Another post to Prof Tao’s blog on 1 Nov 11, recorded here for posterity:

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One more crackpot post, thanks for the indulgence Prof. Tao.

I’d appreciate if someone would comment, even if only to make fun of me. This is the internet after all. 🙂

I know this is a famous problem and it’s bad form to hastily claim a proof (or “near-proof,” as in this case I think) of a famous problem, but since I am not a mathematician, I have no reputation to lose by exposing myself to ridicule for this, and I can easily slink back to the day job when someone pokes a big hole through this.🙂

Let me elaborate on two of the items I’ve claimed without proof in the previous post, and then also show why the ratio of twin pair/primes density to primes/natural numbers density is 1.32, or at least converges and is not less than 1. That will be in the next post to keep this one relatively short.

To find all prime numbers, we need to strip away all composite numbers. We can imagine doing this by progressively removing, or sieving, all prime multiples, working our way from the lowest prime through to infinity.

Let’s start with by removing all multiples of 2, the even numbers, from the set of natural numbers. This strips out half of all natural numbers as prime candidates, leaving only the odd numbers as possible primes. Then remove all multiples of 3, 5, 7 and so on through to Pn, the nth prime, with n approaching infinity.

Consider the second primorial, P2# = 6, after first removing the even numbers, so we have 1, 3 and 5 as prime candidates, and 2, 4 and 6 as confirmed “non-primes” between 1 and 6. Note that this pattern, which is symmetric within 6, repeats ad infinitum, or in other words, after sieving the even numbers, all remaining numbers (mod 6) are either 1, 3 or 5.

Now remove all multiples of 3 from the remaining set of “possible primes.” This removes one possible prime candidate between 1 and 6, leaving 1 and 5 as possible primes. Note again this leaves a symmetric pattern, which, since we have now only removed multiples of 2 and 3, repeats ad infinitum, removing 2/3 of all natural numbers as non-primes and leaving 1/3 for further consideration as possible primes. After having sieved the 2s and 3s, the remaining “possible prime” candidates (mod 6) are all either 1 or 5.

We can continue in the same fashion to remove all remaining multiples of 5, 7, 11, 13, …. to Pn. Each time we do this, we will be left with a set of non-primes and “possible primes” that is symmetric within the respective primorial, and which repeats ad infinitum. The density of “possible primes” within the set of natural numbers, after sieving all primes to some arbitrary Pn, is equal to the number of “possible primes” within the respective primorial divided by the value of the primorial.

The complete set of prime numbers is revealed when we have sieved all possible composite numbers by repeating to infinity each prime, and striking out all these repetitions as non-prime. Therefore, if we continue this process as described, ad infinitum, to remove all multiples of all primes to Pn, where n approaches infinity, then we will be left with only primes. Hence, if we call Nn the number of “possible primes” within Pn#, then Nn/Pn# converges on the density of primes within the set of natural numbers as n approaches infinity, since as n approaches infinity, Nn becomes the set of primes and Pn# becomes the set of natural numbers.

Now let’s examine how many natural numbers are removed as candidates for any given prime, Pn. Choose an arbitrary prime, Pn, and consider the number of multiples of Pn within Pn#. This is clearly P(n-1)#. If no multiples of primes smaller than Pn had yet been removed, then repetitions of Pn to infinity would eliminate Pn prime candidates out of every Pn# natural numbers. However, assuming Pn > 2, and following our sieving process sequentially with monotonically increasing n, by the time we are set to remove remaining candidates that are multiples of Pn, there are a number of multiples that have already been removed, namely all those that are multiples of lower primes < Pn. If you conduct a simple accounting exercise, starting with the smallest prime numbers and working up through a few, you should be able to convince yourself that of the P(n-1)# possible multiples of Pn within Pn#, [P(n-1)# – N(n-1)] have already been eliminated by the multiples of Pn with lower primes, where N(n-1) means the number of possible primes left within the lower primorial, P(n-1)#, after sieving of all primes up to P(n-1), and Nn is equal to, as claimed before:

Nn = (product for all i=1 to n) [Pi – 1] —– eqn [1]

To demonstrate… removal of all even numbers eliminates half of all numbers between 1 and 2 (= P1#), and by extension removes half of all numbers within every multiple of P1# to infinity, or half of all natural numbers.

Removal of all remaining multiples of P2 = 3 between 1 and 6 (= P2#) eliminates the number 3 as a possible prime within subsequent repetitions of P2#. Note that 6 was already removed, since it is a multiple of 2. So there were 2 (= P1#) multiples of 3 within P2#, but 1 (i.e. {P1# – N1}, or {P1# – eqn [1] for n-1, where n = 2} had already been removed by multiples of 3 by 2, leaving only 1 possible prime candidate to be removed at this sieving pass, to leave 2 remaining “possible prime” candidates (1 and 5) within P2#, or 2/6 = 1/3 “possible primes” within the set of all natural numbers after repeating this pattern an infinite number of times. Hence all primes (mod 6) greater than P2 must equal 1 or 5, a well known result.

Now if we remove the remaining multiples of P3 = 5 between 1 and 30, we find that of the 6 (i.e. P2#) possible multiples of 5, 4 had already been removed, leaving only 2 (i.e. {P2# – eqn [1] for n-1 where n = 3}), namely 5 and 25, to be screened by repetitions of 5. Thus of the 10 possible primes within 30 existing after sieving by 2 and 3, two more candidates (5 and 25) have been removed, and all primes (mod 30) greater than P3 must equal 1, 7, 11, 13, 17, 19, 23 or 29 (i.e. exactly 8 possibilities).

Repeating the same logic now for multiples of P4 = 7 up to 210, of the 30 (= P3#) multiples of 7 to be removed, 22 ({P3# – product for i = 1 to 3 of [Pi – 1]}) were already screened by multiples of 7 with 2, 3 or 5, leaving 8 other multiples of 7 to be eliminated at this stage. Thus of 8 x 7 = 56 possible primes within 210 remaining after sieving by 2, 3 and 5, the further sieving by 7 removes 8 other possible prime candidates, leaving 48 “possible primes” within repetitions of 210, and therefore 48 unique values for all primes higher than 7 (mod 210), which I won’t list here.

Now let’s examine why Nn takes the form I’ve claimed, using P4 as a working example.

If we want to count the number of multiples of 7 with 2, 3 and 5, we should expect:

3 x 5 multiples of 2 x 7,

2 x 5 multiples of 3 x 7,

2 x 3 multiples of 5 x 7,

2 multiples of 3 x 5 x 7,

3 multiples of 2 x 5 x 7,

5 multiples of 2 x 3 x 7, and

1 multiple of 2 x 3 x 5 x 7

In order to identify unique multiples of 7 already removed by 2, 3 and 5, we obtain [(3 x 5) + (2 x 5) + (2 x 3)] – (2 + 3 + 5) + 1 = 22, which we can re-organize and re-write as {5 x 3 x 2 – [(5 – 1) x (3 – 1) x (2 – 1)]} = 22. Or, the number of unique multiples of 7 removed within multiples of 210 is 8 = (5 – 1) x (3 – 1). And hence the number of possible primes remaining within multiples of 210 is 8 (from P3#) x 7 – 8, or 48, or 8 x 6, or 8 x (P4 – 1), or (P4 – 1) x (P3 – 1) x (P2 – 1) x (P1 -1).

Hence, Nn = (product for all i=1 to n) [Pi – 1]

Phew. 🙂

I think that is fairly straightforward, and I expect the logic to generate the product formula for Tn is very similar, but this came to me on today’s 9 mile run, and I think I need another good run or two to come around to figuring that one out. 🙂

To wrap this post, I think I’ve defended the prior claimed definition of Nn, the number of “possible prime” candidates within the nth primorial (which by extension serves to define the density of “possible prime” candidates within the set of natural numbers after sieving by primes up to Pn). And, I think I’ve defended the claim that this density of “possible primes” within the nth primorial converges to the density of primes within the set of natural numbers as n approaches infinity.

OK, on to producing the twin prime constant, or something like it, in one more post…