August 24th, 2012, 09:06 AM  #1 
Newbie Joined: Aug 2012 Posts: 20 Thanks: 0  A formula for Goldbach?
After tinkering around, I found something interesting. Now, I'm not a mathematician, so please go easy on me, if I make a dumb mistake, you don't have to be too hostile. Anyway, I found that there are three types of even numbers, 10 and greater. They are in either in the form 6n + 10 6n + 12 6n + 14 6n + 10 even numbers are the sum of primes that are in the form 6x  1 and 6y  1 6n + 12 even numbers are the sum of primes that are in the form 6x + 1 and 6y  1 6n + 14 even numbers are the sum of primes that are in the form 6x + 1 and 6y + 1 Additionally, there are [[n/2]] + 1 ways of expressing 6n + 10 as the sum of two numbers 6x 1 and 6y  1 n + 1 ways of expressing 6n + 12 as the sum of two numbers 6x + 1 and 6y 1 [[n/2]] + 1 ways of expressing 6n + 14 as the sum of two numbers 6x + 1 and 6y + 1 I wanted to examine the ratio between pairs of "6x + 6y" that were prime, and those which weren't. (This is for numbers in the form 6n + 12) I found that by substituting "n" with various powers of numbers, (2,4,8,16) or (3,9,27,81), there became a recurring pattern. For example, take the number 162, which is 6*25 + 12 or 6* (5^2) + 12. It has a ratio of 10 "prime pairs" to "16 nonprime pairs". That is, 10 of the pairs of 6x + 1 and 6y  1 were prime, and 16 were not. That's for a total of 26 total combinations. (The total number of pairs could have been achieved through "n + 1", where n = 25. Anyway, I found that if you add the amount of prime number pairs (10), to the total number of pairs (26), you can achieve a ratio of the number 612 (our starting number) to this number (36). I graphed the ratios for 6*(5^n) + 12 for n = 1,2,3,4... and it turned out that there was a graph that perfectly fit: (4.0045845+0.026466834*x)/(1+0.0051995811*x+(1.7715989E0*x^2) It is a rational function. I also graphed the results for base 2,3,4... and achieved similar results, all of which were rational functions. I think this is pretty neat because it tells me that the ratio of a number 6(p^n) + 12 to its number of makeups, p^n + 1, is always greater than if there were no primes part of these makeups, which means some of the pairs have to be prime. Anyway, I'm gonna go try this out with more even integers in the other two forms, test out all the bases, and see if more patterns emerge. By the way, if you set n to to (1,4,9,16,25...) you see that the number of (prime pairs + total pairs) is a perfect square. So, if it were not for a certain "perfect number of prime pairs", this pattern wouldn't emerge. Lastly, I calculated the number of prime pairs compared to all pairs using a program that I wrote. Please leave feedback, thank you for reading! 
August 24th, 2012, 10:42 AM  #2  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: A formula for Goldbach? Quote:
 
August 24th, 2012, 11:04 AM  #3 
Newbie Joined: Aug 2012 Posts: 20 Thanks: 0  Re: A formula for Goldbach?
Sorry, I just copied the formula from my grapher program. x would be the number we're interested in, 6(5^n) + 12. Also, I found out that for starting values of 0<= n < 5, the formula works fine for most bases. Some bases' formulas can go a little beyond. Nevertheless, it becomes clear that the formula doesn't apply to all values of n. I'm hoping to find out the intervals in which it does satisfy and create a piecewise function that I could predict intervals for. Anyway, it's all a little bit ambitious and hard for me, not being a mathematician, but I'll give it my best. 
August 24th, 2012, 11:18 AM  #4 
Newbie Joined: Aug 2012 Posts: 20 Thanks: 0  Re: A formula for Goldbach?
If anyone was completely confused by what I wrote in the first post, I wouldn't be surprised. I'm pretty bad at conveying math. Hopefully this example will clear it up: A number in the form of 6(5^n) + 12 is made up of two primes in the forms of 6p+1 and 6q1. Granted, not all numbers in these two forms, 6p + 1 and 6q  1, are prime, so I decided to look at how many are, and how many are not. The total number of ways these two numbers, 6p + 1 and 6q 1, add to 6(5^n) is equal to (5^n) + 1. So, if we take a number 762, we can see that it in the form 6(5^3) + 12. There are 126 ways to make up this number using the forms 6p + 1 and 6q  1. This can be seen through (5^3) + 1. Using a program, I've determined that out of these 126 combinations, 30 of them were a pair of prime numbers and 96 were not. I was interested in finding the ratio of prime numbers to the given number, 762. So, I added the amount of primes (30) to the total number of combinations (126) and achieved 156. The result of (762/156) is 4.8846153846153846153846153846154. So, in addition to doing 762, I did 42, 162, 3762, and 18762, all of which are in the same form 6(5^n) + 12, and found their respective ratios. Then, I graphed the numbers (42,162,762...) as "x" and their ratios as "y", and achieving a fitting rational function: (4.0045845+0.026466834*x)/(1+0.0051995811*x+(1.7715989E0*x^2) For numbers in base 10, I achieved (4.5330192+(0.04036329*x)/(1+(0.0085646745)*x+1.4982885E07*x^2) , another rational functions. So far I have layed out all the bases and achieved a rational function for each. It also intrigued me that some of the functions, like those of base 10, could not only be expressed as a rational function, but as a sinusoid. 
August 24th, 2012, 02:04 PM  #5  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: A formula for Goldbach? Quote:
In particular, f(x) is approximately 1493951/x for large values of x. For example, f(10^6) is 2.1150 while the approximation gives 1.493; f(10^9) is 0.0014943 while the approximation gives 0.0014939.  
August 24th, 2012, 02:08 PM  #6 
Math Team Joined: Apr 2012 Posts: 1,579 Thanks: 22  Re: A formula for Goldbach?
One question that spings immediately to mind, if 30 OF 126 pairs are Goldbach pairs (ie both prime), what is the mathematical significance of ADDING 30 to 126?

August 24th, 2012, 02:26 PM  #7 
Newbie Joined: Aug 2012 Posts: 20 Thanks: 0  Re: A formula for Goldbach?
@CRGreathouse I clearly understand what you're saying in the first 2 lines, however the last 2 I lost you. If what you're referring to is that the rational function f(x) goes off track in large values of x, then yes, I agree with you. I wrote in the first post that I'm trying to understand what intervals of x a function f(x) works for. So far it seems to be that after n>4 in 6(p^n) + 12, it seems to go off track, so obviously there isn't much to work with for now. However, if that's not what your saying, then I'm sorry, I don't quite understand. @johnr I've thought the same myself, and since I don't have a definitive answer, (hence, I was "tinkering"), I'm not too sure what to say. I did find it very interesting, however, that there does seem to be some unknown reason for it. 6(1) + 12 = 18, where there are 2 combinations and 2 are prime. 2 + 2 = 4 6(4) + 12 = 36, where there are 5 combinations and 4 are prime. 4 + 5 = 9 6(9) + 12 = 66, where there are 10 combinations and 6 are prime. 10 + 6 = 16 6(16) + 12 = 108, where there are 17 combinations and 8 are prime. 17 + 8 = 25 6(25) + 12 = 162, where there are 26 combinations and 10 are prime. 10 + 26 = 36 6(36) + 12 = 228, where there are 37 combinations and 12 are prime. 37 + 12 = 49 As you can see, adding the total number of prime pairs does do something I find very interesting, and I'm not sure how/why. Keep in mind that while values of 6p + 1 are rising from 7 to 13 to 19 to 25, values of 6q  1 are descending from 6n+5 to 6n  1 to 6n  7. Therefore, all that Goldbach states is that 6n + 1 must be prime and "6(n + 2  p)  1" is also prime. 
August 24th, 2012, 02:46 PM  #8 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 933 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Re: A formula for Goldbach?
So you have a rational function which approximates the number of Goldbach partitions of small numbers. Reasonable enough, of course  the Weierstrass approximation theorem says you can always find a good approximating polynomial on an interval with a sufficiently smooth function, and allowing rational functions only makes it easier. But what does this tell us about the number of Goldbach partitions for large numbers? Or is it supposed to?

August 24th, 2012, 03:13 PM  #9 
Newbie Joined: Aug 2012 Posts: 20 Thanks: 0  Re: A formula for Goldbach?
Yes, I've found that these functions only work for a small range. However, I have found this: On most numbers, (most), only the range 0<= n =<4 in "6(p^n) + 12" can satisfy a given function perfectly. Any number above 4 causes error. The only notable exception I've found to this is 6, which allows for more values of n. 
August 24th, 2012, 03:18 PM  #10 
Newbie Joined: Aug 2012 Posts: 20 Thanks: 0  Re: A formula for Goldbach?
Sorry, I think I didn't word what I last said properly: A function f(x) may give the ratios for 2,4,8,16 perfectly, and 32 not. However, A new function consisting of the next for values of n: 32,64,128,256 will make a new function that satisfies each number. It will not satisfy smaller or larger values of "n". I'm assuming that you could just continue building functions to prove that there is always that ratio, but as far as I've seen, there is absolutely no correlation between functions, making the entire task very hard. It gets even harder considering that the coefficients are so wild and irrational, just as their prime number counterparts. 

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