All Questions
Tagged with binomial-coefficients polynomials
7 questions with no upvoted or accepted answers
8
votes
0
answers
597
views
Does anyone know this sequence of polynomials?
A referee on a paper of mine showed me the following recurrence for polynomials $P_{n,k}\in\mathbb Q[q,q^{-1}]$ for $n\geq 0$ and $0\leq k\leq n/2$.
\begin{align}
P_{0,0}&=1\\
\text{for $n\geq 1$}...
- 18.7k
5
votes
0
answers
183
views
On the polynomials $\sum_{k=0}^n\binom{n+k}k^m q^k$
A sequence of polynomials
$$P_0(q),\ P_1(q),\ P_2(q),\ \ldots$$
with real coefficients is called $q$-log-convex if for each $n=1,2,3,\ldots$ every coefficient of the polynomial $P_{n+1}(q)P_{n-1}(q)-...
- 15.6k
4
votes
0
answers
168
views
How to prove the following equation (which involves binomials and determinant of 2×2 matrices)?
I have tried many ways to prove the following equation, such as the method of induction and expanding all the terms in the summation,but things got more complicated.I could not find an appropriate ...
- 41
4
votes
0
answers
134
views
Irreducibility of polynomials associated to binomial coefficients
Let $n \geq 2$.
Let $M_n$ be the $(n+1) \times (n+1)$ matrix with entries $\binom{l}{k}$ for $0 \leq l,k \leq n$ and $U_n=M_n + M_n^T$ and let $f_n(x)$ denote the characteristic polynomial of $U_n$.
...
- 26.5k
3
votes
0
answers
144
views
Flat polynomials with factors of big height
Let $p(x)$ be a polynomial of degree $n$ with all coefficients in $\{-1,0,1\}$ (such polynomials are sometimes called flat). I am wondering how big the coefficients of a factor of $p$ can be. Call ...
- 13.4k
1
vote
0
answers
183
views
polynomial approximation of hypergeometric function 2F1
I have the following function $T(k_1,k_2)$ resulting from multiphoton transition matrix elements calculations:
$T(k_1,k_2)=\gamma^{-k_2}\sum_{j=0}^{k_1}(j+2)_{l+1}\binom{k_1}{j}(k_1+1)_3(\gamma-1)^{j}{...
- 11
0
votes
0
answers
53
views
Interlaced binomial expansion
For $a_1,a_2,\ldots,a_n < 1$ let
$$f(a_1,a_2,\ldots,a_n) = \sum_{k_1,k_2,...,k_n = 1}^\infty \frac{(k_1 + k_2)!}{k_1! k_2!} \frac{(k_2 + k_3)!}{k_2! k_3!}\ldots \frac{(k_{n-1} + k_{n})!}{k_{n-1}! ...
- 503