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Apr
2
comment Some help with generating functions
See my answer to your previous question: if the relationship between $A(x)$ and $a_n$ is not clear from it, ask again. In short, the relationship is that $A(x) = a_0 + a_1x + a_2x^2 + a_3x^3 + \dots$, so $A(x)$ is a single object that encodes all the number $a_n$: by hanging the $n$th number $a_n$ onto the term $x^n$.
Apr
2
comment Some help with generating functions
Oh and in this current problem, it's not even the case that $a_n = n$. And $a_n$ is not $f(x)$ either. $f(x)$ is one single object that holds (encodes) all the separate $a_n$, for every $n$: it means that $f(x) = a_0 + a_1x + a_2x^2 + \dots$. And yes, you're supposed to answer in terms of $f(x)$.
Apr
2
comment Some help with generating functions
It is not right to write $a_n$ as $1 + 2 + 3 + 4\dots$. The $a_n$ that is given is a single particular fixed finite value, for any given $n$: the formula $a_n = n$ means that $a_0 = 0$ and $a_1 = 1$ and $a_2 = 2$ and $a_3 = 3$ and so on. The infinite sum $1 + 2 + 3 + \dots$ should never appear; you should write down your $A(x) = x + 2x^2 + 3x^3 + \dots$ (which is correct) directly.
Apr
2
answered How many times does a single fair die have to be rolled for a number to repeat
Apr
2
comment How to count the number of types
@eleanora: Oh don't worry about that. One thing you can do as the question asker is to click on the check mark next to some one answer to "accept" it as having answered your question, but it's best to do that after waiting for a day or two, in case you get better answers.
Apr
2
comment Is an SD card a fair coin… to me?
I didn't downvote this, and I thank you for this information and the reference (I've upvoted your comments), but this is not an answer to the question; it's only (as you admit) "a pertinent and interesting piece of information that obviously many people are unaware of". The answer area is for actual answers to the question; you could post this as a comment on the question.
Apr
2
comment Is an SD card a fair coin… to me?
I think you don't have to delete it; often people can benefit from the same thing being said in different ways.
Apr
2
comment How to count the number of types
@eleanora: These are purely analytic bounds, completely ignoring the fact that the $n_i$s must be *integers.* If you take that into account (for instance, the former $k = 1/p$ is not possible when $1/p$ is not an integer, or when that integer does not divide $n$), you can surely get slightly better bounds. But yes, I still do think they are somewhat tight, for sufficiently large $n$, and $p$ in some reasonable range.
Apr
2
answered How to count the number of types
Apr
2
revised minimum number of repetitions in a string
another form, just in case
Apr
2
comment minimum number of repetitions in a string
@user139803: Note that the expression is being raised to the power of $s$. For example, with $s = 4$ and $k = 3$, the first expression is $(1 + z + z^2/2)^4 = z^8/16+z^7/2+2 z^6+5 z^5+(17 z^4)/2+10 z^3+8 z^2+4 z+1$. In this expression, the coefficient of $z^7$ is $1/2$ for instance, so for $n = 7$, the answer is $1 - \frac{7!(1/2)}{4^7} = \frac{1733}{2048}$. Does that make it clear?
Apr
2
answered minimum number of repetitions in a string
Apr
2
comment Question about generating function of kind of fibonacci partial sum
Dividing every term by $x^2$ doesn't change $x^{2n}$ to $x^n$; it only shifts terms by $2$. You have to change $x$ to $\sqrt{x}$ instead.
Apr
2
comment Question about generating function of kind of fibonacci partial sum
@Chris: You have to, actually.
Apr
2
comment How can I find the coefficient of a term in this generating function by using the “old” method?
Nice. :-) But I think from the question that the OP was asking for a combinatorial "counting" method, not via generating functions.
Apr
2
answered Ordinary generating functions problem
Apr
2
comment Question about generating function of kind of fibonacci partial sum
@Chris: Yes, do post your answer.
Apr
2
comment Problem understanding notation
The fact that $1 + x + x^2 + x^3 + x^4 + x^5 + x^6 + \dots = \dfrac{1}{1-x}$ is a formal identity. If you want to plug in actual numbers into it, it's only valid for $|x| < 1$. Try for instance $x = \frac12$. Then $1 + \frac12 + \frac14 + \frac18 + \dots = \frac{1}{1-\frac12} = 2$. (It's the infinite sum on the left-hand side.) And a way to prove the formal identity is to call the sum $S$, and observe that $(1-x)S = (1-x)(1 + x + x^2 + x^3 + \dots) = 1-x \,+\, x-x^2 \,+\, x^2-x^3 \,+\, x^3-x^4 \,+\, \dots = 1$, so as $(1-x)S = 1$, we say $S = \dfrac{1}{1-x}$.
Apr
1
answered What is the number of compositions of the integer n such that no part is unique?
Apr
1
awarded  Generalist