Little investigation into primes

There is a problem with all this! It has nothing to do with primes! The final expression is true for any infinite list of numbers! It's a tautology.

The original idea was to use the Fundamental Theorem of Algebra to partition the natural numbers into those with particular lowest prime factor. But because numbers usually have more than 1 prime factor this is not so simple. Since 3 x 5 = 5 x 3 there is a fundamental symmetry to all this.

Anyway we can construct a list of primes like this:

Ignoring 1 if we take the first non coloured square it is 2. This is prime. We know that every 2nd number is now not a prime because it is divisible or has prime factor of 2. We can colour these red.

The next non coloured square is 3. We know that every 3rd number from 3 is now not prime as it has a prime factor of 3. Let us colour green. The next prime is 5, let us colour that purple etc 

Now of the green squares i.e. the 3 timestable obviously half will have a Lowest Prime Factor (LPF) of 2. That is 2x3,4x3,6x3 etc will all have a LPF of 2 and only half will have a LPF of 3. For 3 timestable beyond 3 then the LPF must be 3 e.g. 3x7 has a LPF of 3. Let us denote the ratio of numbers in timestable T with LPF P as R(T,P). This has values = 0 except for P <= T. For T = 3 the values are R(3,2) = 0.5, R(3,3) = 0.5. The sum of R(T,*) = 1. R is undefined for T = 1 and P = 1.

Let us look at T=5. That is the 5 times table: 1x5,2x5,3x5,4x5,5x5,6x5,7x5,.... the highest LPF is 5 since any multiple of 5 above 5 has a LPF of 5.

R(5,2) = 1/2 as with 3.
R(5,3) = 1/6. That is because half of the 3 times tables has a LPF = 2 as above.
R(5,4) is undefined since LPF of any multiple of 4 is already 2.
R(5,5) = 1/3. We can calculate this as 1-1/2-1/6 but also with the following algorithm.

If the prime at position i in list of primes is Pi then

αi = 1/(P(i-1) -1)+1

And R(Pi,Pi) = R(Pi-1,Pi-1) / αi, Where α1= 1

Once R(Pi,Pi) is known then

R(Pi+1,Pi) = R(Pi,Pi)/Pi

and then

R(Pi+2..∞,Pi) = R(Pi+1,Pi)

As above R(Pi,Pi) = 1 - Sum[ R(Pi,1..Pi-1) ]

So for Pi = 5, i=3

α3 = 1/(3-1)+1=1.5

 R(5, 5) = R(3, 3) / α3 = 0.5/1.5 = 1/3

And for Pi = 7, i=4

α4 = 1/(5-1)+1 = 5/4

R(7, 7) = R(5, 5) / α4 = 1/3 / 5/4 = 4/15

We can work out the rest of the matrix from the diagonal.

R(Pi+1,Pi) = R(Pi,Pi) / Pi

So

R(7..P∞,5) = R(5,5) / 5 = 1/3 / 5 = 1/15

R(11..P∞,7) = R(7,7) / 7 = 4/15 / 7 = 4/105

The rows sum to 1 in the matrix:
Sum(R(Pi,j) , j=P1..P∞) = 1

This is the grid so created:

Each Row represents a timestable and the Columns represent the proportion of numbers with LPF equal to Step. The green squares are derived from the ochre above by dividing by the Step.

The green square is also in the ratio to the ochre square of
1:step-1
where step is the step of the green square.

The green squares sum to 1. That is the ochre squares / step sum to 1.

So R(Pi,Pi) = R(Pi-1,Pi-1) /  [ 1/(P(i-1) -1)+1 ]
And Sum[ R(Pi,Pi) / Pi, {1,∞} ] = 1

Sum[ R(Pi-1,Pi-1)  /  [ Pi/(P(i-1) -1)+Pi ] , {i,1,∞} ] = 1

This simplifies to the series:

Where P1 =2, P2 = 3, P3 = 5, ... are the prime numbers.

Which can also be written:

OR

as by defining the next term:

t(n+1) = t(n) (Pn -1)/P(n+1) where t(1) = 1/P1 and Pn is the nth Prime where P1 = 2

Well that ties all the primes into a single expression. But its not a very nice series as the terms tend to an infinite expression themselves.