Talk:Goldbach conjecture

Expected number of Goldbach representations

I think this sentence is way to long and a run-on sentence also. I am not sure of the meaning by the time I read the end. " Asymptotically, for each of the odd primes (essentially all congruent to 1 or 5 modulo 6) in the lower half, the probability of matching an odd prime in the odd numbers of the upper half is , if is congruent to 0 modulo 6, and , if is congruent to 2 or 4 modulo 6, assuming a random distribution for the odd prime numbers, since, knowing that the first number is chosen to be prime (most likely congruent to 1 or 5 modulo 6,) if is congruent to 0 modulo 6 it is the sum of a prime and an odd number congruent to 1 and 5 modulo 6 or 5 and 1 modulo 6, if is congruent to 2 modulo 6 it is the sum of a prime and an odd number congruent to 1 and 1 modulo 6 or 5 and 3 modulo 6, if is congruent to 4 modulo 6 it is the sum of a prime and an odd number congruent to 1 and 3 modulo 6 or 5 and 5 modulo 6, and the probability of finding a prime among numbers congruent to 1 or 5 modulo 6 is 3 times higher than among all the congruences modulo 6, therefore the expected number of Goldbach representations is asymptotically

if is congruent to 0 modulo 6, and

if is congruent to 2 or 4 modulo 6, assuming a random distribution for the odd prime numbers. "

Also it, "Expected number of Goldbach representations", might be better to look at it the following way:

Except for the cases of where 2 is used, for p or q, we can rule out all but numbers congruent to 1 modulo 2 for being composite to 2. This leaves, when 12 < p < q, two different numbers both of which are congruent to 1, 3, or 5 modulo 6 and both are prime.

Except for the cases of where 2 or 3 is used, for p or q, we can rule out all but numbers congruent to 1 or 5 modulo 6 for being composite to 2 and 3. This leaves, when 12 < p < q, two different numbers both of which are congruent to 1 or 5 modulo 6 and both are prime.

Now because p+q = 2n, 2n must take on all congruent values of modulo 6, so ....

Yes, that might not be as clear as we would like. I will ponder both your proposed version and the current version, but I will also be looking at other things. In the interim, please feel free to edit the article, and don't worry about formatting mathematical expressions, either Dan or I can take care of that. Alonso del Arte 22:36, 23 July 2012 (UTC)

98

If you can explain the near failure of 2*7^2 = 98, then you can nearly prove Goldbach's.

98 has 3 Goldbach representations (see A002375(2n, n = 49)), I don't see a near failure there... — Daniel Forgues 04:17, 23 September 2012 (UTC)

3 + 95 Y

5 + 93 Y

7 + 91 Y

11 + 87 Y

13 + 85 Y

17 + 81 Y

19 + 79 Y

23 + 75 Y

29 + 69 Y

31 + 67 Y

37 + 61 Y

41 + 57 Y

43 + 55 Y

47 + 51 Y

The following will only give already counted [unordered] prime pairs (no point of going past n/2 for the first prime):

53 + 45 Y

59 + 39 Y

61 + 37 Y (but already counted)

67 + 31 Y (but already counted)

71 + 27 Y

73 + 25 Y

79 + 19 Y (but already counted)

83 + 15 Y

89 + 9 Y

97 + 1 Y

A002375(49) gives number of decompositions of 2*49 = 98 as 3, OK. — Daniel Forgues 05:20, 23 September 2012 (UTC)

A002375 From Goldbach conjecture: number of decompositions of 2n into an unordered sum of two odd primes. — Daniel Forgues 04:47, 23 September 2012 (UTC)

Cf. Goldbach Representations for Even Integers 8 to 1000. (It only considers the cases with two distinct odd primes, e.g. 5+5 = 10 is not accounted for.) — Daniel Forgues 04:56, 23 September 2012 (UTC)

I meant using the "Expected number of Goldbach representations" that you have stated in the article. In other words why did so many fail and three remain in respect to congruence at 98?

Also notice that the number of odd primes in the upper half [n/2, n-1] always less than the number of odd primes in the lower half [1, n/2]. This means that the number of primes in each range is asymmetric. This asymmetry is first seen with mod 6. You can think of it like a wheel with the numbers 0, 1, 2, 3, 4, and 5 at each 60 degree angle. If a number mod 6 is 1 put it on 1, if it is 5 put it on 5 else do not place. the wheel will always be leaning towards a point between these two stacks, near 0.

A second note, Ramanujan primes. 97 is the 10th Ramanujan prime. I feel this is a pointer to the solution. By this I mean that ratio of the number of Ramanujan primes to partitions is important because of the ~ with n/(2* log n) is the same as Ramanujan prime. R_n/(2*log(R_n)) ~ n, and R_n/(2*log(R_n)) < n for n>202. John W. Nicholson 01:22, 9 October 2012 (UTC)

For 98 (congruent to 2 (mod 6)) the expected number of Goldbach Representations is

${\frac {3}{4}}{\frac {98}{{(\log 98)}^{2}}}=3.496349\ldots \,$

and the actual number of Goldbach Representations is a(49) = 3. (Pretty much in agreement with expected value for 98!) — Daniel Forgues 05:26, 10 October 2012 (UTC)

A002375 From Goldbach conjecture: number of decompositions of 2n into an unordered sum of two odd primes.

$\scriptstyle n$	$\scriptstyle a(2n)$	$\scriptstyle 2n$	$\scriptstyle 2n{\bmod {6}}$	Expected
9990	906	19980	0	$\scriptstyle {\frac {3}{2}}{\frac {19980}{{(\log 19980)}^{2}}}=\,$ 305.63154475796
9991	328	19982	2	$\scriptstyle {\frac {3}{4}}{\frac {19982}{{(\log 19982)}^{2}}}=\,$ 152.82797964385
9992	326	19984	4	$\scriptstyle {\frac {3}{4}}{\frac {19984}{{(\log 19984)}^{2}}}=\,$ 152.84018669318
9993	670	19986	0	$\scriptstyle {\frac {3}{2}}{\frac {19986}{{(\log 19986)}^{2}}}=\,$ 305.704787054
9994	360	19988	2	$\scriptstyle {\frac {3}{4}}{\frac {19988}{{(\log 19988)}^{2}}}=\,$ 152.86460014533
9995	438	19990	4	$\scriptstyle {\frac {3}{4}}{\frac {19990}{{(\log 19990)}^{2}}}=\,$ 152.8768065482
9996	868	19992	0	$\scriptstyle {\frac {3}{2}}{\frac {19992}{{(\log 19992)}^{2}}}=\,$ 305.77802547127
9997	382	19994	2	$\scriptstyle {\frac {3}{4}}{\frac {19994}{{(\log 19994)}^{2}}}=\,$ 152.90121870767
9998	334	19996	4	$\scriptstyle {\frac {3}{4}}{\frac {19996}{{(\log 19996)}^{2}}}=\,$ 152.91342446433
9999	730	19998	0	$\scriptstyle {\frac {3}{2}}{\frac {19998}{{(\log 19998)}^{2}}}=\,$ 305.85126001126
10000	462	20000	2	$\scriptstyle {\frac {3}{4}}{\frac {20000}{{(\log 20000)}^{2}}}=\,$ 152.93783533161

Why are actual values 2 to 3 times expected values?! — Daniel Forgues 05:26, 10 October 2012 (UTC)

First of all, how many primes are in the upper half? To answer this note the n value to R_n where R_n is Ramanujan prime. Second notice that there are multiple ways to sum 1 mod 6, and 5 mod 6 as to make a complete residue set for 2n. Those are 1+1 = 2 (1 set), 5+5 = 10-6 = 4 (1 set), 1+5 = 5+1 = 6-6 = 0 (2 sets). So, some have more values than others. It is similar to two dice having the greatest number of sums being 7. However, the mod 6 kicks in with higher than 6 values. 98 is an example of the worst case, there is only one pair of mod 6 solution namely 1 mod 6 + 1 mod 6 = 2 mod 6. For, the best case you have to look at all possible ways to sum two modulos of 2, 3, 5 which ends up being 8 modulos of 30, namely for {7, 11, 13, 17, 19, 23, 29, 31=1}. Sub cases are 48 modulos for 2, 3, 5, 7 with 210 and so on.

See http://mathpages.com/home/kmath101.htm .

John W. Nicholson 25 October 2012.

Statements assuming the Goldbach conjecture

From A157931: "Assuming the Goldbach conjecture, this is A001358 intersect (A005843 union A052147), since an odd number n is the sum of two primes iff n-2 is prime. - N. J. A. Sloane, Mar 14 2009" Alonso del Arte 04:12, 9 February 2013 (UTC)

And pending for A089268: "Assuming Goldbach's conjecture that every even number greater than 2 is the sum of two primes, these are the numbers that are the product of two primes but not the sum of two primes. - Michael B. Porter, Feb 8 2013" Alonso del Arte 04:24, 9 February 2013 (UTC)

Ramanjan primes and birthdays

Does the n-th Ramanujan prime, R_n, "not prove the Goldbach conjecture, but it does suggest a counterexample, if it exists, would be very [extremely] difficult to find"? (Quote from 'Bertarnd's postulate' however changed for the generalization.) I think the logic here is *similar* to two people in a classroom of 30 people having the same birthday. However, with a term of probability of failure, P_n, is decreasing for increasing n along with increasing R_n.

(unsigned, presumably from John W. Nicholson) Yes, it is me. John W. Nicholson 21:38, 20 March 2013 (UTC)

I don't think I agree with either of the statements, the sentence from Bertrand's postulate or this one. The Prime Number Theorem plus basic calculus gives, I believe, a heuristic probability of $\exp \left(-{\frac {x}{2\log ^{2}x}}\right)$ that x is a violation of GC. Integrating from 10^9 to infinity gives 10^-505633 or so (close to 0, so no exceptions expected). I don't see a need for R_n.

I don't see the relevance of the birthday paradox either. In Goldbach you're looking only at pairs rather than sqrt-sized groups.

Charles R Greathouse IV 05:36, 20 March 2013 (UTC)

Help me out here, you say "a heuristic probability of $\exp \left(-{\frac {x}{2\log ^{2}x}}\right)$ that x is a violation of GC", but does this heuristic probability of x include even numbers? Multiples of three, five, or seven? As we increase the prime value the prime p_i, the number probable primes after these exclusions increase at a rate of Product((p_i - 1)/p_i). Now for pairs. It seems like there can be twin prime pairs, or Goldbach pairs, or even some other type of pairing which makes the rate of remaindering pairs Product((p_i - 2)? John W. Nicholson 21:38, 20 March 2013 (UTC)

It doesn't make a big difference. If you look at only odd numbers you have half as many chances at twice the probability and those largely wash. You can come up with a more precise estimate if you include these factors but either way the number will be small enough to safely conclude that there can only be counterexamples if there's some conspiracy avoiding the heuristic.

Charles R Greathouse IV 02:36, 21 March 2013 (UTC)

Expected number of representations, redux

As John W. Nicholson pointed out in his recent edits, the expected number is incorrect because it doesn't look at the values mod 3. But more broadly, the expectation needs to take all primes into account. In particular, the expectation should involve a constant which comes from a product over all odd primes.

Charles R Greathouse IV 04:27, 6 October 2013 (UTC)

I suspect that a more mathematically sound formula for the expected number will still fall far short for small numbers, and that as the numbers get larger the formula gives an expectation closer to the actual number. Alonso del Arte 18:58, 6 October 2013 (UTC)

A better way is to use Ramanujan primes. If ${\frac {R_{k}}{2}}<n<R_{k}$, then use the Ramanujan Prime Corollary to 'find the k number of primes'. (I put the single quotes as to indicate that while it is not the most efficient way to find primes, you gain information on primes.)

A204814

Some related sequences can be found by searching for A204814. A173634 is another one. John W. Nicholson 04:43, 13 November 2013 (UTC)

Relation to Twin Primes

Due to work started by User:Mike_Winkler circa Sept 2013, I believe there is a tie-in from GC to the Twin Primes. More shortly after conclusion of ongoing email traffic.--Bill McEachen 15:42, 11 May 2014 (UTC)

'Minus' version

This is the version every even number can be expressed as the difference of 2 odd primes. [ some saying in an infinite number of ways ]. It is interesting because one can find primes above the 'sum' primes. So, 17+13=30, and 30=41-11 as an example.--Bill McEachen 01:52, 20 April 2015 (UTC)

This is called Polignac's conjecture. - Charles R Greathouse IV 12:17, 20 April 2015 (UTC)