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Feb
4
comment Which are the most effective modern intuitive definitions of a vector?
@twirlobite, "direction" is a very ineffective intuition. It makes sense in Cartesian products over fields of characteristic zero, but what would it mean in vector spaces over fields of finite characteristic?
Feb
3
answered Gossip problem why 4?
Jan
25
comment Find three $10\times10$ orthogonal Latin squares.
What's the link to Latin squares?
Jan
25
revised Proof of a Known Claim About Languages
added 489 characters in body
Jan
25
answered Proof of a Known Claim About Languages
Jan
25
comment Proof of a Known Claim About Languages
@xavierm02, $a \notin \{ab\} \implies a \in \overline{\{ab\}} \implies a \in (\overline{\{ab\}})^*$
Jan
25
comment Proof of a Known Claim About Languages
@xavierm02, $a \in (\overline{ab})^*$.
Jan
25
revised Proof of a Known Claim About Languages
added 95 characters in body
Jan
24
comment Consider the smallest number in each of the $n\choose r$ subsets (of size $r$) of $S=\{1,2,\ldots,n\}$…
This might be clearer if you parameterise $\mu$.
Jan
16
comment enumerating in pseudo random order
I think that your amended explanation corresponds to a series of permutation compositions such that if we view the counter $x$ as having digits base 4 of $abcd$ then $$f(x) = (2301)^a A (2301)^b B (2301)^c C (2301)^d D$$ and that you want to ensure that all possible permutations are generated in an order which passes some kind of pseudorandomness test. Does this seem about right?
Jan
15
comment enumerating in pseudo random order
You seem to be trying to find $n^n$ distinct permutations of $n$ items, which is impossible for $n > 1$. Have I misunderstood something?
Jan
15
comment Can 18 consecutive integers be separated into two groups,such that their product is equal?
Although note that Erdős doesn't actually prove it for products of fewer than 100 consecutive numbers: he just refers to a paper by Seimatsu Narumi, Tôhoku Math. Journal, 11 (1917), 128-142 which apparently proves cases up to 202 consecutive numbers by special arguments.
Jan
12
reviewed Close $\mathbb{Q}$ is not isomorphic to $\mathbb{Q}^+$
Jan
12
awarded  Yearling
Jan
11
comment How many combinations does Android pattern have?
Is 12364 possible by backtracking 1236(321)4?
Jan
8
comment Coefficient of $x^n$ in the series
@user103260, Sage can't simplify that sum, so it's not Zeilberger-summable, and since it's a hypergeometric summand that means you're not going to find a hypergeometric closed form.
Jan
7
comment Coefficient of $x^n$ in the series
$\binom{3n-1}{n}$ doesn't even work for $n=1$: it gives $2$ whereas $(1+x+\ldots)^1$ clearly has an $x$-coefficient of $1$.
Jan
3
answered Are the elementary symmetric polynomials “unique”?
Dec
22
comment Kings on a chessboard
@Serkan, we've both been trying to bound the wrong problem. You were trying to bound it for 3 kings and I for 0 to 9 kings, but I see now that the question asks about 6 kings, so the division into $2\times 2$ tiles gives a bound of $\binom{9}{6}4^6$.
Dec
22
comment Kings on a chessboard
Oh. I've been trying to solve the wrong problem.