Can you explain more about how your simulation does exactly what I want the simulation to do. I don't understand what you mean when you talk about smallest cycle number. How does this mimic what I want the simulation to do?

[QUOTE=Awojobi;520648]Can you explain more about how your simulation does exactly what I want the simulation to do. I don't understand what you mean when you talk about smallest cycle number. How does this mimic what I want the simulation to do?[/QUOTE]

Thanks for pointing that out. Sorry for not defining it carefully. The cycle numbers are sorted in increasing prime order. If the prime numbers are {3,11,17,23,29}, cycle 1 would correspond to the prime 3, cycle 2 will correspond to the prime 11 and so on.

Instead of thinking in the cycle by cycle, I am doing it integer by integer. On the explanation about why these two can be made equivalent, observe that if you go from the smallest to the largest cycle number, the time when the integer is covered is at the smallest possible cycle number. If there exist a larger cycle number which can also cover the integer, the integer would have been filtered out by then. If you are still not convinced, I can do the direct method.

Please do the direct method i.e. the way I describe the simulation in previous posts. I will prefer that your output results show how many new arithmetic means are produced for each cycle until N/4 integers are produced in the second quarter. I expect to see that no cycle will not produce any new arithmetic means before N/4 integers are produced in the second quarter. Many thanks.

Here are my results. My intention is to use the direct method proposed by the OP.

Can someone verify that the program is actually working as intended? Thanks in advance. The program is goldbach9.txt and the results file is goldbach9_results.txt.

Note: goldbach9.txt is actually a C++ file, please change the file extension to .cpp before running the program.

Don't stop at OP's limit. Keep going till everything is covered. He needs to see that.

Can you please translate your results in a table e.g. 1st cycle shows how many arithmetic means produced, 2nd cycle shows how many new arithmetic means produced, 3rd cycle shows how many new arithmetic means produced etc.

Okay I can read it now so ignore previous.

Sorry, I can’t do so now. I will get back to axn’s request when I am done/face any issues.

EDIT: OP retracts formatting request.

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 ?

On second thoughts, can you not increase N but double the number of primes used in the first quarter instead. Let's see what happens. Many thanks.

Here are the results.

Note: I took double as c=round(4*ln(N/4)).