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Let $\beta(n)$ represent the number of Goldbach prime pairs that each add up to an even integer $n$.

Observation: If $p$ is a prime, for $n \ge 152$, (ignoring $n$ = powers of $2$)

$$\beta(n) \le \beta(p*n) < p*\beta(n)$$

For powers of $2$, the above inequality is true for $n \ge 128$.

Since any number $m$ can be obtained by multiplying primes, replacing $p$ with $m$ in the above inequality would work as well.

Implication: If we start with some $n = 2p$, we know that $\beta(2p) \ge 1$. And so proving the above inequality would prove GC.

Question: If this inequality is indeed true, how would one go about proving it? What approaches would you take? Any hints? Thanks

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closed as unclear what you're asking by Eric Tressler, Namaste, Daniel W. Farlow, JonMark Perry, Smylic May 24 '17 at 6:18

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  • $\begingroup$ Upto which number did you check the double-inequality ? $\endgroup$ – Peter May 23 '17 at 22:21
  • $\begingroup$ I have access to one million primes.. So whatever the even number is above the last one. $\endgroup$ – sku May 23 '17 at 22:23
  • $\begingroup$ the left inequality is enough to prove GC if it holds true for all primes(integers as well). $\endgroup$ – Ahmad May 23 '17 at 22:24
  • $\begingroup$ I am curious. Why are people trying to close this question. Can someone please explain? Thanks $\endgroup$ – sku May 24 '17 at 3:58
  • $\begingroup$ I wrote my question at the end of this post. Let me know if that is still unclear. If not, could you remove the hold please? Thanks $\endgroup$ – sku May 24 '17 at 20:41
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68 can only be written in two ways as the sum of 2 primes (7 + 61, 31 + 37), however 34 can be written in 4 ways as the sum of 2 primes (17 + 17, 31 + 3, 23 + 11, 29 + 5).

Then $\beta(68) = 2$ and $\beta(34) = 4$ and $68 = 2 * 34$, with 2 being prime...

I am definitely not sure that your inequality is correct. We also have $\beta(152)=4$ and $\beta(76)=5$.

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  • $\begingroup$ Yes, numbers below 128 being small behave funky. I missed this one. Thanks $\endgroup$ – sku May 23 '17 at 22:35
  • $\begingroup$ @sku I found another example, above 128 this time... $\endgroup$ – fonfonx May 23 '17 at 22:43

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