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visits member for 2 years, 3 months
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Oct
16
comment Is This Set a Group? Ring?
What happens when you multiply two linear functions together? Is the result another linear function?
Sep
30
answered Reflection in terms of simple reflections
Sep
23
comment The myth of no prime formula?
@Thursday Thanks, good catch!
Sep
21
comment The myth of no prime formula?
Let $P$ be the set of prime numbers. Then $\pi(x) = \sum_{x \in P} 1$. Useless formulas aren't always as obviously useless as what I just wrote. Drawing the line between such "formulas" and useful/interesting formulas is not easy. (This is discussed in Wilf's paper "What is an answer?")
Sep
19
comment Boolean algebra: Minimizing a product of sums expression?
The only difference between the first two factors is the $b$ and $b'$. Interpreting as propositional variables, if $b$ is true, the truth of the product of the first two terms is reduced to whether $a+c$ is true. Similarly if $b$ is false. Thus, intuitively, the product of the first two factors should be $a+c$. (A similar thing happens with the last two factors.) This can be verified using the Boolean algebra axioms, specifically distributivity, idempotence, $b+b'=1$, and $bb'= 0$. I don't know if there is a minimization procedure, but what I'm suggesting does reduce the number of factors.
Sep
11
comment Determining the number of colourings of regions of a pentagon by Burnside's Lemma
This looks correct.
Sep
10
awarded  Revival
Sep
10
comment Is there a combinatoric identity for the multiplicities of the following set?
Thanks. Those are all good questions. I do suspect special cases like prime powers and products of two distinct primes have a tractable distribution because the cyclotomic polynomials for those have a particular patterned form. Some analysis should be possible. Also, the distributions calculated in the question indicate some patterns exist there.
Sep
9
answered Is there a combinatoric identity for the multiplicities of the following set?
Sep
6
comment Is there a combinatoric identity for the multiplicities of the following set?
Are you sure rounding error from Mathematica isn't skewing the results? For $n=5$, if I understand your notation, $k=1$ corresponds to the vector $u_5^1 + u_5^2 + u_5^3 + u_5^4 + u_5^5$ and $k=2^5$ corresponds to $-u_5^1 - u_5^2 - u_5^3 - u_5^4 - u_5^5$. Both are the zero vector, so there is a vector in $S_5$ that occurs twice.
Aug
29
comment Interesting combinatoral identity
The right hand side is of the form $\left(1/\sqrt{1-4x^2}\right)^N * f(x)$, for some polynomial $f$ and positive integer $N$. This indicates that for fixed $T,q,l$, the right side is a finite, though complicated, manipulation of the sequence $C(2n,n)$. Since the right side is algebraic and hence D-finite/P-recursive, in principle there does exist a (probably complex) combinatorial model via dl.acm.org/citation.cfm?id=2385116. For fixed $T,q,l$, the left side also looks like it satisfies a polynomial recursion, so there may be hope. But it looks really complicated!
Aug
25
comment How to proof equality of del dot a cross b
Neither side is a vector.
Aug
20
comment Regular Group Actions and Functions
A counterexample to your guess: If $G = X = S_3$, and $G$ acts by left multiplication, then left multiplication by $(1 2)$ is a bijection $f:X \rightarrow X$, yet $(1 2) gx = g (1 2)x$ is not true for all $x$ and $g$ in $S_3$. (Any nonabelian group causes problems for the guess.)
Aug
15
comment Is det(A) maximal, if det(A+E) is maximal?
Thank you - sorry the answer (a partial one at that!) is long. The short version is that there are formulas for $\det(A+E)$ involving the matrix $A^c$ obtained by switching 1's and 0's. These can be applied to show that the conjecture is true if $M > 3m - 3$, where $M$ and $m$ are the maximal 1-2 and 0-1 determinants, respectively. When that bound doesn't hold, we've seen that $A^c$ can take on different values, so the possibility is there for the conjecture to fail. I thought I should give at least some justification for my claims, though!
Aug
14
answered Is det(A) maximal, if det(A+E) is maximal?
Jul
5
awarded  Yearling
Jun
24
comment bijective function $h:\mathbb{N}\rightarrow A\cup B$ from bijective functions $f:\mathbb{N}\rightarrow A$, $g:\mathbb{N}\rightarrow B$
The answer in the question you linked to is correct up to a minor detail/possible typo.
Jun
24
comment Prove that a function is bijective
You mean $h(n+1) = g(k)$, where $k$ is the smallest natural number such that $g(k) \not\in \{h(1),\ldots,h(n)\}$ (if $h(n) = f(m)$ for some $m$)? After all, the sets $A$ and $B$ need not be well-ordered themselves.
Jun
4
comment Relations counting in two sets
Your 2^15 should be 2^{15}. (This was my first attempted outside edit - I guess the site requires more "substantial" edits!)
Jun
3
comment angles between simple roots are obtuse, problem with proof
The negation of $(\beta,\alpha) \leq 0$ is that $(\beta, \alpha) > 0$, so you should change those weak inequalities to strict ones in your argument. The argument works nonetheless.