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Jan
25
comment Proof of a Known Claim About Languages
@xavierm02, $a \in (\overline{ab})^*$.
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
14
comment Trains with cuisenaire rods
Surely the circular version is less complicated, not more? For a start, you only need to explicitly create subtrains up to length $N(N+1)/4$ because if you have a subtrain of length $x$ then by removing just that subtrain from the original circle you leave a subtrain of length $N(N+1)/2 - x$.
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 $1$.
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.
Dec
22
comment Kings on a chessboard
A more relevant OEIS sequence is A018807.
Dec
22
comment Kings on a chessboard
@Ross, only if they're at (4,5) and (5,4).
Dec
22
comment Kings on a chessboard
@Serkan, I'm not sure what you're counting there. The more obvious criticism is that the last sentence of my comment talks about "reducing" to a problem class which already contains the original problem.
Dec
20
comment Kings on a chessboard
Not a complete answer, but it might help: divide the board into 9 $2\times 2$ tiles. Each such tile can contain at most one king, so this gives an easy (but very loose) upper bound of $5^9$. Moreover, this approach reduces the problem to a 2D analogue of combinatorics on words: counting grids which avoid forbidden subgrids.
Dec
15
comment Calculate how many ways to get change of 78
What's wrong with the various programs that people supplied you in answer to your earlier question?
Dec
10
comment combinatorics - fixed point permutations
Wolfram Alpha suggests not. If there were a closed form then Petkovšek's algorithm should find it, and I would be very surprised if Alpha doesn't implement it.
Dec
7
comment Polynomials with rational zeros
In addition to saying what you've tried - homework is given to help you learn by doing - it would help to say where you encountered this. What theorems did your course just cover?
Dec
4
comment Closed form expression for unusual sum of binomial coefficients
The way computer algebra systems derive the closed form expression is by using knowledge about what the answer looks like. Specifically, if your expression has an indefinite sum then it's the term multiplied by a rational polynomial, and it's possible to bound the degrees of the numerator and denominator. See Gosper's algorithm
Dec
2
comment Determining Ambiguity in Context Free Grammars
An even simpler ambiguous string is $()$.