[QUOTE=Awojobi;520589]Those who read every word of my proof with understanding will see why Goldbach's conjecture is true. I have selected primes that are well spaced apart and therefore enables me to get a much smaller upper bound of repetitions than my previous failed proof. So every cycle will produce new arithmetic means until all N/4 integers are produced.[/QUOTE]

1. Let's now go back to the fundamentals. The last sentence of the above quote is the claim, correct?

2. Now, let's break up the claim. Let statement A be the statement where each of the c = round(2*ln(N/4)) cycles will produce new (at least 1 yet to be covered) arithmetic means (in the second quarter). Let statement B be the statement where when all c cycles are done, all N/4 arithmetic means are produced.

3. Hence, the claim is equivalent to both statement A and statement B being true.

4. I am going to generate empirical data by simulating your proof. Just one counterexample found during the simulation is enough to prove that your proof is incorrect, even though it is overkill, as it was pointed out.

5. However, if the empirical data supports your proof, it does not mean that your proof is correct, but rather the claim is probably correct. You need to prove your claim rigorously since the claim is an assumption rather than an proven statement.

6. The probabilistic simulation that I did resulted in counterexamples, showing only about 96.3% of all integers in the second quarter (coverage) produced for large N. This shows that there exist at least one counterexample (since I did the simulation with replacement), hence proving statement B wrong.

7. Since statement B is wrong, it is impossible that both statements A and B are true, hence disproving your claim.

8. However, the simulation did not test the truth of statement A since it did not check for the presence of new arithmetic means produced each cycle.

[QUOTE=Awojobi;520573]Your latest simulation has not disproved my proof. I don't believe your probabilistic simulation shows me anything. In fact it doesn't model my proof. Computer modelling of my proof must compare results of ALL previous cycles with the current cycle. Yours does not since you have said you didn't take each cycle into consideration. I am now more convinced than ever that my proof is correct. Can I ask if the simulations you did prior to your latest simulation actually compared results of ALL previous cycles with the current cycle. If your answer is 'no' then maybe the biggest value of N you used before ( approximately 1billion) is large enough to confirm my proof if a computer program is written the way I describe.[/QUOTE]

9. In the quoted statement above, you wanted me to test the truth of statement A. Even though the truth of statement A is irrelevant to whether your claim is correct, you wanted me to redo the simulation and hopefully, both A and B miraculously become true, validating your 'proof'.

10. Since the coverage was near 100%, you were more confident that your proof was right since by changing this part of the program, the coverage looks likely to become 100%.

11. I repeated the simulation with your suggestion put into place and got the following results:
2^ 7: 7 cycles, 30 out of 32 (93.8%) covered, 2 counterexample arithmetic mean found, number of cycle counterexamples: 2, smallest counterexample cycle number = 5
2^ 8: 8 cycles, 62 out of 64 (96.9%) covered, 2 counterexample arithmetic mean found, number of cycle counterexamples: 2, smallest counterexample cycle number = 6
2^ 9: 10 cycles, 124 out of 128 (96.9%) covered, 4 counterexample arithmetic mean found, number of cycle counterexamples: 1, smallest counterexample cycle number = 9
2^10: 11 cycles, 252 out of 256 (98.4%) covered, 4 counterexample arithmetic mean found, number of cycle counterexamples: 0
2^15: 18 cycles, 7948 out of 8192 (97.0%) covered, 244 counterexample arithmetic mean found, number of cycle counterexamples: 0
2^20: 25 cycles, 251987 out of 262144 (96.1%) covered, 10157 counterexample arithmetic mean found, number of cycle counterexamples: 0
2^25: 32 cycles, 8084942 out of 8388608 (96.4%) covered, 303666 counterexample arithmetic mean found, number of cycle counterexamples: 0
2^30: 39 cycles, 258234632 out of 268435456 (96.2%) covered, 10200824 counterexample arithmetic mean found, number of cycle counterexamples: 0

12. According to the new simulation, statement A turns out to be probably true for large N, but statement B is definitely false (when N=2^30, there exists over 10.2 million counterexamples, which means over 10.2 million ways to disprove the statement), hence disproving your claim.

Please let me know the numerical index of the points you disagree with, if any.

The major reason I retracted my initial proof was because I overlooked the fact that a term in the equation was supposed to be c^2 and not c. This meant that the number of repetitions could be larger than the number of arithmetic means which would immediately put holes in my proof. I ingeniously decided to limit the number of primes in the first quarter because I know that not all the primes in the first quarter are need to produce all N/4 integers in the second quarter. No one has properly disproved my proof because you can see that for sufficiently large N, the first term in my equation will always be greater than the second term of my equation. Everyone who is experienced in prime number theory knows that one doesn't need exact values to prove some conjectures. Approximate values would suffice e.g. the proof that the number of primes less than N is approximately N/log e N. No one has come up with an exact value of this number. No formula exists. Basically I am fully convinced that if a computer program is written the way I described in a previous post, it will show that first of all every cycle will produce new arithmetic means until all N/4 integers are produced and secondly this will be done quickly i.e. the number of cycles will not be much. For instance the logarithm of a very large number is not so large, case in point, log e 10^100 is just 230. Unless someone can come up with a good mathematical reason that my proof is false or come up with undeniable counterexamples using convincing computer simulations (by showing that there are cycles that don't produce new arithmetic means), I stand by my proof.

[QUOTE=Awojobi;520632]The major reason I retracted my initial proof was because I overlooked the fact that a term in the equation was supposed to be c^2 and not c. This meant that the number of repetitions could be larger than the number of arithmetic means which would immediately put holes in my proof. I ingeniously decided to limit the number of primes in the first quarter because I know that not all the primes in the first quarter are need to produce all N/4 integers in the second quarter. No one has properly disproved my proof because you can see that for sufficiently large N, the first term in my equation will always be greater than the second term of my equation. Everyone who is experienced in prime number theory knows that one doesn't need exact values to prove some conjectures. Approximate values would suffice e.g. the proof that the number of primes less than N is approximately N/log e N. No one has come up with an exact value of this number. No formula exists. Basically I am fully convinced that if a computer program is written the way I described in a previous post, it will show that first of all every cycle will produce new arithmetic means until all N/4 integers are produced and secondly this will be done quickly i.e. the number of cycles will not be much. For instance the logarithm of a very large number is not so large, case in point, log e 10^100 is just 230. Unless someone can come up with a good mathematical reason that my proof is false or come up with undeniable counterexamples using convincing computer simulations (by showing that there are cycles that don't produce new arithmetic means), I stand by my proof.[/QUOTE]

You are looking for statement A, I see. Now, I will have to vary c accordingly to cover 100% of arithmetic means, is that what you mean?

Which statement did you disagree with (with reference to my previous post)?

Since you are so convinced that your proof is correct, please write a paper and submit to a reputable peer-reviewed journal. You will gain fame in the news for a short while, but I’m 99% certain that your proof will be rejected.

Is it ok if someone else independently writes another program to prove/disprove my point? Thanks in advance.

Let N=10^100+82401702

The smallest prime p such that N-p is also prime is p=27791 (3034th prime). That means any range that contains the above N (or N/2 in OP's arithmetic mean range) requires at least 3033 "cycles".

[QUOTE=2M215856352p1;520633]Is it ok if someone else independently writes another program to prove/disprove my point?[/QUOTE]
If you "sieve" the range in question with small primes starting from 3, 5, 7, ... and keep track of number of unprocessed numbers, you'll see that they keep counting down. For the most part, each cycle will hit some as-yet-unprocessed numbers. However as the number of remaining unprocessed ones keep decreasing, eventually reaching really small count (say < 100), you'll start seeing cycles that don't generate any new hits. When it reaches really small (< 10) the misses keep increasing. It is very hard to kill off that last few survivors.

[QUOTE=axn;520636]If you "sieve" the range in question with small primes starting from 3, 5, 7, ... and keep track of number of unprocessed numbers, you'll see that they keep counting down. For the most part, each cycle will hit some as-yet-unprocessed numbers. However as the number of remaining unprocessed ones keep decreasing, eventually reaching really small count (say < 100), you'll start seeing cycles that don't generate any new hits. When it reaches really small (< 10) the misses keep increasing. It is very hard to kill off that last few survivors.[/QUOTE]

That was for the old failed proof. What the OP wants is for the case of evenly spaced primes between 3 and N/4.

EDIT: The idea is still the same, but at c=2*ln(N/4), that point of ‘very hard to kill off last points’ would not have manifested yet. You will probably need c in the order of (ln(N))^2 to disprove statement A, referring to the post where I defined statements A and B. The idea works for any valid cycle permutation.

Any convincing simulation should find the first set of arithmetic means using the first prime in the first quarter. The second prime in the first quarter produces a new set of arithmetic means i.e. repetitions from the first set must be filtered out by the computer simulation. The third prime in the first quarter produces a new set of arithmetic means i.e. repetitions from the second and third sets must be filtered out by the computer simulation. This carries on until all N/4 integers in the second quarter are produced. Of course, I could learn to use MATLAB and write it myself to verify but I couldn't be bothered because I can't see any reason why I won't get the results I am looking for based on my sound equation (the first term is always bigger than the second term for large enough N).

Concerning getting my proof published in a reputable mathematics journal, obviously the moderators wouldn't even bother to read it especially when they see the title and the name of who they would consider a crank author. In their heads they wouldn't believe that such a famous unsolved mathematics problem could be solved using basic secondary school (high school) mathematics. They would just squeeze up my proof and toss it into the trash can. The main reason I am posting in a public forum is, if one of these top mathematicians use my proof in the future, I will be able to challenge them to acknowledge me.

Any convincing simulation should find the first set of arithmetic means using the first prime in the first quarter. The second prime in the first quarter produces a new set of arithmetic means i.e. repetitions from the first set must be filtered out by the computer simulation. The third prime in the first quarter produces a new set of arithmetic means i.e. repetitions from the first and second sets must be filtered out by the computer simulation. This carries on until all N/4 integers in the second quarter are produced. Of course, I could learn to use MATLAB and write it myself to verify but I couldn't be bothered because I can't see any reason why I won't get the results I am looking for based on my sound equation (the first term is always bigger than the second term for large enough N).

Concerning getting my proof published in a reputable mathematics journal, obviously the moderators wouldn't even bother to read it especially when they see the title and the name of who they would consider a crank author. In their heads they wouldn't believe that such a famous unsolved mathematics problem could be solved using basic secondary school (high school) mathematics. They would just squeeze up my proof and toss it into the trash can. The main reason I am posting in a public forum is, if one of these top mathematicians use my proof in the future, I will be able to challenge them to acknowledge me.

Just made a correction 'first and second'.

[QUOTE=2M215856352p1;520638]That was for the old failed proof. What the OP wants is for the case of evenly spaced primes between 3 and N/4.[/QUOTE]

I don't know whether we're talking at cross-purposes.

I did a program which checked a range of even numbers (in the above case, 10^100-10^100+10^8) by sieving with p=3, 5, 7, ...

Whenever I get a hit (an N where N-p is prime), I mark that location, if there was no previous hit there. Invariably, it takes much more than log(N) primes to completely cover the range. The number of hits per cycle is roughly proportional to the number of unmarked location. So as the number of unmarked locations comes down, the hits per cycle also comes down. Eventually we start getting cycles without any hits.

You could say, "well, if you increase the range you're sieving, you're more likely to hit some unmarked location". True, but also missing the point. You're merely pushing the point where the cycles start getting no hits. Larger range means you need _more_ cycles. Eventually any range will have the unmarked count dwindling down as you progress through the primes, and you will start to get no hit cycles.

[QUOTE=Awojobi;520640]Any convincing simulation should find the first set of arithmetic means using the first prime in the first quarter. The second prime in the first quarter produces a new set of arithmetic means i.e. repetitions from the first set must be filtered out by the computer simulation. The third prime in the first quarter produces a new set of arithmetic means i.e. repetitions from the second and third sets must be filtered out by the computer simulation. This carries on until all N/4 integers in the second quarter are produced. Of course, I could learn to use MATLAB and write it myself to verify but I couldn't be bothered because I can't see any reason why I won't get the results I am looking for based on my sound equation (the first term is always bigger than the second term for large enough N).
[/QUOTE]

You are just restating your point. Please explain and elaborate more. The issue I ran with my simulation is that c=2*ln(N/4) cycles are not enough to cover all the N/4 arithmetic means. If I can increase c, I will, but then there will exist a cycle such that no new arithmetic means are produced. This happens even before all arithmetic means are produced. Hence, I don’t exactly know how to make the simulation work.

My simulation does the same as what you explain, but indirectly. For each integer in the second quarter, it finds the smallest cycle number which ‘covers’ the integer. Directly following your method could be about two times slower, but manageable.

Your equation looks reasonable but your result is off based on my simulation. And this time, you chose to defend your proof.