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This question is related to https://mathoverflow.net/questions/61842/about-goldbachs-conjecture

Let $ y(x) $ be a function such that $ \alpha_{n}=O(y(n)) $ . taking $ y(x) : =x^{1/2}\log^{2} x $ leads to $ y'=\dfrac{1}{2}x^{-1/2}\log^{2}x+2x^{-1/2}\log x $ . As $1/ \log x $ is roughly equal to $ \pi(x)/x $ , is there any reason to believe that $ \alpha_{n}=O(z(n)) $ where the function $ z(x) $ fulfills the differential equation $ z'=\dfrac{z}{2\Delta}+z\dfrac{\pi(x+\Delta)-\pi(x-\Delta)}{\Delta} $ where $ \Delta : =\Delta(x) $ is such that $ \Delta(x)\leq (1+o(1))x $?

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  • $\begingroup$ why ? ${}{}{}{}{}$ $\endgroup$ – reuns Mar 15 '17 at 12:11
  • $\begingroup$ Because. In case you wouldn't have noticed, I think induction can be used in mathematics to lead to discovery. $\endgroup$ – Sylvain Julien Mar 15 '17 at 12:15
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    $\begingroup$ Goldbach is extremely complicated to analyze. You didn't define $\alpha_n$. Your differential equation is a mess : do you really think we can do anything better than replacing $\pi(x+a)-\pi(x)$ by $\frac{a}{\ln x} + O(a^\theta)$ and "hope" it will give something acceptable ? $\endgroup$ – reuns Mar 15 '17 at 12:57
  • $\begingroup$ $ \alpha_{n} $ is an error term defined in the link I gave. $\endgroup$ – Sylvain Julien Mar 15 '17 at 13:06
  • $\begingroup$ In a more rigorous setting, one can consider the sequence of differential equations $ y'_{k}=\dfrac{y_{k}}{2x^{1-1/k}}+y_{k}\dfrac{\pi(x+x^{1-1/k})-\pi(x-x^{1-1/k})}{x^{1-1/k}} $ and then take the limit as $ k $ tends to $ \infty $ . $\endgroup$ – Sylvain Julien Mar 15 '17 at 13:44

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