View Single Post
Old 2019-12-21, 11:06   #5
mart_r's Avatar
Dec 2008
you know...around...

23·5·17 Posts
Default "Kein Schwein ruft mich an, keine Sau interessiert sich für mich..."

_No objections from anyone, so I have to throw in my own concerns.

_Even if Ri'(x) is as precise as possible for describing the local average probability that x is prime, I still have a few second or third thoughts about \varphi(q). Especially when q is a primorial, the actual prime count may vary measurably from the one predicted by the formula, a phenomenon I'm still trying to work out some details to (see Of course it's certainly a negligible effect for the q's I'm looking at, but it's there, and I was wondering if an error bound can be obtained.

_I'm well aware this all sounds nitpicky. But when collecting data about extraordinarily large gaps, a "fair and square" measure of the gaps should be of the essence.

_In other news, and as not even WolframAlpha could give an answer to my satisfaction (i.e. one I was hoping to find) for the series expansion at x=1, I've worked out my own

\frac{\arctan[\frac{\pi}{\log(1+x)}]}{\pi}\hspace{2}=\hspace{2}\frac{1}{2}+\sum_{n=1}^\infty (-1)^nx^n\sum_{k=1}^{\lfloor\frac{n+1}{2}\rfloor}\frac{(-1)^{k+1}\hspace{1}[2(k-1)]!\hspace{1}s(n,2k-1)}{n!\hspace{1}\pi^{2k}}

_where s(n,k) are Stirling numbers, and, for x>e^\pi,

\frac{\arctan(\frac{\pi}{\log x})}{\pi}-\frac{1}{\log x}\hspace{2}=\hspace{2}\sum_{n=1}^\infty \frac{(-1)^n\hspace{1}\pi^{2n}}{(2n+1)\hspace{1}\log^{2n+1}x}
mart_r is offline   Reply With Quote