Timeline for A conjecture based on Wilson's theorem

Current License: CC BY-SA 3.0

17 events

when toggle format	what		by	license	comment
Dec 10, 2015 at 22:24	history	edited	Włodzimierz Holsztyński	CC BY-SA 3.0	removed superfluous ( ) in the numerator
Dec 9, 2015 at 9:56	history	edited	martin	CC BY-SA 3.0	added 9 characters in body
Dec 7, 2015 at 12:14	answer	added	Alexey Ustinov		timeline score: 11
Dec 7, 2015 at 12:13	vote	accept	martin
Dec 7, 2015 at 12:06	answer	added	Jan-Christoph Schlage-Puchta		timeline score: 35
Dec 7, 2015 at 12:05	comment	added	Sean Eberhard		With @YCor's notation, if it's reasonable to assume that $i$ and $c_i$ behave roughly independently then one would have $\frac1p \sum i c_i \approx \left(\frac1p \sum i\right) \left(\frac1p \sum c_i\right) \approx p^2/4$, so this heuristically justifies the $1/8$.
Dec 7, 2015 at 11:58	history	edited	martin	CC BY-SA 3.0	added 70 characters in body
Dec 7, 2015 at 11:58	comment	added	barak manos		It might help to note that $\lim\limits_{n\rightarrow\infty}\dfrac{\sum\limits_{k=2}^{n/2}k(n-k)-1}{n^3} = \frac{1}{12}$.
Dec 7, 2015 at 11:53	history	edited	martin	CC BY-SA 3.0	added 70 characters in body
Dec 7, 2015 at 11:49	comment	added	YCor		Also the original formulation with $(k-1)/p_n$ was natural because this quotient is indeed an integer. In the new formulation, the $-1$ in $k-1$ seems artificial and can be removed since its contribution tends to 0. Second, the sum $\sum_A k$ is indexed by a quotient set of $\{2,\dots,p-2\}$ (modded out by the inverse involution). An alternative is to denote by $c_i$ the inverse of $i$ mod $p$ in $\{1,\dots,p-1\}$ and write the sum as $\frac12\sum_{2\le i\le p-2}ic_i$.
Dec 7, 2015 at 11:49	comment	added	YCor		Because as a set, $\{12,12,45,56\}=\{12,45,56\}$.
Dec 7, 2015 at 11:47	comment	added	martin		@YCor ok, will alter, but what should I call $A_p$ if not a set?
Dec 7, 2015 at 11:46	comment	added	YCor		You can't say $A$ is a set, since you take multiplicities into account. Also you should denote it as $A_p$ and write $A_{p_n}$ instead of $A$ in the conjecture.
Dec 7, 2015 at 11:39	history	edited	martin	CC BY-SA 3.0	deleted 49 characters in body
Dec 7, 2015 at 11:29	comment	added	barak manos		You might as well write it as $\lim\limits_{n\rightarrow\infty}\dfrac{\sum\limits_{k \in A}k-1}{(p_n)^3}\approx\frac18$.
Dec 7, 2015 at 11:26	comment	added	barak manos		You might as well take that $P_n$ outside the $\sum$.
Dec 7, 2015 at 11:23	history	asked	martin	CC BY-SA 3.0

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