All Questions
146
questions
4
votes
3
answers
70
views
$(1-x)^{n+a} \sum_{j=0}^\infty \binom{n+j-1}{j}\binom{n+j}{a} x^j = \sum_{j=0}^a \binom{n}{a-j}\binom{a-1}{j} x^j$
Let $n$ and $a$ be natural numbers. How to prove the following for $x \in [0, 1)$?
$$
(1-x)^{n+a} \sum_{j=0}^\infty \binom{n+j-1}{j}\binom{n+j}{a} x^j = \sum_{j=0}^a \binom{n}{a-j}\binom{a-1}{j} x^j
$$...
- 9,748
6
votes
2
answers
267
views
Counting bit strings with given numbers of higher-order bit flips
Background information
Bit flips
Given a bit string, we say that bit flip happens when $0$ changes to $1$ or $1$ changes to $0$.
To find bit flips, we can shift the string by $1$ and xor that new ...
6
votes
2
answers
221
views
Is this identity I found playing around with generating function with coefficients $I(n) = \int_{0}^\pi \sin^n(x) dx$ useful and or reducible?
Let $I(n) = \int_{0}^\pi \sin^n(x) dx$ , using $\sin^2(x) = 1-\cos^2(x)$ and integrating by parts we get.
$$ \begin{align} I(n) = \dfrac{n-1}{n} I(n-2) \end{align} $$
With $I(0) = \pi$ and $I(1) = 2$ ...
- 61
1
vote
2
answers
184
views
Value of $\sum_{k=0}^n \binom{n}{k}^2$ using analysis.
I'm trying to solve this following question:
By deriving $f(x) = x^n (1+x)^n$ $n$ times, determine the value of $\sum_{k=0}^n\binom{n}{k}^2$.
My attempt on this was to express $f(x)$ in 2 ways: $f(x)...
- 187
1
vote
2
answers
71
views
Alternating hockey stick sum $\sum_{m=k}^M (-1)^m \binom{m}{k}$
I'd like to calculate the following sum efficiently
$$S(k, M) = \sum_{m=k}^M (-1)^m \binom{m}{k}$$
I've tried to use generating functions and write
$$
S(k,M) = \sum_{m=k}^M [x^k] (-1-x)^m
= [x^k] \...
- 9,748
1
vote
1
answer
101
views
Demonstrating a Binomial Identity? #2 (The exclusion)
$$
\sum_{m=1}^{\lfloor j/(k+1) \rfloor}(-1)^m\binom{n}m\binom{j-m(k+1)+n-1}{n-1}
= \sum_{m=1}^{j-k} {j-k-1 \choose m-1}{n \choose m}m
$$
(Actually $\not =$ see edit end of post)
Is there a simple ...
- 526
0
votes
1
answer
72
views
Why is $\sum_{n=0}^{\infty}{ n+1 \choose n }x^{n} = \frac{1}{(1-x)^2}$ when computing recurrences with generating functions?
Like the title says. I don't know how one gets from $G(x) = \frac{2}{1-2x} - \frac{1}{(1-x)^2}$ to $G(x) = 2\sum2^nx^n - \sum{n+1 \choose n}x^n$. I know this gives $a_{n} = 2^{n+1}-n-1$. I get the ...
- 11
1
vote
2
answers
89
views
How to prove $\sum^{m}_{k=0}(-2)^k\binom{2n-k}{n}\binom{m}{k}=0$ for $m=1,3,5,\dots,2n-1$?
I would like to prove that
$$
\sum^{m}_{k=0}(-2)^k\binom{2n-k}{n}\binom{m}{k}=0,\quad m=1,3,5,\cdots,2n-1,n\in\mathbb{N}^*.
$$
Currently I have no idea of this: expanding $\displaystyle\binom{2n-k}{n}\...
- 1,923
0
votes
1
answer
114
views
Finding the coefficient of a generating function $f$
Finding the coefficient of a generating function
$$f(x)=\dfrac{(1+x^4)(1+x^{4^2}) \cdots (1+x^{4^n})\cdots}{1-x}$$
I once searched for the problem when replacing the number $4$ with the number $2$. At ...
- 817
0
votes
0
answers
43
views
Chu-Vandermonde-like expression
For given $n,k$, I am looking to simplify
$$f(i,j)=\sum_{\ell=1}^n \left(\begin{array}{c}\ell \\ i\end{array}\right)\left(\begin{array}{c}\ell \\ j\end{array}\right)\left(\begin{array}{c}n-\ell \\ k-i\...
- 490
1
vote
0
answers
45
views
Help evaluating the series ${n/2 + k}\choose{k}$
I'm trying to find an alternate way of writing this series here:
https://en.wikipedia.org/wiki/Minimal_polynomial_of_2cos(2pi/n)
I feel like I could do it if I can come up with a nice formula for:
$$\...
- 4,284
0
votes
1
answer
102
views
Conjectured recurrence/generating function for a binomial sum
Computing a special case of a question by another user, I have noticed that apparently the OEIS A202349 sequence can be expressed as:
$$a(n)=\sum_{k=0}^{n}{\binom{n}{k}(2^k \bmod 7)} \tag{1}$$
This ...
- 2,103
2
votes
1
answer
71
views
Find the value of $\sum_{d=0}^\infty\binom{d+m-1}{m-1}\frac{x^d}{d!}$. [closed]
It is known that the generating function of the number of multisets is given by
$$
\sum_{d=0}^\infty\binom{d+m-1}{m-1}{x^d}=\frac{1}{(1-x)^m}.
$$
Then, does a similar series
$$
\sum_{d=0}^\infty\binom{...
- 21
2
votes
1
answer
102
views
Proving the identity $\sum_{k=0}^{m} \binom{n-k}{m-k}= \binom{n+1}{m}$ .
Today we started talking about generating functions in class. I came across this identity:
$$\sum_{k=0}^{m} \binom{n-k}{m-k}= \binom{n+1}{m}$$
Here is my proof for this identity (If it is wrong please ...
- 633
3
votes
3
answers
113
views
How to prove $\sum_{k=p}^n (-1)^k\binom{n}{k}\binom{k}{p} = (-1)^p$ by using generating functions
I am learning about generating functions and I tried to prove some identity using it:
$\sum_{k=p}^n (-1)^k\binom{n}{k}\binom{k}{p} = (-1)^p$ for $n=p$ and $=0$ for else.
Let $n =p$, then I don't ...
- 165