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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 $()$.
Nov
25
comment Prove that every practical number is either a power of two or a power of two times a non-trivial polygonal number
@half-integerfan, I've added lists of 1000 and 1000000 practical numbers to my personal site at cheddarmonk.org/maths/practical_numbers/… and obvious substitution. I will refresh my memory on the current comment submission process for OEIS at a later date.
Nov
8
comment Representing everywhere a camera can see as a matrix
you can test a point $Q=(x_Q, y_Q, z_Q)$ against the near and far planes by simply comparing $z_Q$ to the near and far clipping distances; you can clip against the side planes by comparing $x_Q / z_Q$ to $\pm\tan (\theta_w/2)$ where $\theta_w$ is the angular width; and you can clip against the top and bottom planes by comparing $y_Q / z_Q$ to $\pm\tan (\theta_h/2)$ where $\theta_h$ is the angular height.
Nov
8
comment Representing everywhere a camera can see as a matrix
@Imray, the frustrum is bounded by 6 planes. Testing which side of a plane a point is on comes down to looking at the sign of a dot product. (E.g. if the plane is defined by a point $P$ in the plane and its normal $N$ then you can test which side a point $Q$ is by looking at the sign of $(Q-P).N$). It wouldn't surprise me, though, if actual implementations take a different approach. If you transform the world such that your camera is positioned at the origin and looks along the $z$-axis, with the up vector pointing along the $y$-axis, then (cont)
Nov
7
comment Probabilty to win in die rolling game
Thanks for editing to add your ideas. The point at which you're going wrong is to interpret the question as asking for a probability which relates to a single die roll. It's actually asking for the probability that you lose on the first die roll (which you correctly state to be $k/N$), or that you lose on your next die roll, or a subsequent one. I hope that makes it clearer what you should be recursing on.