[QUOTE=Awojobi;520807]Maybe your switch is due to the fact that you are frustrated that your computer simulations backed up my proof within its limits i.e. you couldn't model for sufficiently large N. Your model went for a maximum N of slightly over 1 billion which is minuscule. However, it achieved 100% and number of cycles required was less than the number of primes you used in the 1st quarter. In answer to some of the questions you raised, if only 1 integer was left in the 2nd quarter, the next cycle would have produced it. [/QUOTE]

You are wrong. I am fine if the simulation backs up your proof. I will just tell you to prove that my simulation will back up your proof for large N. My frustration is that you came up with excuses that your proof is correct despite repeatedly finding multiple counterexamples (only 1 is enough to invalidate your proof). Your excuse is that N is not large enough, and you did not justify why when N is larger, your assumption is likely to work. I’m sorry that I lost my patience on you over this. Even a genuine attempt or a heuristic would be ok.

Regarding the last sentence, please explain why the next cycle would have produced the 1 integer. This involves proving that something is prime. Please why the relevant number is prime.

Furthermore, apart from my simulations, you are generally not receptive to any other feedback by other people in the forum, including me, ignoring most of the criticism regarding the proof. I tried to help but you generally do not listen. If this is the case, who would like to help you? It is like helping you for nothing. I suspect that this is why many users decided to ignore you in the thread which you sent a failed proof of the Twin Prime Conjecture.

Sorry for calling you a crank and a troll and my tone was not very nice in post #132. However, this is what some serious mathematicians in this forum call you. axn referred to you as a troll and RDS sent a PM to me calling you a crank.

[QUOTE=Uncwilly;520808]Go here and rate your score:
[url]https://primes.utm.edu/notes/crackpot.html[/url][/QUOTE]

I strongly recommend you go there to understand what not to do in the field of mathematics.

[QUOTE=Awojobi;520715] !00% achieved (SIC!)[/QUOTE]
Correct!
[QUOTE=Awojobi;520715]2 out of 3 predictions of my proof[/QUOTE]
You even don't realize how ridiculously that sounds... You make a proof sound like a prophecy or something... :davar55:

As somebody already said, the fact that some fancy "simulation" leaves the possibility to crush such a "proof", says everything about the value of the "proof".

And don't worry, we do not believe that you are a crank, as you said,
[QUOTE=Awojobi;520807]a crank is someone who uses secondary school maths to prove an age old conjecture[/QUOTE]
but for now, you are not a crank. You didn't prove anything and you didn't use any math. :rofl:

[QUOTE=Awojobi;520715]!00% achieved with cycles less than the number of primes in the first quarter even though some latter cycles produced no knew arithmetic means. So 2 out of 3 predictions of my proof have occurred.[/QUOTE]

On second thought, turns out that "some latter cycles produced no new arithmetic means" seems to be irrelevant to the proof. I don't see how you assumed that property in the equation. Your equation can predict 0.25 more arithmetic means covered in the next cycle, hence it is possible that the actual number of arithmetic means produced in the next cycle is 0 (even if the accuracy of the equation is 99.99%). Therefore, my previous simulation actually completely backed up the 'proof'.

I did another simulation to determine whether the bounds used in the equation are really upper bounds. I used c0 = round(ln(N)^2). I used the variable name c0 to denote the initial intended number of cycles as indicated in your proof (to avoid confusion with c later on in the post).

Here is an overview of the results.

2^12: 69 out of 69 cycles done, 1023 out of 1024 (99.9023438%) covered, 1 counterexample arithmetic mean found, number of cycle counterexamples found: 44

For larger values of N (up to 2^32), I got 100% coverage. Hence, this is likely due to 2^12 being too small.

Here are the results for the first part of the inequality:
2^15: c=50, total arithmetic means produced=81386, c*(N/2)/ln(N/2)≈84418 (ratio: 0.96408)
2^20: c=120, total arithmetic means produced=4680321, c*(N/2)/ln(N/2)≈4777185 (ratio: 0.97972)
2^25: c=216, total arithmetic means produced=214657959, c*(N/2)/ln(N/2)≈217839657 (ratio: 0.98539)
2^30: c=294, total arithmetic means produced=7753891780, c*(N/2)/ln(N/2)≈7852243265 (ratio: 0.98747)
2^32: c=302, total arithmetic means produced=29809243278, c*(N/2)/ln(N/2)≈30182113898 (ratio: 0.98765)

That's quite a tight upper bound, but you still need to rigorously show why this is really an upper bound for all N. See post #143 from this thread.

[QUOTE=Awojobi;520807]By the way, you said you wanted to prove me wrong on something. Still waiting.[/QUOTE]

Here it is.

Here are the results for the second part of the equation:
2^12: c=69, number of repetitions=16817, 2*(c^2)*ln(N/4)≈66001 (ratio: 0.2548)
2^13: c=37, number of repetitions=15386, 2*(c^2)*ln(N/4)≈20876 (ratio: 0.73701)
2^14: c=59, number of repetitions=47758, 2*(c^2)*ln(N/4)≈57908 (ratio: 0.82472)
2^15: c=50, number of repetitions=73194, 2*(c^2)*ln(N/4)≈45055 (ratio: 1.6246)
2^20: c=120, number of repetitions=4418177, 2*(c^2)*ln(N/4)≈359327 (ratio: 12.296)
2^25: c=216, number of repetitions=206269351, 2*(c^2)*ln(N/4)≈1487616 (ratio: 138.66)
2^30: c=294, number of repetitions=7485456324, 2*(c^2)*ln(N/4)≈3355121 (ratio: 2231.1)
2^32: c=302, number of repetitions=28735501454, 2*(c^2)*ln(N/4)≈3793068 (ratio: 7575.8)

Clearly, 2*(c^2)*ln(N/4) does not look like an upper bound for N≥2^15, hence causing the value of c obtained from the equation to be very suspicious.

Detailed summary: goldbach11_summary.txt
Detailed results: goldbach11_results.txt

[QUOTE=Awojobi;520807]Your next question concerning how I got the expression for the upper bound can be seen if I use an example. If 10 green bottles are in a row spaced approximately equally apart, is it not clear that an upper bound for the number of a particular spacing, say the spacing between the 2nd and 5th green bottles is 10 green bottles? Now, the total number of repetitions form an arithmetic series added together and this is where the c^2 shows up.[/QUOTE]

I see the flaw in logic here. You might have assumed that the number of new arithmetic means produced each cycle is constant, rather than at least one, until all arithmetic means are produced. Based on my previous simulations, the number of new arithmetic means produced in one cycle generally decreases (barring a few spikes) as the number of cycles completed increases.

Additional Note: Hopefully, I can be back to my original helpful self but please help yourself.

How did you work out your total number of repetitions?. The values are ginormous.

[QUOTE=Awojobi;521051]How did you work out your total number of repetitions?. The values are ginormous.[/QUOTE]

My guess is, the poster's code actually ran it that many times. Yes, billions and billions of repetitions, just for you. (Did you say thanks? :grin:)

More and more simulations, but not a proof in sight. :loco:

[QUOTE=CRGreathouse;521066]My guess is, the poster's code actually ran it that many times. Yes, billions and billions of repetitions, just for you. (Did you say thanks? :grin:)[/QUOTE]

Yes, that’s right! That many times! I used the OP’s direct method except that the arithmetic means are not filtered out after they are covered.

[QUOTE=retina;521068]More and more simulations, but not a proof in sight. :loco:[/QUOTE]

Simulations can be used to disprove a hypothesis by finding counterexamples, but they cannot form a proof unless all the cases are considered.

[QUOTE=2M215856352p1;521069]Simulations can be used to disprove a hypothesis by finding counterexamples, but they cannot form a proof unless all the cases are considered.[/QUOTE]

[url]https://en.wikipedia.org/wiki/Hypercomputation[/url]
[url]https://en.wikipedia.org/wiki/Supertask[/url]

[QUOTE=CRGreathouse;521099][url]https://en.wikipedia.org/wiki/Hypercomputation[/url]
[url]https://en.wikipedia.org/wiki/Supertask[/url][/QUOTE]

Thanks for the references. I wasn’t aware of them.