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Oct 19 at 0:27 vote accept T. Amdeberhan
Oct 17 at 18:59 comment added Abdelmalek Abdesselam Don't close. This is a good question.
Oct 17 at 18:57 answer added Abdelmalek Abdesselam timeline score: 7
Oct 17 at 1:19 comment added T. Amdeberhan @AbdelmalekAbdesselam: N is allowed to increase freely.
Oct 16 at 23:21 review Close votes
Oct 21 at 3:03
Oct 16 at 23:05 history edited Andrés E. Caicedo
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Oct 16 at 22:29 comment added Abdelmalek Abdesselam @T.Amdeberhan: Ok I think I understand, except what is $N$. It is nowhere defined. I also have eyes to see an absolute value is taken on the LHS.
Oct 16 at 21:27 comment added T. Amdeberhan @AbdelmalekAbdesselam: $N$ and $n$ are different. The formulation is correct. On the LHS, we take the absolute value.
Oct 16 at 21:14 comment added T. Amdeberhan @PietroMajer: no, it is not a typo.
Oct 16 at 20:50 comment added Pietro Majer Also, is $D\in C$ a typo?
Oct 16 at 20:08 comment added Abdelmalek Abdesselam BTW the question (even in need of clarification) looks very interesting to me, because lots of estimates of this kind occur in rigorous statistical mechanics. The main topic here is Moebius inversion in the partition lattice, following Rota, Groeneveld, Penrose,...
Oct 16 at 20:07 comment added Abdelmalek Abdesselam I believe the sum is over partitions $\sigma$ which are finer than $\rho$. There are lots of problems with the current formulation though. Is $N$ the same as $n$? Should the product be over blocks $C$ of $\rho$ instead of $\sigma$ so then the sum over $D$ is over $\sigma$ blocks inside the block $C$ of $\rho$? Is $|D|$ the number of points in $D$? For the estimate to make sense are we taking the max of the left-hand side over $\rho$?...
Oct 16 at 19:29 history edited T. Amdeberhan CC BY-SA 4.0
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Oct 16 at 19:09 history edited T. Amdeberhan CC BY-SA 4.0
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Oct 16 at 18:57 history edited T. Amdeberhan CC BY-SA 4.0
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Oct 16 at 18:48 comment added Chris McDaniel What are you summing over here?
Oct 16 at 18:40 history asked T. Amdeberhan CC BY-SA 4.0