Peter Taylor
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 Mar16 comment Mathematical Analysis Question: Cauchy sequences proof You must surely have tried something. If you show us what you've tried and why it doesn't work, you'll be able to get help which actually helps you. Mar15 comment The irreducibility of polynomials for specific cases In your second paragraph, irreducible should say reducible. Mar9 comment Selection from cliques of a graph in polynomial time Are you given that the cliques are maximal? Mar6 comment Examples of non-hamiltonian decomposable graphs Either I'm misunderstanding your notation or $C_2 \times C_3 = K_6$, which is 5-regular. Note also the statement in the second MathWorld page I linked that every Hamiltonian vertex-transitive graph with no more than 31 vertices has a Hamilton decomposition except the line graphs of the Petersen and Coxeter graphs. The next smallest known non-decomposable vertex-transitive graph has 48 vertices, so if you're looking for examples which can be checked by hand you almost certainly want to look at non-vertex-transitive graphs. Mar5 comment Examples of non-hamiltonian decomposable graphs Any non-hamiltonian graph, for a start. Or any graph whose number of edges isn't an integer multiple of its number of vertices. Can you narrow down a bit more what you're looking for? PS Mathworld's page on Hamilton decomposition may have what you want. Mar5 comment Probability of get 3 cards out of 4 One million tries seems somewhat overkill given that there are only 270725 ways of choosing 4 cards from a 52-card deck: it would probably be easy and just as fast to adapt your program to calculate the exact value you seek. But I'm not clear what it is you're calculating. What do you mean by "off-suited and off-ranked"? Feb27 comment Asymptotic for binomial coefficients @BrianM.Scott, you've lost a factor of $k^k$ in the last step of the first approximation. Feb26 comment Multinomial theorem: Number of elements where all coefficients have even powers.. An alternative way of phrasing what you're after is the elements of $(a_1^2 + \ldots + a_n^2)^{r/2}$. Feb26 comment Ball-of-wacks combinations "Never share an edge or vertex" is equivalent to "Never share a vertex", since two faces which share an edge also share two vertices. In this particular case, the structure of the faces is such that we can ignore the vertices of order 3 and the problem is equivalent to edge-colouring a regular icosahedron, which may be an easier problem to think about. Feb12 comment Find two numbers whose sum is 20 and LCM is 24 Depending on the conventions adopted for an LCM involving negative numbers, -4 and 24 could be another solution. Feb12 comment Find two numbers whose sum is 20 and LCM is 24 @AlexR, it's a question of conventions, and I'm sure there are some people who would argue that an lcm involving a negative number should be negative. But since you say it's 24, that's a counterexample to your earlier claim. Feb12 comment Find two numbers whose sum is 20 and LCM is 24 @AlexR, that depends on what you consider to be the value of $\textrm{lcm}(-4, 24)$. Feb10 comment Is the language “substrings of an even-lengthed regular language” also regular? @j6m8, I think he's answering the question in the title rather than the question in the body. Maybe you should edit the title to something like "Is the language of 50%-prefixes of words in a regular language also regular?" Feb6 comment A simple graph problem that seems NP complete An equivalent formulation is to select a set of vertices and then the score is the number of edges between two vertices in the set minus the total costs of the vertices. This suggests that it might be interesting to look at cliques, which maximise the edges/vertices ratio. Feb5 comment Of fibonomials, pellonomials, and tribonomials, etc @TitoPiezasIII, I would guess that being figurate numbers is a consequence of the figurate numbers being binomials; I can see how binomial numbers might be related to the number of different values of $\prod_i\beta_i{}^{\lambda_i}$ in the case that the $\beta_i$ are sufficiently independent. Feb1 comment $f(xy)=\frac{f(x)+f(y)}{x+y}$ Prove that $f$ is identically equal to $0$ I think the missing step is that since $f: \mathbb{R} \to \mathbb{R}$, $f(0)$ must be defined. Then $\forall x \ne 0: f(0) = \frac{f(x)+f(0)}{x}$, and in particular $f(0) = f(1)+f(0)$ whence $f(1) = 0$. Jan31 comment $f(xy)=\frac{f(x)+f(y)}{x+y}$ Prove that $f$ is identically equal to $0$ $f$ identically equal to $0$ satisfies the functional equation, so you seem to be asking the impossible. Did you mean "Prove that there exists an $f$ satisfying this functional equation which isn't identically equal to $0$"? Also, what do you mean by $f : R ->$? Jan31 comment What is the number $p(n)$ of partitions of an abundant number $n$ into distinct, proper divisors of $n$? @Travis, yes, it does. $f(n, m, k)$ counts the number of partitions of $m$ into distinct divisors of $n$ each of which is at least $k$, so the first term on the RHS counts the ones where $k$ is not in the partition, and the second term counts the ones where $k$ is in the partition - but obviously it can only be in the partition if it's a divisor of $n$. Jan30 comment Why the space of all permutations of a vector (n!) is smaller than the space of all possible permutations of a sorting network? Ah, I've just seen your earlier question. If this question is also about abusing sorting networks to generate permutations, it would help a lot to edit the question and make that clear. I've been assuming that you were trying to use the sorting network to sort. You may find this thread on another site in the StackExchange network useful. Jan30 comment Why the space of all permutations of a vector (n!) is smaller than the space of all possible permutations of a sorting network? If you have a 3-gate sorting network for 3 elements, that's 6 possible inputs and (if I understand you correctly) 8 "solutions". That's small enough that you can draw the whole lot on a piece of paper and identify the interesting ones.