75	115	135	175	195	235	295
36	56	76	116	136	176	196	236	296
37	57	77	117	137	177	197	237	297
38	58	78	118	138	178	198	238	298
39	59	79	119	139	179	199	239	299

Once again you must select a prime number with its exponent value for each and every prime number, (make one selection from each column), then push a button to create the number and you have your number.

Let's change it up a bit and see what would happen if I restricted your choice of exponents for the prime number 3.

If I restricted the choice of exponents for the prime number 3, so that you can only select 3⁰, what numbers could you create?

30	50	70	110	130	170	190	230	290
	51	71	111	131	171	191	231	291
	52	72	112	132	172	192	232	292
	53	73	113	133	173	193	233	293
	54	74	114	134	174	194	234	294
	55	75	115	135	175	195	235	295
	56	76	116	136	176	196	236	296
	57	77	117	137	177	197	237	297
	58	78	118	138	178	198	238	298
	59	79	119	139	179	199	239	299

The only numbers you could create are numbers in the soe3_rem set.
The subset of all natural numbers that only contains numbers with prime factors greater than the prime number 3, with the exception that the number 1 is always in the remainder set.
( 1, 5, 7, 11, 13, 17,... )

If I then changed that up and restricted the choice of exponents for the prime number 3, so that you can only select integer powers of 3 and you could never select 3⁰, what numbers could you create?

	50	70	110	130	170	190	230	290
31	51	71	111	131	171	191	231	291
32	52	72	112	132	172	192	232	292
33	53	73	113	133	173	193	233	293
34	54	74	114	134	174	194	234	294
35	55	75	115	135	175	195	235	295
36	56	76	116	136	176	196	236	296
37	57	77	117	137	177	197	237	297
38	58	78	118	138	178	198	238	298
39	59	79	119	139	179	199	239	299

Now you could not create any numbers of the soe3_rem set, the only numbers you could create are numbers of the soe3 set.
The subset of all natural numbers that have a smallest common prime factor of 3.
( 3, 9, 15, 21, 27, 33, 39,... )

But what is this telling us?

It is telling us this, the numbers of the soe3 set are composed exactly of each power of 3 times the numbers of the soe3_rem set.

$${\displaystyle 3^1\times (1,5,7,11,13,17,...)}$$ $${\displaystyle 3^2\times (1,5,7,11,13,17,...)}$$ $${\displaystyle 3^3\times (1,5,7,11,13,17,...)}$$ $${\displaystyle ⋮}$$ $${\displaystyle 3^n\times (1,5,7,11,13,17,...)}$$

This directly implies that the soe3 set is equal to the 3 geometric series times the soe3_rem set. We should literally be able to write the infinite series of all numbers of the soe3 set as equal to the product of the 3 geometric series times the infinite series of the soe3_rem set and this should preserve every term of the infinite series of the soe3 set:

$$3+9+15+21+27+33+39+45+51+\cdots =\left(\begin{array}{c}\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\\ \end{array}\right)\times (1+5+7+11+13+17+19+\cdots )$$

Since we are examining the factorization of the soe2_rem set (the odds) into the soe3 and soe3_rem set, the above equation only accounts for the soe3 set and is missing the soe3_rem set. Working with the entire soe2_rem set (odds) as a series, we can write:

Re-order the terms of the soe2_rem series (odds) into the soe3 subseries plus the soe3_rem subseries: $$1+3+5+7+9+11+13+15+17+\cdots =(3+9+15+21+27+33+39+\cdots )+(1+5+7+11+13+17+19+\cdots )$$ Re-write the right hand side of the equation into a product, the 3 geometric series times the soe3_rem subseries. $$1+3+5+7+9+11+13+15+17+\cdots =\left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times (1+5+7+11+13+17+19+\cdots )$$ You will notice in the product on the right hand side of the above equation, we have included 3⁰ as the first term of the 3 geometric series. This maintains all of the terms of the soe3_rem subseries and then the integer powers of the 3 geometric series maintains all of the terms of the soe3 subseries, the entirety of which is equal to the soe2_rem (odds) series on the left hand side of the equation.

We can now return to equation $\text{(3)}$, where the second term of the product on the right hand side is the soe2_rem (odds) subseries, and re-write that subseries into the right hand side of the equation we just derived above.

Equation 3: $$1+2+3+4+5+6+7+8+9+10+11+\cdots =\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\\ \end{array}\right)\times (1+3+5+7+9+11+13+15+17+19+\cdots )$$

Re-writing the second term of the product on the right hand side (soe2_rem subseries) into a product.

$$\begin{array}{}\text{(4)}& 1+2+3+4+5+6+7+8+9+10+11+\cdots =\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\\ \end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\\ \end{array}\right)\times (1+5+7+11+13+17+19+\cdots )\end{array}$$

Where is this all going:

If we continue to recursively examine the factorization of each 'Sieve of Eratosthenes' set and remainder set, we will see that each soe set and soe_rem set have a similar relationship as follows:

Each soe set factor wise, is simply that prime number's geometric series times the soe_rem set (remainder set), expressed as a series.

So we can add to our 'Sieve of Eratosthenes' definitions.

Sieve of Eratosthenes set definitions:

For each P (prime),

Each soe set can be defined as: 

The subset of all natural numbers that have a smallest common prime factor of P.

Each soe_rem set can be defined as: 

The subset of all natural numbers that only contains the number 1 and the natural numbers with prime factors greater than the prime number P.

Each soe set can be expressed as a product:
Where the first term is the prime number P's geometric series with positive integer exponents and the second term is the infinite series of all of the elements of the soe_rem set.

Or speaking purely in the context of series, the infinite series of all natural numbers:

For each P (prime),

Each soe subseries can be defined as: 

The subseries of all natural numbers that have a smallest common prime factor of P.

Each soe_rem subseries can be defined as: 

The subseries of all natural numbers that only contains one and the natural numbers with prime factors greater than the prime number P.

Each soe subseries can be expressed as a product:
Where the first term of the product is the prime number P's geometric series with positive integer exponents, and the second term of the product is the infinite subseries of all natural numbers that only contains one and the natural numbers with prime factors greater than the prime number P. (This is the property that we are concerned with for the 'Sieve of Eratosthenes'.)

With the definition of this new property of the 'Sieve of Eratosthenes', we can use the following logic to pursue the equality we are looking for, namely 'the infinite series of all natural numbers has an equality as an infinite product of prime geometric series'.

The 'Sieve of Eratosthenes' performed on the infinite series of all natural numbers:

$$\begin{array}{}\text{(1a)}& x=1+2+3+4+5+6+7+8+9+10+\cdots \end{array}$$

Sieving pass for the prime number 2.
We can reorder the infinite series on the right hand side, to be the soe2 infinite subseries (the evens) + the soe2_rem infinite subseries (the odds).
$$x=\underset{\text{soe2}}{\underbrace{(2+4+6+8+10+\cdots )}}+\underset{\text{soe2\_rem}}{\underbrace{(1+3+5+7+9+\cdots )}}$$

Since each soe subseries can be expressed as a product of the prime number's geometric series times the soe remainder subseries, we can write.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \underset{\text{soe2\_rem}}{\underbrace{(1+3+5+7+9+\cdots )}}$$

Sieving pass for the prime number 3.
We can re-order the terms of the last term of the product on the right hand side, sieving by the next prime number, 3.
This gives us an soe3 subseries and an soe3_rem subseries.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \underset{\text{soe3}}{\underbrace{((3+9+15+21+27+\cdots )}}+\underset{\text{soe3\_rem}}{\underbrace{(1+5+7+11+13+\cdots ))}}$$

Since each soe subseries can be expressed as a product of the prime number's geometric series times the soe remainder subseries, we can write.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \underset{\text{soe3\_rem}}{\underbrace{(1+5+7+11+13+\cdots )}}$$

Sieving pass for the prime number 5.
We can re-order the terms of the last term of the product on the right hand side, sieving by the next prime number, 5.
This gives us an soe5 subseries and an soe5_rem subseries.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \underset{\text{soe5}}{\underbrace{((5+25+35+55+65+\cdots )}}+\underset{\text{soe5\_rem}}{\underbrace{(1+7+11+13+17+\cdots ))}}$$

Since each soe subseries can be expressed as a product of the prime number's geometric series times the soe remainder subseries, we can write.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \left(\begin{array}{c}\\ 5^0\\ +\\ 5^1\\ +\\ 5^2\\ +\\ 5^3\\ +\\ 5^4\\ +\\ 5^5\\ +\\ ⋮\\ 5^n\end{array}\right)\times \underset{\text{soe5\_rem}}{\underbrace{(1+7+11+13+17+\cdots )}}$$

Sieving pass for the prime number 7.
We can re-order the terms of the last term of the product on the right hand side, sieving by the next prime number, 7.
This gives us an soe7 subseries and an soe7_rem subseries.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \left(\begin{array}{c}\\ 5^0\\ +\\ 5^1\\ +\\ 5^2\\ +\\ 5^3\\ +\\ 5^4\\ +\\ 5^5\\ +\\ ⋮\\ 5^n\end{array}\right)\times \underset{\text{soe7}}{\underbrace{((7+49+77+91+119+\cdots )}}+\underset{\text{soe7\_rem}}{\underbrace{(1+11+13+17+19\cdots ))}}$$

Since each soe subseries can be expressed as a product of the prime number's geometric series times the soe remainder subseries, we can write.

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \left(\begin{array}{c}\\ 5^0\\ +\\ 5^1\\ +\\ 5^2\\ +\\ 5^3\\ +\\ 5^4\\ +\\ 5^5\\ +\\ ⋮\\ 5^n\end{array}\right)\times \left(\begin{array}{c}\\ 7^0\\ +\\ 7^1\\ +\\ 7^2\\ +\\ 7^3\\ +\\ 7^4\\ +\\ 7^5\\ +\\ ⋮\\ 7^n\end{array}\right)\times \underset{\text{soe7\_rem}}{\underbrace{(1+11+13+17+19\cdots ))}}$$

This process can be continued for every prime number and the result is:

$$x=\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \left(\begin{array}{c}\\ 5^0\\ +\\ 5^1\\ +\\ 5^2\\ +\\ 5^3\\ +\\ 5^4\\ +\\ 5^5\\ +\\ ⋮\\ 5^n\end{array}\right)\times \left(\begin{array}{c}\\ 7^0\\ +\\ 7^1\\ +\\ 7^2\\ +\\ 7^3\\ +\\ 7^4\\ +\\ 7^5\\ +\\ ⋮\\ 7^n\end{array}\right)\times \left(\begin{array}{c}\\ 7^0\\ +\\ 7^1\\ +\\ {11}^2\\ +\\ {11}^3\\ +\\ {11}^4\\ +\\ {11}^5\\ +\\ ⋮\\ {11}^n\end{array}\right)\times \left(\begin{array}{c}\\ {13}^0\\ +\\ {13}^1\\ +\\ {13}^2\\ +\\ {13}^3\\ +\\ {13}^4\\ +\\ {13}^5\\ +\\ ⋮\\ {13}^n\end{array}\right)\times \left(\begin{array}{c}\\ {17}^0\\ +\\ {17}^1\\ +\\ {17}^2\\ +\\ {17}^3\\ +\\ {17}^4\\ +\\ {17}^5\\ +\\ ⋮\\ {17}^n\end{array}\right)\times \left(\begin{array}{c}\\ {19}^0\\ +\\ {19}^1\\ +\\ {19}^2\\ +\\ {19}^3\\ +\\ {19}^4\\ +\\ {19}^5\\ +\\ ⋮\\ {19}^n\end{array}\right)\times \cdots$$

Substituting for the value of x in equation $\text{(1a)}$ we get:

The infinite series of all natural numbers has an equality as an infinite product of prime geometric series:

$$1+2+3+4+5+6+7+8+9+10+\cdots =\left(\begin{array}{c}\\ 2^0\\ +\\ 2^1\\ +\\ 2^2\\ +\\ 2^3\\ +\\ 2^4\\ +\\ 2^5\\ +\\ ⋮\\ 2^n\end{array}\right)\times \left(\begin{array}{c}\\ 3^0\\ +\\ 3^1\\ +\\ 3^2\\ +\\ 3^3\\ +\\ 3^4\\ +\\ 3^5\\ +\\ ⋮\\ 3^n\end{array}\right)\times \left(\begin{array}{c}\\ 5^0\\ +\\ 5^1\\ +\\ 5^2\\ +\\ 5^3\\ +\\ 5^4\\ +\\ 5^5\\ +\\ ⋮\\ 5^n\end{array}\right)\times \left(\begin{array}{c}\\ 7^0\\ +\\ 7^1\\ +\\ 7^2\\ +\\ 7^3\\ +\\ 7^4\\ +\\ 7^5\\ +\\ ⋮\\ 7^n\end{array}\right)\times \left(\begin{array}{c}\\ 7^0\\ +\\ 7^1\\ +\\ {11}^2\\ +\\ {11}^3\\ +\\ {11}^4\\ +\\ {11}^5\\ +\\ ⋮\\ {11}^n\end{array}\right)\times \left(\begin{array}{c}\\ {13}^0\\ +\\ {13}^1\\ +\\ {13}^2\\ +\\ {13}^3\\ +\\ {13}^4\\ +\\ {13}^5\\ +\\ ⋮\\ {13}^n\end{array}\right)\times \left(\begin{array}{c}\\ {17}^0\\ +\\ {17}^1\\ +\\ {17}^2\\ +\\ {17}^3\\ +\\ {17}^4\\ +\\ {17}^5\\ +\\ ⋮\\ {17}^n\end{array}\right)\times \left(\begin{array}{c}\\ {19}^0\\ +\\ {19}^1\\ +\\ {19}^2\\ +\\ {19}^3\\ +\\ {19}^4\\ +\\ {19}^5\\ +\\ ⋮\\ {19}^n\end{array}\right)\times \cdots$$

Summarizing the 'Sieve of Eratosthenes' Process

This 'Sieve of Eratosthenes' Process can be applied to the natural numbers, the inverse natural numbers, and the rational numbers.

In the inverse natural numbers, the sieving process is the same, except we are working with 0 and negative integer exponent values. We even see the 'Sieve of Eratosthenes' sets and remainder sets employed in Euler's Theorem 7, Euler's infinite product over the primes. 

In the rational numbers, the sieving process is the same, except that we need to sieve in both the numerator and denominator for each iteration, and the prime geometric series then includes exponents 0, positive integers, and negative integers. The property where each soe subseries can be expressed as a product of the prime number's geometric series times the soe remainder subseries is also true in the rationals.

Example: The first iteration of sieving all rationals, we would sieve all rational numbers with a factor of 2 in the numerator or the denominator and our remainder set would then have only rational numbers with prime factors of 3 or greater in the numerator or denominator. From that first remainder set then we would sieve out all rational numbers with a factor of 3 in the numerator or denominator and the new remainder set would then have only rational numbers with prime factors of 5 or greater in the numerator or denominator. I can provide an additional post to step through this is needed.

In the inverse natural numbers the prime geometric series is convergent and we have an algebraic solution to prove that the series of inverse natural numbers has an equality as an infinite product of prime geometric series, à la Euler's infinite product over the primes. This however does not mean that the natural numbers and rational numbers do not have the same type of equality, it is just that the prime geometric series in the naturals and the rationals are not convergent and this prevents them from having an algebraic solution similar to Euler's Theorem 7.

The Sieve of Eratosthenes - Backing Up

I have realized much of the information I have provided requires a greater level of context and simplification. This is the beginning of the effort to provide better context with a much simpler foundation. My idea or proposal that the infinite series of all natural numbers has an equality as an infinite product of prime geometric series, just as Euler's product is an infinite product of inverse prime geometric series, requires a simpler foundation and mathematical logic. That simpler foundation and mathematical logic is the 'Sieve of Eratosthenes'.

I will use the explanation for the the 'Sieve of Eratosthenes' from Wikipedia, https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes.

Overview

A prime number is a natural number that has exactly two distinct natural number divisors: the number 1 and itself.

To find all the prime numbers less than or equal to a given integer n by Eratosthenes' method:

1. Create a list of consecutive integers from 2 through n: (2, 3, 4, ..., n).
2. Initially, let p equal 2, the smallest prime number.
3. Enumerate the multiples of p by counting in increments of p from 2p to n, and mark them in the list (these will be 2p, 3p, 4p, ...; the p itself should not be marked).
4. Find the smallest number in the list greater than p that is not marked. If there was no such number, stop. Otherwise, let p now equal this new number (which is the next prime), and repeat from step 3.
5. When the algorithm terminates, the numbers remaining not marked in the list are all the primes below n.

The main idea here is that every value given to p will be prime, because if it were composite it would be marked as a multiple of some other, smaller prime. Note that some of the numbers may be marked more than once (e.g., 15 will be marked both for 3 and 5).

SKopp at German Wikipedia, CC BY-SA 3.0, via Wikimedia Commons

For the data that we are interested in from the 'Sieve of Eratosthenes', we want the sets of numbers that are the result of each iteration, and only the numbers that have not been previously marked. We also want to know what the remainder set is after each iteration. The list of integers that we begin with also starts with the number 1 and therefore the number 1 will always be in each remainder set.

Example: For the prime number 3 iteration, the original sieve marks out 3,6,9,12,15,18,21,24,27,... However 6,12,18,24 have already been marked out by the previous prime number 2 iteration. We only want the set of numbers from each iteration that have not been previously marked out. Also, we want to include the prime number in each iteration as a member of that iteration's sieve set. The easiest method to accomplish this is to only mark out multiples of the next prime number in the previous iteration's remainder set.

Here is a table with the 'Sieve of Eratosthenes' sets we are looking for marked out.

1	2	3	4	5	6	7	8	9	10	11	12	13	14	15	16	17	18	19	20
21	22	23	24	25	26	27	28	29	30	31	32	33	34	35	36	37	38	39	40
41	42	43	44	45	46	47	48	49	50	51	52	53	54	55	56	57	58	59	60
61	62	63	64	65	66	67	68	69	70	71	72	73	74	75	76	77	78	79	80
81	82	83	84	85	86	87	88	89	90	91	92	93	94	95	96	97	98	99	100
101	102	103	104	105	106	107	108	109	110	111	112	113	114	115	116	117	118	119	120
121	122	123	124	125	126	127	128	129	130	131	132	133	134	135	136	137	138	139	140
141	142	143	144	145	146	147	148	149	150	151	152	153	154	155	156	157	158	159	160
161	162	163	164	165	166	167	168	169	170	171	172	173	174	175	176	177	178	179	180
181	182	183	184	185	186	187	188	189	190	191	192	193	194	195	196	197	198	199	200
201	202	203	204	205	206	207	208	209	210	211	212	213	214	215	216	217	218	219	220

Legend:

Prime	Set Name	SOE Sieve Set	Remainder Name	SOE Sieve Remainder Set
2	soe2	2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36,38,40, 42,44,46,48,50,52,54,56,58,60,62,64,66,68,70,72,74,76,78, 80,82,84,86,88,90,92,94,96,98,100,102,104,106,108,110,112, 114,116,118,120,122,124,126,128,130,132,134,136,138,140, 142,144,146,148,150,152,154,156,158,160,162,164,166,168, 170,172,174,176,178,180,182,184,186,188,190,192,194,196, 198,200,202,204,206,208,210,...	soe2_rem	1,3,5,7,9,11,13,15,17,19,21,23,25,27,29,31,33,35,37,39, 41,43,45,47,49,51,53,55,57,59,61,63,65,67,69,71,73,75,77, 79,81,83,85,87,89,91,93,95,97,99,101,103,105,107,109,111, 113,115,117,119,121,123,125,127,129,131,133,135,137,139, 141,143,145,147,149,151,153,155,157,159,161,163,165,167, 169,171,173,175,177,179,181,183,185,187,189,191,193,195, 197,199,201,203,205,207,209,...
3	soe3	3,9,15,21,27,33,39,45,51,57,63,69,75,81,87,93,99,105,111, 117,123,129,135,141,147,153,159,165,171,177,183,189,195, 201,207,...	soe3_rem	1,5,7,11,13,17,19,23,25,29,31,35,37,41,43,47,49,53,55,59, 61,65,67,71,73,77,79,83,85,89,91,95,97,101,103,107,109,113, 115,119,121,125,127,131,133,137,139,143,145,149,151,155,157, 161,163,167,169,173,175,179,181,185,187,191,193,197,199,203, 205,209,...
5	soe5	5,25,35,55,65,85,95,115,125,145,155,175,185,205,...	soe5_rem	1,7,11,13,17,19,23,29,31,37,41,43,47,49,53,59,61,67,71,73, 77,79,83,89,91,97,101,103,107,109,113,119,121,127,131,133, 137,139,143,149,151,157,161,163,167,169,173,179,181,187,191, 193,197,199,203,209,...
7	soe7	7,49,77,91,119,133,161,203,...	soe7_rem	1,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83, 89,97,101,103,107,109,113,121,127,131,137,139,143,149,151, 157,163,167,169,173,179,181,187,191,193,197,199,209,...

Interestingly, the soe sets and remainder sets are all periodic and symmetric. 

This is easy to notice starting in the soe3_rem remainder set.

soe3_rem: {1,5} for the primorial length of 6.

{1,5} +6n, where n = 0→∞

*** This is where we get the statement that every prime number greater than 3, must be of the form 6n ±1

soe5 set: {5,25} for the primorial length of 30. 

{5,25} +30n, where n = 0→∞

soe5_rem set: {1,7,11,13,17,19,23,29} for the primorial length of 30.

{1,7,11,13,17,19,23,29} +30n, where n = 0→∞

*** The soe5_rem set is the foundation of Gary Croft's: 'The Prime Spiral Sieve' https://www.primesdemystified.com

soe7 set: {7,49,77,91,119,133,161,203} for the primorial length of 210. 

{7,49,77,91,119,133,161,203} +210n, where n = 0→∞

soe7_rem set: {too big to fit here, see below} for the primorial length of 210.

{1,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,121,127,131,137,139,143,149,151,157,163,167,169,173,179,181,187,191,193,197,199,209} +210n, where n = 0→∞

*** The soe7_rem set is the foundation of a recent paper entitled “The Periodic Table of Primes" [4]. Where the authors argue that primes can be predicted using a periodic table-like structure." 

https://phys.org/news/2024-04-breakthrough-prime-theory-primes.html

https://papers.ssrn.com/sol3/papers.cfm?abstract_id=4742238

These 'Sieve of Eratosthenes' sets and remainder sets are also the basis for James McCanney's work, 'New Definition of Prime Numbers with Sppn Tables and Proofs by Induction'.

https://papers.ssrn.com/sol3/papers.cfm?abstract_id=4800997#:~:text=James%20McCanney,-jmccanneyscience%2C%20LLC&text=The%20prime%20numbers%20are%20calculated,with%20unique%20ancestors%20and%20descendants.

These 'Sieve of Eratosthenes' sets and remainder sets are also the basis for 'Liu Fengsui's Prime Formula' and subsequent papers.

https://www.primepuzzles.net/problems/prob_037.htm

The 'Sieve of Eratosthenes' sets and remainder sets have a lot of interesting properties. However we are interested in a different property of these sets and that will be the topic of my next post.

Continuation: The 'The Sieve of Eratosthenes - Part 2