All Questions
Tagged with algebraic-combinatorics binomial-coefficients
7
questions
4
votes
1
answer
131
views
Is there a $q$-analog for the product of binomial coefficients?
The $q$-analog of the binomial coefficient $\binom{n}{k}$ may be defined as the coefficient of $x^k$ in $\prod_{i=0}^{n-1}(1+q^ix)$.
Classical arithmetic identities tend to have $q$-analogs. I am ...
- 1,913
0
votes
0
answers
26
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Proving Pascal's Identity for more than two terms [duplicate]
This involves a slightly different version of Pascal's Identity but is still based on it.
Prove that
$$
{a \choose k} + {a+1 \choose k} + \dots {a+n \choose k} = {a+n+1 \choose k+1}
$$
I have tried to ...
0
votes
1
answer
129
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generating function of finite sum involving $\left(\!\!\binom{n}{k}\!\!\right)$ [closed]
Notation:
$$
\left(\!\! \binom{n}{k}\!\!\right)={n+k-1 \choose k}=\frac{(n+k-1)!}{k!(n-1)!}
$$
where $n!$ is the factorial, i.e. $1\cdot 2\cdots n.$
Let $n,N\in \mathbb{Z}_{>0}.$ I'm stuck at ...
- 53
3
votes
1
answer
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Find the closed form of $\sum\limits_{k=0}^n \frac{1}{n\choose k}$: Incomplete Beta function in a combinatoric question
Recently I asked a question about the sum of $\sum_{k=0}^n {n\choose k}^p f\left(k\right)$. Then, I was thinking of the case when $p=-1, f\left(x\right)=1$, which is $\sum_{k=0}^n \dfrac{1}{n\choose k}...
- 4,043
5
votes
3
answers
391
views
"Binomial coefficients" generalized via polynomial iteration
This is a question I will answer myself immediately by repeating one of my old AoPS posts, since the latter post no longer renders on AoPS.
Convention. In the following, whenever $A$ is a ...
- 17.9k
12
votes
0
answers
393
views
How to prove this $p^{j-\left\lfloor\frac{k}{p}\right\rfloor}\mid c_{j}$
Let $p$ be a prime number and $g\in \mathbb{Z}[x]$.
Let $$\binom{x}{k}=\dfrac{x(x-1)(x-2)\cdots(x-k+1)}{k!} \in \mathbb{Q}[x]$$ for every $k \geq 0$.
Fix an integer $k$. Write the integer-valued ...
- 459
14
votes
6
answers
1k
views
Combinatorial interpretation of an alternating binomial sum
Let $n$ be a fixed natural number. I have reason to believe that $$\sum_{i=k}^n (-1)^{i-k} \binom{i}{k} \binom{n+1}{i+1}=1$$ for all $0\leq k \leq n.$ However I can not prove this. Any method to prove ...
- 547