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


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

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 $\beta(76)=5$.

  • $\begingroup$ 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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