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Nov
4
comment Counting all possible board positions in Quoridor
@lameK, that's right. There are 3344 ways to order one row.
Oct
31
comment Polynomial factorisation - absolute value of coefficients
Ah, that explains it. Oops.
Oct
31
comment Polynomial factorisation - absolute value of coefficients
You could also have $x^4+x^3-x^2-1 = (x-1)(x^3+2x^2+x+1)$. If I'm understanding arxiv.org/abs/0904.3057 correctly it claims that this is the unique quartic with height 1 and an irreducible factor of height greater than 1, in which case your example is a correction.
Oct
31
comment Does $A193201$ count the partitions of $n$ of arbitrary dimension?
I'm not quite sure what I'm supposed to be looking at in the example you give of partitions of 4. For fixed layout, the preformatted text (the button that looks like {} in the editor, or the result of prepending 4 spaces to each line) is a bit easier than MathJAX.
Oct
30
comment Math puzzle: 10 digit strings generations
Independent set problem
Oct
30
comment Math puzzle: 10 digit strings generations
Not quite practical as a computational solution, but this can be reduced to the independent set problem on a graph of $10!$ nodes and $\frac{10! 7!}{2}$ edges.
Oct
15
comment Number of unlabeled simple graphs with $n$ nodes even for all $ n\ge 5$?
Ah, it doesn't have simple in the title so my search didn't find it. Thanks for adding the link.
Oct
15
comment Number of unlabeled simple graphs with $n$ nodes even for all $ n\ge 5$?
Are you referring to A005470 (in which case you seem to be missing the very important word planar from the question) or to a different sequence?
Oct
12
comment Counting all possible board positions in Quoridor
@lameK, I got 55 by writing a short computer program, but as I used an obscure language to do it I didn't see any benefit to sharing it. I'll add some examples to the last paragraph.
Oct
6
comment Can anyone extend my findings for Langford Pairings?
I don't actually see a question. If this is an attempt to follow David Eppstein's advice that for your ideas to be included in the Wikipedia page on Langford pairing you should first get them published, then you should be aware that he was talking about publication in a peer-reviewed journal, not on a Q&A site.
Sep
22
comment No simple closed form for Bell numbers
The premise of your question is somewhat unclear, because "closed form" is a somewhat variable quantity. Unless you allow factorials in a closed form, I can't think of any basic combinatorial quantity which has one. Allowing them lets in binomial coefficients and therefore Catalan numbers, but what else? Such basic combinatorial quantities as Stirling numbers and the partition function don't have well-known closed forms.
Aug
19
comment The Day Camp Stacking Game
Is rule 4 a repetition of rule 2 or is it trying to say something additional? (I assume that "clockwise" and "left" mean the same thing here).
Aug
6
comment Deformable circle from a cubic Bezier approximation
What do you mean by "smoothness"? Do you want C2 continuity? G2? Something else?
Jul
31
comment Lower bound for a relative of the central binomial coeff
@Raphael, can you explain why there's a singularity at $-2+\sqrt{5}$? I see singularities at $-2-\sqrt{5}$ and $\frac{1}{4}$.
Jul
30
comment Lower bound for a relative of the central binomial coeff
@GerryMyerson, the central binomial coefficient has a rather neat product representation which allows an easy ad hoc proof of the bound. The only way I can see to get a similar product representation is to factor out a central binomial coefficient from each of the binomial coefficients in the second sum. Then bounding with $\frac{1}{\sqrt{m-2j-1}}\ge\frac{1}{\sqrt{m}}$ I get a non-Gosper-summable hypergeometric, and crudely bounding the terms to get a summable geometric sequence gives a worse bound than simply taking the first term of the original sum.
Jul
30
comment Prove equivalence of Diffie-Hellman shared secret
@codeomnitrix, $a \equiv b\pmod c$ is a shorthand notation for $\exists k : a - b = ck$
Jul
23
comment Transforming a latin square into a sudoku
@EwanDelanoy, that was my first thought for a counterexample, but just permuting the rows to give leading column $147258369$ suffices.
Jul
23
comment Transforming a latin square into a sudoku
And similarly for the 9x9 case it should suffice to fix the top-left 6x6 block.
Jul
18
comment What is the count of the strict partitions of n in k parts not exceeding m?
The answer to "How many different sets $X_1, \ldots, X_m$...?" would seem to be $m$...
Jul
16
comment “I have found a dead body on my car.”
This seems to be more a question about the English language than about mathematical logic.