Peter Taylor
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 Feb 27 comment Asymptotic for binomial coefficients @BrianM.Scott, you've lost a factor of $k^k$ in the last step of the first approximation. Feb 26 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 $(a_1^2 + \ldots + a_n^2)^{r/2}$. Feb 12 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. Feb 12 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. Feb 12 comment Find two numbers whose sum is 20 and LCM is 24 @AlexR, that depends on what you consider to be the value of $\textrm{lcm}(-4, 24)$. Feb 10 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?" Feb 6 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. Feb 5 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. Feb 1 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 $f(1) = 0$. Jan 31 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 $f : R ->$? Jan 31 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 $n$. Jan 30 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. Jan 30 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. Jan 29 comment Why the space of all permutations of a vector (n!) is smaller than the space of all possible permutations of a sorting network? I'm not quite sure what the question is, but perhaps you would find it helpful to work through the 6 permutations of 3 elements with a 3-gate sorting network. Jan 27 comment Placing $4n$ non-attaking queens of in a $4n \times 4n$ chessboard. @tone, the Wikipedia article linked by hardmath shows solutions which do place a queen in a corner of the board. Jan 19 comment Counting rings of order $p^3$ R. Raghavendran, Finite associative rings, Compositio Mathematica vol 21 no 2 (1969) p 195-229 (referenced from the OEIS page you link) claims that there are 11 rings with identity of order 8 and 12 rings with identity of order $p^3$ for odd prime $p$, so the discrepancy of $3(p-1)$ rings of order $p^3$ between the two sources presumably relates to rings which don't have an identity. Jan 17 comment Simplification of recursive polynomials Is any of the stuff in this paper useful? Jan 15 comment how to count possible planar bipartitions? Ah. At present the question says "any bipartition at all is a valid solution", so if you really want each half of the partition to be connected you should edit the text to clarify that. Jan 14 comment how to count possible planar bipartitions? It seems to me that the number of ways a graph can be bipartitioned is $2^{|V|-1}$ unless you add some constraints. Have I missed a constraint? Jan 3 comment How to denote sum over partitions? @ruadan, it depends. If you're just referring to frequencies and total then you can use the standard notation for the frequency representation and subscript the sum with $(1^{a_1}2^{a_2}\ldots)\vdash n$. If you're referring to both parts and frequencies you need to consider whether to subscript as $\lambda = (1^{a_1}2^{a_2}\ldots)\vdash n$ or just to explain in the accompanying text how the $\lambda_i$ relate to the $a_j$.