n)$ also an upper bound?

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Apr 7, 2017 at 22:08	comment	added	H A Helfgott		@PeterHumphries Yes, but I need the other two for applications.
Apr 7, 2017 at 14:48	comment	added	Peter Humphries		@HAHelfgott, well, you've studied $\sum_{n \leq x} \frac{\mu(n)}{\sigma(n)} \log \frac{x}{n}$ and $\sum_{n \leq x} \frac{\mu(n)}{n} \log \frac{x}{n}$. We have that $\frac{\mu(p)}{p} = -\frac{1}{p}$ and $\frac{\mu(p)}{\sigma(p)} = -\frac{1}{p + 1}$. You could also consider $\frac{\mu(p)}{\varphi(p)} = -\frac{1}{p - 1}$.
Apr 7, 2017 at 6:52	comment	added	H A Helfgott		$x\geq 1465993117$. I think I'll stop here.
Apr 7, 2017 at 6:06	comment	added	H A Helfgott		@PeterHumphries: no, but I can. Should I? Why? Also, make that $x\leq 1465991461$. When, oh when will we get processors working at IEEE quadruple precision...
Apr 7, 2017 at 5:47	comment	added	Peter Humphries		@HAHelfgott, have you also looked at $\sum_{n \leq x} \frac{\mu(n)}{\varphi(n)} \log \frac{x}{n}$?
Apr 7, 2017 at 5:46	comment	added	H A Helfgott		Make that $x\leq 1465990562$.
Apr 6, 2017 at 5:40	comment	added	H A Helfgott		It's interesting to see what happens if we impose the condition that $n$ run over the odd numbers only. The limit is then $\pi^2/4$. It is an upper bound for much longer: it is valid for $x<=1465984277$ (and possibly a bit further).
Apr 5, 2017 at 21:24	comment	added	H A Helfgott		30. The gap should be easy to remove by the use of higher-precision arithmetic. The calculation I'm doing uses an interval-arithmetic package, and thus is rigorous, but the package is based on merely double-precision arithmetic (for speed). Hence the (small) gap, even after some careful optimizations.
Apr 5, 2017 at 18:12	history	edited	Peter Humphries	CC BY-SA 3.0	added 5 characters in body
Apr 5, 2017 at 17:04	history	edited	Peter Humphries	CC BY-SA 3.0	added 27 characters in body
Apr 5, 2017 at 15:44	comment	added	Gerhard Paseman		30, or 20? Gerhard "Or Nineteen Or Twenty Nine?" Paseman, 2017.04.05.
Apr 5, 2017 at 15:19	history	edited	Peter Humphries	CC BY-SA 3.0	added explanation of bias
Apr 5, 2017 at 15:13	comment	added	H A Helfgott		Good call. A more precise computation shows that the proposed inequality is false for x = 65371730 (and possibly for some other x> 65371719; for x<=65371719, it is true)
Apr 5, 2017 at 9:31	vote	accept	H A Helfgott
Apr 5, 2017 at 8:51	comment	added	Greg Martin		Very nice. If you assume RH and the linear independence of $\{\gamma>0\colon \zeta(\frac12+i\gamma)=0\}$ over $\Bbb Q$, can you prove that the logarithmic density of those $x$ for which the sum exceeds $\frac{\pi^2}6$ is actually $\frac12$? Or is there a bias-causing term somewhere in the explicit formula? (I don't see one....)
Apr 5, 2017 at 1:45	history	answered	Peter Humphries	CC BY-SA 3.0

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