Timeline for How to count fixed-sized subsets of pairwise co-prime numbers less than a prime, satisfying an additional ‎constraint‎?

Current License: CC BY-SA 3.0

14 events

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Apr 13, 2017 at 12:57	history	edited	CommunityBot		replaced http://mathoverflow.net/ with https://mathoverflow.net/
Aug 4, 2016 at 9:26	history	edited	Toughee	CC BY-SA 3.0	edited body
Jul 18, 2016 at 16:47	history	edited	Toughee	CC BY-SA 3.0	added 5 characters in body
Jul 12, 2016 at 6:05	history	edited	Toughee	CC BY-SA 3.0	improve formatting
Jul 9, 2016 at 14:50	history	edited	Toughee	CC BY-SA 3.0	added 5 characters in body
Jun 29, 2016 at 9:27	history	edited	Toughee	CC BY-SA 3.0	added 10 characters in body
Jun 25, 2016 at 20:03	comment	added	Greg Martin		I recommend grouping these subsets according to the product of their $k$ largest elements. For every $m<p$, the number of ordered factorizations of $m$ into $k$ divisors is roughly $\frac1{(k-1)!}(\log m)^{k-1}$ on average, and restricting these factorizations to being pairwise relatively prime won't change this significantly. For each such factorization, where $s$ is the smallest of the $k$ factors, we would then have about $Cs^{n-k}$ ways to fill out the $k$-tuple into an $n$-tuple, for some constant $C$ related to values of the zeta function and $\phi(m)/m$....
Jun 25, 2016 at 14:26	comment	added	Toughee		For the sake of clarification, I reworded the question. Thanks to @Paseman, I also think that there ‎are enough approaches now, and I need to improve the existing bounds.‎
Jun 25, 2016 at 14:25	history	edited	Toughee	CC BY-SA 3.0	deleted 197 characters in body; edited tags
Jun 20, 2016 at 20:30	history	edited	Toughee	CC BY-SA 3.0	edited body
Jun 20, 2016 at 15:26	history	edited	Toughee	CC BY-SA 3.0	added 9 characters in body
Jun 19, 2016 at 18:58	answer	added	Gerhard Paseman		timeline score: 1
Jun 19, 2016 at 16:48	history	edited	Toughee	CC BY-SA 3.0	edited title
Jun 19, 2016 at 15:41	history	asked	Toughee	CC BY-SA 3.0