20180303, 20:12  #1 
Feb 2018
2^{2}×3^{2} Posts 
Counting Goldbach Prime Pairs Up To...
Goldbach Prime Pairs Up To 8282 = (91squared+1), both numbers on number line (1+n30).
Others from (13+n30) + (19+n30) not counted. (Primes 7 to 61, each minus 2, multiplied together) divided by (Primes 7 to 61, multiplied together) × ((Primes 67 to 89(91 isn't prime), each minus 1, multiplied together) divided by (Primes 67 to 89) × (82821)÷60) = 28.2098. There are 28 pairs of prime numbers on number line (1 + n30) that when added = 8282 Or, (5/7 × 9/11 × 11/13....×(prime n1 minus2)/(prime n)) × (66/67 × 70/71....×(prime n minus1)/prime n) × (N  1)/60 which equals 0.22126 × 0.92387 × 138 = 28.20980 N = 8282. n = prime preceding square root of(N  1)/60 which is 89. n1 = prime preceding square root of (N  1)/120 which is 61. Right but wrong? 
20180303, 21:02  #2  
"Forget I exist"
Jul 2009
Dumbassville
2^{4}×3×5^{2}×7 Posts 
Quote:
Oh , and n=11 not 89 and n1= 7 based on those definitions. Last fiddled with by science_man_88 on 20180303 at 21:40 

20180303, 22:45  #3 
Aug 2006
5987_{10} Posts 
Thank you for attempting to explain.
I assume that by "Goldbach partitions" you mean pairs (p,q) of primes which add up to a given even number. When you say congruent to 1 mod 30, do you mean that both primes are 1 mod 30? And then when you write "<8282=28", what does that mean? Something like "There are 28 Goldbach partitions of 8282 where both primes are 1 mod 30"? (I count 26, though.) Steve has some sort of product which has the \(\prod\) parts you suggest, but it looks like he has 8281/60 instead of 90^2/60. I find it somewhat inscrutable because the product with 8281/60 gives 29.239147 while the product you have gives 28.600059 and the product using his partial work "0.22126 × 0.92387 × 138" suggests 8280 which gives 29.235615 (all numbers rounded to six digits). Probably this is all a mess because Steve doesn't know what he's doing. But if we pretend he does we could ask: what is this supposed to mean? It seems to be some sort of prediction of the density or count of the primes on some interval, but which and what exactly is it? More little mistakes showing that he doesn't know what he's doing (or is trolling, I suppose). 
20180303, 23:25  #4  
"Forget I exist"
Jul 2009
Dumbassville
20D0_{16} Posts 
Quote:


20180304, 16:41  #5 
Aug 2006
5987_{10} Posts 
OK, but you don't know, for example, what region this formula is supposed to count primes over?

20180304, 16:48  #6 
"Forget I exist"
Jul 2009
Dumbassville
2^{4}·3·5^{2}·7 Posts 
No, because the thread title and first sentence, don't align with a statement made later. Thread title and first sentence say up to, other part says 28 just for 8282

20180306, 19:09  #7  
Feb 2018
2^{2}·3^{2} Posts 
Quote:
Not, n = prime preceding square root of(N  1)/60 which is 89. n1 = prime preceding square root of (N  1)/120 which is 61. But, Should be n = ..................................(N  1) Should be n1 = ................................(N 1)/2 Interesting, that you didn't notice that. But then again maybe not. Your ganging up is quite funny. I don't actually dislike notation. I just don't know it and didn't appreciate making out that it was godly and putting down my simplistic notation. You guys remind me of my days playing Pool for a living on the road, The Colour Of Money/Hustler style in America in the early nineties. People like you would show off with your 2,000 dollar+ cues, then l'd get one out of the rack and batter you with it, leaving you pennyless. Do you get the analogy? Probably not. Regardless of how crap l've written everything, the fundamentals are correct. It's called intelligence, except the Collatz proof, that's just pure genius. Whereas yours is called being able to remember what somebody else told you, then just regurgitating it. Wow! That's amazing! 

20180306, 19:18  #8  
"Forget I exist"
Jul 2009
Dumbassville
20D0_{16} Posts 
Quote:


20180306, 19:20  #9  
Aug 2006
5,987 Posts 
Quote:


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