# ARupinski

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## Registered User

 Name ARupinski Member for 2 years Seen yesterday Website Location Birmingham AL Age 29
Doctorate in Math from U. of Pennsylvania, specializing in Linear Algebra and Representation Theory.
 May15 asked Picturing a Certain Torus and Klein Bottle May14 awarded ● Organizer May14 revised Simplifying an algebraic integer expressionfixed tex formatting, changed tags to be more appropriate to question May11 revised Does this Linear Algebra Construction have a Name?Added additional thoughts May8 comment Square and reversed integer@Gerry Myerson: thanks for clearing that up. Obviously I read the formulation through too quickly without thinking about what it was asserting. May6 revised Does this Linear Algebra Construction have a Name?clarified formula expansion May6 comment Square and reversed integerYou claim that this is true for $m < 10^8$, but why does for example $m = 32$ work? I have that $f(32\times 32) = 4201 \neq f(32)\times f(32) = 529$ even though for 32 one has $a_0,a_1\in\{0,1,2,3\}$. Is there an extra assumption that is missing here? May6 asked Does this Linear Algebra Construction have a Name? May4 comment $n$-in-a-row game on $\mathbb{R}^2$@Benjamin: I had thought about that sort of approach too, but how do you ensure that your 4-in-a-rows so that they align with each other (i.e. why can P2 not see this strategy and play to block the 4-in-a-rows that you need, forcing you to form one not aligned with your previous 4-in-a-rows)? Is there a good way to force P2 to only block certain potential 4-in-a-rows so that P1 eventually gets a 5-in-a-row? May2 comment $n$-in-a-row game on $\mathbb{R}^2$Also by playing his points so that intersections of any of the lines he forms do not coincide with any of P2s already played points, it is obvious that by move 5 P1 has too many double 3-in-a-rows lined up for P2 to stop them all with just one point. May2 comment $n$-in-a-row game on $\mathbb{R}^2$@Ricky: So maybe he needs one more move then... if P1 always plays his points such that they are not collinear with any pair of points already on the plane, then even if P2 has blocked 2 of the potential 3-in-a-rows on the Fano configuration, P1s fifth point will form several new partial Fano configurations; by this point there are 10 different lines formed by the pairs of P1s points; each of these lines intersects one another somewhere in the plane and at most 3 of P2s points are on any of these lines (P2s first point does not help P2 in any way because of P1s non-collinearity strategy). May2 answered $n$-in-a-row game on $\mathbb{R}^2$ May2 comment $n$-in-a-row game on $\mathbb{R}^2$So no matter how P2 responds now, at least one of the edge points of this Failed Fano configuration is open, so P1 forms two 3-in-a-rows simultaneously, thereby foiling any further attempt by P2 to win. I don't see any good way to extend this approach to $n>4$ (on account of the rather special nature of the Failed Fano configuration), but maybe this sparks an insight by someone else for the general case... May2 comment $n$-in-a-row game on $\mathbb{R}^2$For $n=4$ a similar approach should work. P1 can always play his 2nd and 3rd points into a triangle with at most one of its edges containing one of P2s first two points. Now no matter how P2 chooses to block one of these edges, P1 can find a point such that none of the lines connecting that point to his previous 3 points contain any of P2s points thusfar and which is not collinear with any pair of his previously placed points. Note that P1s points now form the 3 vertices and center point of the Failed Fano configuration. Jan30 comment Factorization of permutations.Your question seems to be covered by the comments and answers to [mathoverflow.net/questions/62088/…. And unless I misunderstand what you are asking in (2), there is no way $\Phi$ can be surjective for an entire range of $\alpha$'s if it fails to be surjective for some $\alpha$ in the given range. Jan30 revised Edge-coloring of the complete graph without any rainbow pathsadded 123 characters in body Jan29 revised Edge-coloring of the complete graph without any rainbow pathsadded 21 characters in body; deleted 21 characters in body Jan29 answered Edge-coloring of the complete graph without any rainbow paths Jan23 comment Confirm/refute $f(x)$ where $f(x) =$x-th Mersenne prime ($M_p$) where $x$ is [1,2,3,4,5,6,7…]See primes.utm.edu/notes/faq/NextMersenne.html and some of the linked pages; in particular there are conjectures about the log distributions of Mersennes and some supporting evidence. Jan21 comment Does this system of equations admits a solution?The covering sequence can then be taken as the $a_i$ and the sequence of cofactors can be taken as the $b_i$ which then gives a solution to your particular problem. Jan21 comment Does this system of equations admits a solution?@Antisha: Ronald Graham's 1964 paper "A Fibonacci-like sequence of composite numbers" shows how to construct a linearly recurrent sequence all of whose entries are composite numbers using the Fibonacci recurrence. By modifying his approach it should be possible to find a covering sequence for this recurrence, (although there may be some complications I don't immediately see due to the fact that here the associated polynomial $x^2-3x+2$ factors while for the Fibonacci sequence the associated polynomial $x^2-x-1$ does not factor). Jan20 comment Does this poset have a unique minimal element?There also seem to be examples with infinite degree vertices, so perhaps total classification might be out of reach, but classification of those with only finite degree vertices might be doable. I am definitely going to think about this some more. Jan20 comment Does this poset have a unique minimal element?@Pietro: I was actually thinking about that exact thing a few days ago. I was able to come up with some infinite families, but I didn't think about it enough to come up with a conjecture as to what a complete list would look like. The first infinite family I was able to come up with consists of taking the doubly infinite path and attaching 3 single leafs to nodes which are not evenly spaced; the other families I had thought of were variations of this with multiple infinite branches. Jan20 comment Automorphism of finite groups and Hurwitz spacesIts definitely true that if you choose the regular representation of $G$ then every automorphism extends to an inner automorphism in $S_{|G|}$ (so $G$ has at least one transitive representation with the desired property), although in your case it seems you need the result to hold for a far wider class of transitive representations. Jan20 comment Does this poset have a unique minimal element?On reading through, everything seems to make sense with your fix in place. I wish it were possible to accept two answers in situations like this where two posters both come up with beautiful answers. Jan20 comment Does this poset have a unique minimal element?Ok, I've had time to read through this carefully, and I'm convinced by each step. The only part I had to think about for awhile to convince myself it was true was your parenthetical remark that $l_2$ is the special leaf with respect to $v$ and $T$; aside from that part, the other parts of the proof all look good as far as I can tell. Jan20 accepted Automorphism of finite groups and Hurwitz spaces Jan20 answered Automorphism of finite groups and Hurwitz spaces Jan19 comment Does this poset have a unique minimal element?I dont have any particular references. As far as motivation, I was thinking about some problems related to antichains in posets and was tired of thinking about some common examples such as $\mathbb{N}$ and the poset of Young Diagrams of partitions, so I was trying to come up with a more complicated poset to think about in context (and it seems I did; while $\mathbb{N}$ and the set of Young Diagrams each obviously have a unique minimal element, proving this is also true in $\mathcal{AFT}$ is apparently much more difficult) Jan18 awarded ● Yearling Jan17 comment Apollonian gasket and the degree of convergence I don't have mathscinet access right now so I can't look in the paper, but is that critical value special to the (1,2,2)-gasket in the question or is it independent of the specific set of curvatures in the gasket? Jan17 comment Apollonian gasket and the degree of convergence For what its worth, while in principle one has a recursive formula for the curvatures, you are not alone in being unable to effectively use it. My advisor in grad school spent some time looking at the question of 'given starting curvatures of a gasket what curvatures can/cannot appear in the gasket?' without doing an exhaustive depth search. As far as I know this question is still open (the best one can do is find some modular restrictions), so short of finding a better answer, you probably wouldn't be able to calculate the critical $\alpha$ without a computer like in Gerald's answer. Jan16 comment Does this poset have a unique minimal element?Wow. On my first read through, everything in here looks reasonable and believable. I will continue thinking about it to make sure I understand it all before accepting, but it looks good so far. Jan16 comment Does this poset have a unique minimal element?But for what its worth, I agree that for local rings you are probably right as far as largest would suffice (if people actually thought about what 'largest' is constrained to mean in context). Jan16 comment Does this poset have a unique minimal element?For other instances such as local rings, I might venture to guess that the usage might have something to do with largest not always being equivalent to unique maximal. For example, a given group may have several 'largest' proper subgroups all of the same size (and so not unique, think $G = (\mathbb{Z}/2\mathbb{Z})^n$ with $n\geq 2$) and likewise it may have maximal subgroups which are not largest in size among all subgroups (for example the Monster has a maximal subgroup with only 1640 elements, far smaller than many of its non-maximal subgroups). Jan16 comment Does this poset have a unique minimal element?@Tom: interesting meta-question. At least in this case I suppose describing it this way I meant to make the focus of the present question whether colloquially 'all roads lead to Rome' or, translated into this problem, 'all paths in $\mathcal{AFT}$ lead to $E_7$'. Jan5 comment For which triples of cycle structures $\alpha,\beta,\gamma$ are there permutations $x,y$ with $C(x),C(y),C(xy)=\alpha,\beta,\gamma?$(Since its a little ambiguous looking now that I reread it, above $|\alpha|$ refers to the size of conjugacy class of structure $\alpha$ in $S_n$) Jan5 comment For which triples of cycle structures $\alpha,\beta,\gamma$ are there permutations $x,y$ with $C(x),C(y),C(xy)=\alpha,\beta,\gamma?$The second observation, which is probably more useful is in counting the number of elements with cycle structures $\alpha$, $\beta$, and $\gamma$ in some fixed $S_n$. If in this $S_n$ $|\alpha|\cdot|\beta| < |\gamma|$ then there are also no solutions since all elements of structure $\gamma$ are conjugate and hence by conjugacy would lead to decompositions of all such elements. This is probably useful for determining lower bounds on which $S_n$'s such permutations could live in, although I couldn't use it to come up with any counterexamples to Stefan's question (yet). Jan5 comment For which triples of cycle structures $\alpha,\beta,\gamma$ are there permutations $x,y$ with $C(x),C(y),C(xy)=\alpha,\beta,\gamma?$Some thoughts which turned out not to apply in any useful way when I was thinking about Stefan's followup question, but which might be useful here. Firstly, there are obvious constraints on the total number of elements moved by the C's; if the sum of the number of elements moved by cycle structure $\alpha$ and the number moved by $\beta$ is not at least the number moved by $\gamma$, then certainly there are no solutions. Jan5 comment Does this poset have a unique minimal element?@Pietro: I will unaccept it if you really prefer, but I think the ideas are nevertheless useful for getting to a final proof or at least avoiding some cases. In particular, as you note, for any minimal element, removal of a leaf leads to a unique non-trivial automorphism of order 2 on the resulting sub-tree has led me to start thinking about trees with this latter property and how one might do something with them to build up to a proof of the original conjecture. Dec27 comment Does this poset have a unique minimal element?We conclude that some branch of $N$ must have a sub-path from $N$ of length at least 3. Again by considering automorphisms fixing $N$, some other branch must have a sub-path from $N$ of length at least 2 (otherwise all other branches from $N$ have length 1 leading to an automorphism of the tree), and finally since every branch has a sub-path from $N$ of length at least 1, we find that $E_7$ is a subgraph of every element of $\mathcal{AFT}$. Dec27 comment Does this poset have a unique minimal element?@Per: unfortunately, this set is empty. Any element of $\mathcal{AFT}$ contains a node $N$ of degree at least 3. From $N$, consider all branches emanating from it; if each has maximal length 2 to any leaf then it is a star graph (a hub node with $k$ spokes for some $k\geq 1$) with one of its spokes' terminus equal to $N$. Clearly any star branch with 3 or more spokes will admit an automorphism fixing the hub; but there are only 2 possible star branches with fewer than 3 spokes; thus any choice of these for the $\geq 3$ branches of $N$ leads to an automorphism of the tree fixing $N$. Dec26 comment Does this poset have a unique minimal element?@Pietro: thanks for the answer... I am digesting the details now, but so far everything seems to fit nicely. Dec25 awarded ● Nice Question Dec24 revised Does this poset have a unique minimal element?updated definition Dec24 asked Does this poset have a unique minimal element?