[QUOTE=Awojobi;520664]I really appreciate what you are doing concerning my proof. I looked at your largest N i.e. 2^30 and saw that each of the 39 cycles produced new arithmetic means, which is what my proof expects for sufficiently large N, even though all possible cycles needed to be exhausted and still 100% wasn't achieved. I still insist that if N is sufficiently large, all N/4 integers in the 2nd quarter will be produced before the last prime number in the first quarter is reached. Is there no way you can get results for N = 10^30 ?[/QUOTE]

"I still insist" = not a proof. Not even a little bit.

I mean, I insist Goldbach's Conjecture is true, and nobody yet has shown me any counterexample to my insistence. My proof is just as good as yours, presently.

[QUOTE=Awojobi;520664]Is there no way you can get results for N = 10^30 ?[/QUOTE]

Do you have any idea how large that is? :confused:

If every person on the planet had a 100-core, 10 GHz computer it would take them years just to *list* the numbers, let alone do any calculations with them.

[QUOTE=retina;520032]Let's see. On average it takes three throws of a coin to get both a head and a tail as outcomes (obviously some outcomes appear more than once). So if we throw a coin ten times we should be absolutely guaranteed to get both a head and a tail, always. Right? [B]Or if you disagree then perhaps we throw it one hundred times, that should guarantee it. Right?[/B] Or if you still disagree, how many times will it take?[/QUOTE]
This has been posted quite a while ago here.
And it is a simple and relevant demonstration of the failure of "the proof".

"The simulations and/or cycles" = "throws of a coin".
"c^2 etc (large number)" = "we have very, very many throws of a coin".
Proof (without quotes) = "we are guaranteed to have a head and tail".

If one reads this with understanding - they will see.

I really appreciate your computer simulations and the results its churning out. I am mainly interested in 2^30 which gives 99.8% and new arithmetic means are formed for each cycle. Please can you try another simulation and double the number of primes in the first quarter one more time for N = 2^30 only as this is the only result I have been concentrating on. Many thanks.

Are you just going to keep running simulations until everyone dies of old age?

Simulations prove nothing. Simulations serve to bolster faith that something is true, but they won't ever prove it.

Show a valid mathematical [u]proof[/u].

If it pleases the court, I will proceed with the demonstration how probabilistic proofs fail.
Are you ready? Ok, lets'go.

A restatement of the [URL="https://mersenneforum.org/showthread.php?p=520704#post520704"]GOLDBACH'S CONJECTURE[/URL]
[YOUTUBE]rQOBwQbssgo[/YOUTUBE]
and of course, the proof --
[YOUTUBE]OsH_JszzicE[/YOUTUBE]

[QUOTE=Awojobi;520684]I really appreciate your computer simulations and the results its churning out. I am mainly interested in 2^30 which gives 99.8% and new arithmetic means are formed for each cycle. Please can you try another simulation and double the number of primes in the first quarter one more time for N = 2^30 only as this is the only result I have been concentrating on. Many thanks.[/QUOTE]

I have done a simulation with c = round(a*ln(N/4)) for a = 2,4,8,12 and 16.
Here is a summary of my results:

a=2: 39 out of 39 cycles done, 258234632 out of 268435456 (96.1998969%) covered, 10200824 counterexample arithmetic means found, number of cycle counterexamples found: 0
a=4: 78 out of 78 cycles done, 267898335 out of 268435456 (99.7999068%) covered, 537121 counterexample arithmetic means found, number of cycle counterexamples found: 0
a=8: 155 out of 155 cycles done, 268433857 out of 268435456 (99.9994043%) covered, 1599 counterexample arithmetic means found, number of cycle counterexamples found: 0
a=12: 233 out of 233 cycles done, 268435451 out of 268435456 (99.9999981%) covered, 5 counterexample arithmetic means found, number of cycle counterexamples found: 10
a=16: 258 out of 311 cycles done, 268435456 out of 268435456 (100.0000000%) covered, 0 counterexample arithmetic means found, number of cycle counterexamples found: 27

Note: For a = 16, only 258 cycles are needed to cover all the arithmetic means. There exist 27 cycles such that no new arithmetic means are produced before all possible arithmetic means are produced.

Results for a=2,4,8,12 and 16: goldbach10_results.txt
Results for a=4 (doubling the original value of c): awojobi_results.txt

Many thanks again for the latest simulations. !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. So I stand by my proof if N is large enough. The only thing I will change in my proof is increasing the proportion of primes chosen in the 1st quarter to maybe the proportion that produced 100% just for the sake of testing with a computer. I know I might now be becoming a pain, but hey, this is a maths forum that discusses and explores maths concepts. Please can you simulate N = 2^40 with the same proportion of primes in the first quarter that produced 100%. or will this crash your computer?

[QUOTE=Awojobi;520715]Many thanks again for the latest simulations. !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. So I stand by my proof if N is large enough. The only thing I will change in my proof is increasing the proportion of primes chosen in the 1st quarter to maybe the proportion that produced 100% just for the sake of testing with a computer. I know I might now be becoming a pain, but hey, this is a maths forum that discusses and explores maths concepts. Please can you simulate N = 2^40 with the same proportion of primes in the first quarter that produced 100%. or will this crash your computer?[/QUOTE]

That will either crash my computer by taking a lot of memory or dramatically slow down my program. It takes hours to days to finish and I don’t really have the time. No matter how large N is, I conjecture that I can only get 2 out of 3. Since you stand by your proof, why not just prove that your equation works rigorously and/or prove my conjecture wrong? Once you are done, please show me your updated proof. All the best.

As for the proportion, c=(ln(N/4))^2 should be ok. I tried with smaller values of N and a tends to increase as N increases.

Note: your value of c should match the empirical data I had for my simulations, at least in the order of growth, otherwise your equation is probably wrong.

I don't need to rewrite my proof. Nothing more needs to be added. Your computer simulation backs up my proof i.e. 100% for 2^30 and number of cycles is less than the number of chosen primes in the first quarter. The fact that the some cycles produced no new arithmetic means is due to the fact that N = slightly over 1 billion, is not sufficiently large. Some other values of N that you used produced new arithmetic means for each cycle though. Consider the equation in my proof i.e. c(N/2)/log e (N/2) – 2(c^2)log e (N/4) = N/4 and consider sufficiently large N = 10^30, say. Log e (10^30) = 230 which is so tiny in comparison to 10^30. Substitute these values into the equation. Is it not absolutely clear that since c(N/2)/log e (N/2) is way much greater than 2(c^2)log e (N/4), then cycle after cycle will produce N/4 integers in the 2nd quarter eventually?

[QUOTE=Awojobi;520753]I don't need to rewrite my proof. Nothing more needs to be added. Your computer simulation backs up my proof i.e. 100% for 2^30 and number of cycles is less than the number of chosen primes in the first quarter. The fact that the some cycles produced no new arithmetic means is due to the fact that N = slightly over 1 billion, is not sufficiently large. Some other values of N that you used produced new arithmetic means for each cycle though. Consider the equation in my proof i.e. c(N/2)/log e (N/2) – 2(c^2)log e (N/4) = N/4 and consider sufficiently large N = 10^30, say. Log e (10^30) = 230 which is so tiny in comparison to 10^30. Substitute these values into the equation. Is it not absolutely clear that since c(N/2)/log e (N/2) is way much greater than 2(c^2)log e (N/4), then cycle after cycle will produce N/4 integers in the 2nd quarter eventually?[/QUOTE]

Yes, but the bounds are randomly generated out of nowhere.

I predict that your (2c^2)(ln(N/4)) bound is wrong. I can do a simulation to prove/disprove that. That will be the last or second last one, since whatever I throw to you will not convince you anyway.

For my simulation, I either got new arithmetic means for each cycle or 100% coverage, sometimes neither, but never both at the same time, which [B]completely invalidates your proof[/B].

I am now 200% convinced that Awojobi is a crank. And yes, I am also a crank to some extent.

Awojobi might not be satisfied with the value of N I reach, no matter how high I get for N, given that he/she is probably such a noob who understands nothing about proof to all serious mathematicians. Since N=2^30 does not satisfy Awojobi, he/she demands 10^30. The likelihood is that regardless of what N he/she picks, it will not work: see post #104 by axn. [B]At the tail, the likelihood is that there exists a cycle where no new arithmetic means are produced, imagine when only 1 arithmetic mean is left to be covered. How can you be so sure that the cycle will definitely cover the last arithmetic mean? The tail just before all arithmetic means are produced always exists; it cannot be truncated somewhere, otherwise the Goldbach conjecture can never be proven.[/B] Suppose I actually did the simulation to 10^30, my result will probably be consistent with the one at 2^30 given axn’s reasoning. Probably due to his/her insistence on the correctness of his/her proof, he/she may demand 10^300. And the cycle repeats, with N faster than exponentially higher each time. And when I can’t get any higher, he/she does the handwaving by dismissing the result, saying that N is too small.

P.S. RDS has sent me a PM asking me to stop arguing with Awojobi since it is a waste of resources and I can never convince him/her. I strongly agree with him on that, but I still wish to share my pieces of evidence. I am aware that I can never win a fight against RDS.