Why is Goldbach conjecture that every even = a prime + a prime becomes

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  • #1
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more and more likely to be true the bigger the even? Primes become more rare, so it seems to me this notion is counter intuitive. :confused: A few recent papers all point to that Goldbach becomes more and more likely the higher up you go.

A very large even can be the sum of two large odds or 1 small odd and 1 very large odd. The first case, at least by intuition, should eventually fail for a very large even, since both large odds would increase and thus decrease both their chances of being prime. In the second case, suppose a small odd is kept fixed at a prime, the very big odd would have to increase. For instance, if you fix one of the odds at 3, the other odd must increase as the even increases, which decreases the chance of the other odd of being a prime. Again, at least by intuition, the second case should eventually fail for a very large even.

The average gap between two consecutive primes grows very slowly. As I saw from the data I worked with, by the time the first 240 million primes were reached, the average gap between two consecutive primes only grows to about 20 to 30 or so, even though gaps of over 300 between consecutive primes have already occurred. Perhaps, in my opinion, when an even surpasses 10 to the power of 1000, one cannot find two primes to add to that even.
 
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  • #2
mathman
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Intuition isn't much help. Opposing your intuition is the intuition that as even numbers get larger, there are more primes to pair up to possibly equal the even number in question.
 
  • #3
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Intuition isn't much help. Opposing your intuition is the intuition that as even numbers get larger, there are more primes to pair up to possibly equal the even number in question.
Wow, so somehow the increase in the number of pairings between primes would increasingly more than offset the decrease in density of primes as the even number in question becomes larger, i.e. the number of ways that an even can be written as the sum of two primes actually increases as the even is increased. :eek: Looks like this conjecture could take longer than 350 years to solve and surpass even Fermat's Last Theorem in notoriety. Looks like the testing shall continue. :biggrin:
 
  • #5
jbriggs444
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That means for n there are about [itex](\frac n {\ln n})\times(\frac n {\ln n} - 1)[/itex] possible pairs.
Nitpick: Since addition is commutative, the relevant pairs are not ordered. There is no prohibition on pairing a prime with itself so no minus one is required. After accounting for the problem of double-counting self-pairs, the result would be, more accurately, [itex]\frac {\frac n {\ln n}\times(\frac n {\ln n} + 1)} 2[/itex] possible pairs.

[However, a factor of 2 one way or the other is unimportant to the asymptotic behavior]

Now, these pairs each sum to some value in the range from 1 to 2n. The figure we care about is the probability that at none of those sums is equal to n. Denote the number of pairings by N. After accounting for the fact that almost all of the possible sums would be even, a naive estimate of that probability would be [itex](1-\frac 1 n)^N[/itex] ~= [itex]e^{-\frac N n}[/itex]

The likelihood of finding a counterexample to Goldbach's conjecture could be roughly estimated as the sum of this probability for n going from 4 to infinity.
 
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  • #6
Borek
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Nitpick:
Good point.

[However, a factor of 2 one way or the other is unimportant to the asymptotic behavior]
Another good point.

The difference between them is that I was aware of the latter, but I missed the first one
grumpy_borek.png
 

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