20211118, 12:32  #23  
Jun 2003
Oxford, UK
2^{5}×3^{2}×7 Posts 
Quote:
For N = 10 and N=34 , c1=0 

20211118, 17:21  #24 
Jun 2003
Oxford, UK
2^{5}·3^{2}·7 Posts 
Coming back to Goldbach.
It is proved that there is a prime between N/2 and N (Bertrand Chebyshev), with N even Using this prime residue technique, is it correct to paraphrase Goldbach to say it is conjected that there exists at least 1 prime p(i) between N/2 to N where N mod p(i) == a prime, p(ii) with p(ii) =< p(i) ? I think this is equivalent to stating that there are 2 primes which add up to N. It should be possible to list N in terms of the prime mod primes discovered between N/2 and N (c2) to find out the factors of N where the count of c2 is very low compared to the size of N. I have done this to N=200000, and the lowest are as follows: My interpretation of the data is that the lowest number of prime pairs that add up to N are predominately found when N factors as 2^x.prime, x from upwards. In this range, there are no small factors other than 2. In this example, the smallest factor is 263. Code:
N p >n/2 c2 ratio factors 198728 8347 1008 0.12076195 2.2.2.24841 182012 7692 929 0.120774831 2.2.45503 144128 6217 751 0.120797812 2.2.2.2.2.2.2.2.563 181904 7685 929 0.120884841 2.2.2.2.11369 199328 8366 1013 0.121085345 2.2.2.2.2.6229 196754 8272 1004 0.121373308 2.98377 191834 8073 980 0.121392295 2.95917 195368 8215 998 0.121485088 2.2.2.2.4421 158428 6791 826 0.121631571 2.2.39607 198916 8353 1016 0.121632946 2.2.223.223 167438 7127 867 0.121650063 2.83719 173948 7381 898 0.121663731 2.2.43487 191024 8039 979 0.121781316 2.2.2.2.11939 188882 7964 970 0.121798091 2.94441 168068 7151 871 0.121801147 2.2.42017 179618 7610 927 0.121813403 2.89809 189872 8004 976 0.12193903 2.2.2.2.11867 196628 8266 1008 0.121945318 2.2.49157 176678 7493 914 0.121980515 2.88339 183916 7771 948 0.121992022 2.2.45979 174362 7400 903 0.122027027 2.87181 176024 7448 909 0.122046187 2.2.2.22003 199846 8387 1024 0.122093716 2.99923 189578 7993 976 0.122106843 2.94789 188834 7963 973 0.122190129 2.263.359 Code:
N p >n/2 c2 ratio factors 2310 152 114 0.7500000000 2.3.5.7.11 660 54 41 0.7592592593 2.2.3.5.11 270 25 19 0.7600000000 2.3.3.3.5 78 9 7 0.7777777778 2.3.13 96 9 7 0.7777777778 2.2.2.2.2.3 300 27 21 0.7777777778 2.2.3.5.5 840 65 51 0.7846153846 2.2.2.3.5.7 144 14 11 0.7857142857 2.2.2.2.3.3 34 5 4 0.8000000000 2.17 42 5 4 0.8000000000 2.3.7 84 10 8 0.8000000000 2.2.3.7 168 16 13 0.8125000000 2.2.2.3.7 240 22 18 0.8181818182 2.2.2.2.3.5 390 33 27 0.8181818182 2.3.5.13 180 17 14 0.8235294118 2.2.3.3.5 48 6 5 0.8333333333 2.2.2.2.3 126 12 10 0.8333333333 2.3.3.7 630 49 41 0.8367346939 2.3.3.5.7 60 7 6 0.8571428571 2.2.3.5 66 7 6 0.8571428571 2.3.11 150 14 12 0.8571428571 2.3.5.5 330 28 24 0.8571428571 2.3.5.11 420 35 30 0.8571428571 2.2.3.5.7 90 10 9 0.9000000000 2.3.3.5 120 13 12 0.9230769231 2.2.2.3.5 10 2 2 1.0000000000 2.5 16 2 2 1.0000000000 2.2.2.2 36 4 4 1.0000000000 2.2.3.3 210 19 19 1.0000000000 2.3.5.7 Last fiddled with by robert44444uk on 20211118 at 17:26 
20211118, 20:25  #25 
Feb 2017
Nowhere
12357_{8} Posts 
The Prime Number Theorem (PNT) gives the asymptotic estimate N/log(n)  (N/2)/log(N/2) which is approximately (1/2)*N/log(N)) for the number of primes between N/2 and N.
The HardyLittlewood conjectural asymptotic formula for the number of representations of N as the sum of two odd primes is 2*C_{2}*N/log^{2}(N)* where the product is over the odd prime factors of N and C_{2} = 0.66016... is the "twin primes constant." This is of smaller order than the number of primes in (N/2, N) by a factor of log(N). Your observation about small prime factors is perfectly in line with the last factor in the HardyLittlewood formula, a product over the odd prime factors of N. That factor is biggest when N has a lot of small prime factors, and is smallest if N is a power of 2 (when the factor is the empty product, or 1). For N = 2^{x}*p, p an odd prime, the factor is (p1)/(p2). I'm still making no further progress with the residues N%p for p < N/2. As we have already seen, you can carve out intervals N/(k+1) < p < N/k for which N%p is guaranteed to be composite, but that's not nearly enough to get the count of prime residues down to something of order N/log^{2}(N). EDIT: I did come up with a handwaving argument that for large even N the number of prime residues N%p, 2 < p < N/2, is of order N/log^{2}(N). By the Prime Number Theorem, the number of primes x is asymptotic to x/log(x). So if p is a "large" prime, about 1/log(p) of the numbers < p are prime. Now for sqrt(N) < p < N/2, log(p) only varies between (1/2)*log(N) and log(N)  log(2). So, assuming, by wave of hand, that for a given p < N/2 not dividing N the residues 1 to p1 N%p are all "equally likely," we guess that the number of prime residues for sqrt(N) < p < N/2 is of order 1/log(N)*N/log(N) or N/log^{2}(N). There are obviously no more than sqrt(N) primes p < sqrt(N), and sqrt(N) is of lower order than N/log^{2}(N). Last fiddled with by Dr Sardonicus on 20211119 at 20:03 Reason: as indicated 
20211120, 13:55  #26  
Jun 2003
Oxford, UK
2^{5}×3^{2}×7 Posts 
Quote:
The data below was obtained by running my program for about a week. Quote:
Last fiddled with by robert44444uk on 20211120 at 14:00 

20211120, 14:55  #27  
Feb 2017
Nowhere
23×233 Posts 
Quote:


20211121, 11:00  #28 
Jun 2003
Oxford, UK
2^{5}×3^{2}×7 Posts 
Yes, the last of these is the lowest ratio, N1 prime, seen once small N are taken out of the consideration.
Last fiddled with by robert44444uk on 20211121 at 11:16 
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