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Feb
7
comment Number of ways to partition $40$ balls with $4$ colors into $4$ baskets
I hope the formatting is better now. I would be interested in seeing what can be done for five colors. How long does it take to get to order 30? Unfortunately, I suspect that one would have to go to much higher order than that to see something for five colors.
Feb
7
comment Number of ways to partition $40$ balls with $4$ colors into $4$ baskets
$$\begin{aligned}&\frac{1}{(1-x)^{10} (1+x)^5 \left(1+x+x^2\right)^3\left(1+x^2\right)}\\ &\times\left(x^{18}+6 x^{17}+58 x^{16}+213 x^{15}+646 x^{14}+1415 x^{13}+2515 x^{12}+3554 x^{11}+4296 x^{10}\right.\\ &\quad\left.+4248 x^9+3578 x^8+2452 x^7+1421 x^6+628 x^5+240 x^4+61 x^3+12 x^2-x+1\right).\end{aligned}$$
Feb
7
comment Number of ways to partition $40$ balls with $4$ colors into $4$ baskets
According to some of the linked entries on the OEIS page you cite, there is a rational generating function in the cases of two or three colors, and the denominators are products of relatively small powers of the first few cyclotomic polynomials. In hopes that that pattern continues, I tried multiplying the generating function for four colors (to the order given in the OEIS) by powers of the first four cyclotomic polynomials, and it appears to be the case that the generating function is
Feb
6
reviewed Close unsure how the 1/2 gets in this problem
Feb
6
reviewed Close Door Prizes - Probability
Feb
6
reviewed Close What is the meaning of $ \mathbb{R}^n$ to $\mathbb{R}^{n+1}$?
Feb
6
reviewed Leave Open Can you find the maximum or minimum of an equation without calculus?
Feb
6
reviewed Leave Open Number of monomials of degree $m$
Feb
6
reviewed Leave Open Second eigenvector of double eigenvalue matrix
Feb
6
reviewed Leave Open an edge coloring of $k_{16}$ with no monochromatic triangle
Feb
5
comment Understanding the hamming bound
No, $(q-1)^k$ is correct for the number of words that differ from $w$ in exactly $k$ letter positions. Suppose the alphabet is $\{0,1,2,3\}$. Then there are $3^2=9$ words that differ from the word $000000$ in the first two positions, not $3\cdot2$ such words. These nine words are $110000$, $120000$, $130000$, $210000$, $220000$, $230000$, $310000$, $320000$, $330000$. And for any of the $\binom{6}{2}$ choices of two letter positions, you will have nine words that differ from $000000$ in those positions.
Feb
5
answered Understanding the hamming bound
Jan
31
comment Using generating functions to answer how many bit strings of length N have no 000
Your work seems fine so far. The roots of the cubic in the denominator are not very nice, but can be found in principle, and then the usual partial fraction methods can be used to get a closed form involving powers of these roots. Your combinatorial analysis implies a very simple recurrence, reminiscent of the Fibonacci recurrence, which can be used to compute the coefficients in the generating function as high as you like. Maybe this isn't the kind of solution you had in mind, however.
Jan
31
comment I need a ratio of numerator below 36 to get to closest the following ratio of 38/45 =0.84444444
Well, the word "closest" was in the title. As I read it, it's a practical question, and a pretty clear one: find the whole number ratio closest to $38/45$ using numerators between $10$ and $36$. Presumably there's some constraint relating to the number of teeth on the available gears.
Jan
31
comment I need a ratio of numerator below 36 to get to closest the following ratio of 38/45 =0.84444444
@fleablood: "Closest" is not the same as equal.
Jan
31
answered I need a ratio of numerator below 36 to get to closest the following ratio of 38/45 =0.84444444
Jan
30
answered Binomial distribution, explanation formula
Jan
30
comment Prove Divisibility In Fibonacci Sequence Over A Prime Number
One should also have a look at the answers to this linked question.
Jan
30
revised Prove Divisibility In Fibonacci Sequence Over A Prime Number
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Jan
29
revised Prove Divisibility In Fibonacci Sequence Over A Prime Number
added 500 characters in body