20201116, 23:50  #1 
Nov 2020
2^{2} Posts 
Polignac's Conjecture Question
Are there flaws in any of the considerations below?
A) All Prime Numbers greater than 3 are contained within the Sets 6x1 and 6x+1 for x>0 [1]. B) The Set 6x1 contains an infinite number of Prime Numbers [2]. C) P_(6x1) (x), the probability that a given number in the Set 6x1 is a Prime Number, is:  1) equivalent to the Prime Number Density, π_(6x1)(x)/x (by definition) [3]  2) always greater than zero (because there are an infinite number of Prime Numbers in the Set 6x1) D) The Set 6x+1 contains an infinite number of Prime Numbers [2]. E) P_(6x+1) (x), the probability that a given number in the set 6x+1 is a Prime Number, is  1) equivalent to the Prime Number Density, π_(6x+1)(x)/x (by definition) [3]  2) always greater than zero (because there are an infinite number of Prime Numbers in the Set 6x+1) F) The probability of finding a gap of size g is  1) P_(g=2) = P_(6x1) (x) * P_(6x+1) (x)  2) P_(g=2) = P_(6x1) (x) * P_(6x+1) (x1)  3) P_g is as per Table 1 here when g is 6, 8, 10, 12 ... G) Because P_(6x1) (x)>0 and P_(6x+1) (x)>0, then P_g (x)>0 for any value of x or g; therefore, there are infinitely many cases of two consecutive prime numbers separated by any given gap. H) Based on the considerations above and the Prime Number Theorem, predictions can be made for Prime Gap Densities for any value of x (as has been done here , where the Actual Prime Gap Densities agrees well predictions). 
20201117, 00:21  #2 
"Curtis"
Feb 2005
Riverside, CA
1000111111111_{2} Posts 
The probabilities you use as inputs tend asymptotically to zero. You treat them as finitely positive, which isn't correct.
You also treat the probabilities as independent, but they are not (necessarily). Your argument amounts to "there are probably an infinite number of prime pairs of a particular gap, because the individual probabilities don't vanish to zero". It's not a proof more of a summary of a heuristic. 
20201117, 03:26  #3 
Feb 2017
Nowhere
1000001000011_{2} Posts 
A Survey of Results on Primes in Short Intervals indicates that, in order to insure
in accord with PNT, y has to be a larger function of X than you might hope. 
20201118, 03:38  #4 
Nov 2020
2^{2} Posts 
Thanks for responding. I am not so much interested in a proof at this point, but rather more interested in accurate predictions of Prime Gap Densities. I would greatly appreciate your assistance with some further questions:
A) Dr. Sardonicus, from PNT (1) π(n) ~ n/log(n)define n in terms of x for x>0 (2) n = 6x+1and define the Prime Counting Function, π(n), in terms of the two sets containing all Primes except the numbers 2 and 3 when x>0; namely the Prime Counting Functions of the sets 6x1 and 6x+1, which are denoted as π_(6x1)(x) and π_(6x+1)(x), respectively. (3) π(n) = π_(6x1)(x) + π_(6x+1)(x) + 2then substituting (2) and (3) into (1) (4) π_(6x1)(x) + π_(6x+1)(x) + 2 ~ (6x+1)/log(6x+1)Assumption #1 (5) π_(6x1)(x) = π_(6x+1)(x)then by substituting (5) into (4) (6) 2[π_(6x1)(x)] ~ (6x+1) / log(6x+1)and when x>>0 (7) π_(6x1)(x) ~ 3x / log(6x+1) (8) π_(6x+1)(x) ~ 3x / log(6x+1)When you say that "in order to ensure π(X+y)  π(X) ~ y/log(X) (X → ∞) y has to be a larger function of X than you might hope", I am not sure what I am supposed to be hoping for. Eq. (7) and Eq. (8) are just restatements of PNT (i.e. Eq. (1)) in terms of x and making use of Assumption #1. Though Assumption #1 is not justified or proven at this point, it is not impossible and I can't see that restating PNT in other terms through some algebraic manipulation requires any further work to be true or that anything is missing or invalid. Can you please clarify your statement? B) VBCurtis, If one has 2 Nsided dice; A and B, the probability that one rolls a 1 with A (i.e. P_A(1)) is 1/N and the probability that one roles a 1 with B (i.e. P_B(1)) is 1/N. The probability of rolling 1s with both A and B in the same turn (i.e. P_AB(1,1) is the product of the probabilities. So, (9a) P_A(1) = 1/N (10a) P_B(1) = 1/N (11a) P_AB(1,1) = P_A(1) * P_B(1) = 1/N*1/N = 1/N^2Probabilities in Eqs. (9a) and (10a) both tend asymptotically to zero, but Eq. (11a) correctly describes the probability of both dice coming up 1s. I can't see anything incorrect about this. Analogously, for the case of the two Prime Number Sets (9b) P_(6x1)(x) = π_(6x1)(x) / x (10b)P_(6x1)(x) = π_(6x1)(x) / x (11a) P_g=2 = P_(6x1)(x) * P_(6x+1)(x) = [π_(6x1)(x)+π_(6x1)/x)]/x^2Probabilities in Eqs. (9b) and (10b) both tend asymptotically to zero, but Eq. (11b) correctly describes the probability of both numbers being Prime if, as you correctly point out, the probabilities are independent (Assumption #2). I can't see anything incorrect about this either. Could you please clarify what is incorrect about this? (Regarding Assumption #1  independence of the probabilities in (9b) and (10b)  there is no proof presented here that the probabilities are independent, but, like Assumption #1, it is not impossible for Assumption #2 to be correct.) C) If the assumptions and considerations of the original question and the questions above are incorrect, can you please explain the close agreement between the predictions of these considerations and the actual outcomes. Specifically with respect to the attached figures. 
20201118, 04:36  #5 
"Curtis"
Feb 2005
Riverside, CA
17×271 Posts 
Well, I think both your numbered assumptions are incorrect. I think the probabilities are *not* independent, and if they're not the rest of your reasoning doesn't achieve anything.
This article may help it shows independence is not a valid assumption between consecutive primes. https://www.quantamagazine.org/mathe...racy20160313/ 
20201118, 22:45  #6 
Nov 2020
100_{2} Posts 
You're right that if assumptions are incorrect, we can't achieve accurate results. So, I suggest that we test the assumptions and see how well the predicted results match actual results. If the assumptions are incorrect, then the agreement between the predictions and results will be poor and your opinion will be validated.
Assumption #1: π_(6x1)(x) = π_(6x+1)(x) Empirically n/log(n) is a lower limit on Prime Counting Function in that it underestimates the number of prime numbers. So, (1) π(n) ≥ n/log(n)and then based on Assumption #1 (2) π_(6x1)(x)/x ≥ 3 / log(6x+1) (3) π_(6x+1)(x)/x ≥ 3 / log(6x+1)Figure 2a and 2b from the last post show empirically that Assumption #1 is pretty accurate. Not only that, but Eq. (2) and (3) are also accurate, but could be improved if we used a more accurate estimator of π(n) than n/log(n) (but we can stick with n/log(n) as there is a proof of PNT using this function). Assumption #2: P_(6x1)(x) and P_(6x+1)(x) are independent Twin Primes will occur when 6x1 and 6x+1 are both prime. Using Assumption 2 the probability of finding Twin Primes for a given value of x is the product of P_(6x1)(x) and P_(6x+1)(x). So, (4) P_g=2 = P_(6x1)(x) * P_(6x+1)(x) = π_(6x1)(x)/x * π_(6x+1)(x)/xand substituting (2) and (3) into (4) (5) P_g=2 ≥ 3 / log(6x+1) * 3 / log(6x+1)This can be extended for Cousin Primes, Sexy Primes, etc. as described in the first post and P_g can be evaluated for various values of g and x. Figure 3 from the last post shows that Assumption #2 is also pretty accurate as it shows excellent agreement between predicted and actual values in both trend and values. The empirical evidence doesn't show that Assumptions #1 & #2 result in poor agreement between predicted and actual results. On the contrary, predictions based on these assumptions seem pretty accurate. So, to persist in thinking that the assumptions are incorrect requires some explanation to resolve the paradox as to why the assumptions, despite being incorrect, make accurate predictions. I hope to hear that you have a solution to this paradox. Otherwise, I would suggest that the assumptions need to be treated as correct. 
20201119, 00:14  #7 
"Curtis"
Feb 2005
Riverside, CA
10777_{8} Posts 
Did you read the article?
Or perhaps I'm missing what you are trying to achieve; it appears you're rephrasing longknown heuristics and trying to then claim that you've reached a new conclusion because you phrase them all in terms of primes being 1mod6 or 5mod6. I now fear we're talking right past each other maybe you just like that the heuristics work nicely, while I'm trying to point out that you are no closer to a demonstration of infinite numbers of twin primes than anyone else who reads the heuristic formula for the number of twin primes below a particular bound. "my heuristics work for these small numbers, see?" doesn't advance a claim. Maybe you haven't looked at bigenough numbers; maybe, like in the article, the correlations between consecutive primes are subtler than you realize. 
20201119, 18:09  #8  
Dec 2008
you know...around...
2^{2}×5×31 Posts 
Quote:
If I had done some of those calculations and found out how primes (in base 10) with the same end digits seemed to "repel each other", I would've done some quick modular restriction maths and shaken it off as something rather trivial and not really worth mentioning, and I'm almost dead sure many other people have done so before. Working out the details of course is a task for those more in the know than me. Actually I stumbled upon this kind of problem myself a couple of years ago (for my constant y such that floor(p#*y) is always prime), and found a similar solution. 

20201119, 22:22  #9 
Nov 2020
2^{2} Posts 
Yup, read the article. It was interesting.
I just like that the heuristics work nicely. Nothing beyond that except to see how closely this method matches reality when the conjecture is formulated this way. If you want to predict the number of "sexy" primes between 0 and 60,001. How would you do it? 
20201119, 22:42  #10 
"Curtis"
Feb 2005
Riverside, CA
17·271 Posts 
Yep, I misunderstood you. I thought you were claiming you had created something more rigorous than the twin primes conjecture to "prove" the infinitude of twin primes.
I don't even know what a sexy prime is, so I would have quite a lot of work to do to enumerate them! 
20201120, 20:13  #11  
May 2020
2×13 Posts 
Quote:
(primes with a gap of six: p and p+6 are prime) 

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