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S Apr 22, 2017 at 15:47 history bounty ended CommunityBot
S Apr 22, 2017 at 15:47 history notice removed CommunityBot
Apr 17, 2017 at 15:07 comment added Kevin Smith I have no idea! If it is then an explanation of how this is so would be a welcome answer.
Apr 17, 2017 at 14:31 comment added Sylvain JULIEN Isn't the conjectured asymptotics moreorless equivalent to GRH ?
Apr 17, 2017 at 5:24 answer added Max Alekseyev timeline score: 4
Apr 17, 2017 at 0:01 history edited Kevin Smith CC BY-SA 3.0
included some additional information on motivation and reason for asking
S Apr 14, 2017 at 14:29 history bounty started Kevin Smith
S Apr 14, 2017 at 14:29 history notice added Kevin Smith Draw attention
Apr 14, 2017 at 14:28 history edited Kevin Smith CC BY-SA 3.0
Added some new information. Explained reasons for bounty.
Apr 10, 2017 at 7:01 history edited Kevin Smith CC BY-SA 3.0
spelling, grammar, English language.
Apr 9, 2017 at 19:19 comment added Kevin Smith @GregMartin: Please would you include a little more on how to proceed with the summation as indicated in your comment?
Apr 9, 2017 at 19:03 comment added მამუკა ჯიბლაძე Not sure if this can be useful but I find it sort of amusing - the denominator of $S(X)$ is$$\prod_{\text{prime $p\leqslant X$}}p^{\lfloor\log_p(n)\rfloor+1}$$
Apr 9, 2017 at 19:02 comment added Greg Martin You could also try the functional equation: for even characters $L(0,\chi)$ vanishes, while for odd characters $L(0,\chi) = -\frac i{\sqrt\pi} \tau(\chi) L(1,\bar\chi)$. Indeed, I think this gives $O(X^{1+\varepsilon})$ right away, and there's even some hope for an asymptotic formula.
Apr 9, 2017 at 17:48 history asked Kevin Smith CC BY-SA 3.0