12
votes
Accepted
A matrix identity related to Catalan numbers
After unpacking the equation $$\left( {{C(i+j,k+2)}} \right)_{i,j = 0}^{n - 1}=A_{n}G_{n,k} A_{n}^T$$
we see that we want to prove the identity
$$C(i+j,k+2)=\frac{k+2}{(2i+2j+k+2)}\binom{2i+2j+k+2}{i+...
- 85.6k
8
votes
Hankel determinant of incomplete gamma functions
Your quantity is
$$ P(n,\alpha)=\det_r A_{i,j}(n,\alpha),$$
with
$$ A_{i,j}(n,\alpha)=\int_0^\alpha t^{n+r-i-j}e^{-t}dt.$$
By the Andreief identity, this is
$$ P(n,\alpha)=\frac{1}{r!}\int_0^\alpha e^{...
- 2,552
8
votes
Determinant of identity matrix plus Hilbert matrix
One thing you can say is that your determinant is the sum of determinants of the Cauchy matrices $C_S$ for subsets $S$ of $\{1,\ldots, n\}$, where $(C_S)_{i,j} = 1/(S_i + S_j - 1)$ for $1 \le i,j \le |...
- 54.2k
7
votes
Accepted
Hankel determinants of harmonic numbers
I prove your identity $$\sum_{j=0}^n (-1)^j\binom{n}{j}\binom{n+j}{j} H_j= 2(-1)^n H_{n}$$
which you claim to imply the result.
The method is the same as here.
At first, use $(-1)^k\binom{n+k}k=\...
- 109k
7
votes
Accepted
a Hankel matrix of involution numbers
As in arXiv:0902.1650 it suffices to show that $a(n,0)=I_n$ if $a(n,j)$ satisfies $a(n,j)=a(n-1,j-1)+a(n-1,j)+(j+1)a(n-1,j+1)$ with $a(n,-1)=0$ and $a(0,j)=[j=0]$.
But it is easily verified that $a(n,...
- 5,612
7
votes
Accepted
An identity related to Hankel determinants of $\sum_{k=1}^n \frac{2^k}{k}$
Define the sequence $a_n(x):=\sum_{j=0}^n(-2)^{n-j}\binom{n}j\binom{n+j}jx^j$ so that
$r(n)=\int_0^2\frac{a_n(x)-a_n(1)}{x-1}dx$.
We need the following fact which follows from the Vandermonde-Chu ...
- 43.2k
6
votes
Accepted
Hankel determinant evaluation of special lattice paths
The $d(n):=D(n,n)$ is OEIS sequence A005773. The Hankel determinant property is given in the sequence entry. Also the recursion $\;nd_{n}=2nd_{n-1}+3(n-2)d_{n-2}$. A proof could come from a similar ...
- 2,784
5
votes
Accepted
An interesting Hankel determinant
Denote $a_n=a(n,t)$ and $b_n=b(n,t)$. To help avoiding the min function, write
$$b_n=\binom{n+3}3t^n+\sum_{j=0}^{n-1}\binom{3+j}3\left[t^{2n-j}+t^j\right].$$
Notice that $a_n=\frac{nt^{n+1}-(n+1)t^n+1}...
- 43.2k
5
votes
Some curious Hankel determinants
The identity is clearly invariant to scaling $a$ by a constant. (One has to check the constant appears in the determinant to power $n(n+1)$). So wlog we may assume $a(1)=1$.
We can expand $\log a$ ...
- 148k
5
votes
Accepted
Some more binomial coefficient determinants
Johann Cigler and I have posted a proof of many of these observations on arXiv:
"An interesting class of Hankel determinants", arXiv:1807.08330.
Let $d_r(N)=\det\left({2i+2j+r\choose i+j}\right)_{i,...
- 1,593
5
votes
Determinant of identity matrix plus Hilbert matrix
If you look at $H=\frac 1{(i+j-1)\pi}$ instead, then $\det(1 + H)\sim n^{3/8}$, as $n\to\infty$. This is basically proven in arXiv:1808.08009, arXiv:1905.03154
- 51
4
votes
Hankel determinants of harmonic numbers
As asked by Fedor Petrov I sketch the missing details.
If $a(n)$ is any sequence with $a(0)=1$, such that all Hankel determinants $M_n=\det\left(a(i+j)\right)_{i,j=0}^n$ are $ \neq 0$, define a ...
- 5,612
4
votes
Accepted
Some nice polynomials related to Hankel determinants
See Example 4 in Section 3.1.6 of https://arxiv.org/abs/1409.2562 (Federico Ardila, "Algebraic and geometric methods in enumerative combinatorics").
It explains that your $g_n(k)$ has an ...
- 24.2k
4
votes
Bijective proof of deteminant formula for Hankel matrix of central binomial coefficients
As Ira Gessel points out in his comment, the result follows easily via a weighted version of the standard LGV proof one uses for Hankel determinants of Catalan numbers. Continued fractions aren’t ...
- 19.7k
4
votes
3
votes
Accepted
Number of bounded Dyck paths with negative length as Hankel determinants
Here's how I think this can be proved based on what Richard Stanley already did in your previous question.
If we take the network in Section 3.1.6, Example 4 part (a) of https://arxiv.org/abs/1409....
- 24.2k
3
votes
Some nice polynomials related to Hankel determinants
This approach uses "Number Walls".
Given a sequence of elements
$\,a:\mathbb{Z}\to F\,$ in a field $\,F.\,$ Define the infinite matrix
of $\,n\times n\,$ Hankel determinants
$$ W_{n,m} :=\...
- 2,784
3
votes
Hankel determinants of harmonic numbers
We propose a proof (somewhat different from Fedor's) for the crucial relation
$$\sum_{j=0}^n (-1)^j\binom{n}{j}\binom{n+j}{j} H_j= 2(-1)^n H_{n}.\tag1$$
To this end, define the polynomials
$$P_n(x):=\...
- 43.2k
3
votes
Hankel determinants of binomial coefficients
This is not an answer.
Although there might be no "closed formula" for the case $p=4$, I wish to add the following to the case $p=2$ for which the determinants evaluate to $2^{n-1}$.
Suppose we ...
- 43.2k
2
votes
Hankel determinants of harmonic numbers
Identities involving harmonic numbers that are of interest for physicists, Utilitas Mathematica 83 (2010), 291-299, H. Prodinger.
This paper contains the identity (1) as well.
Now starts Johann ...
2
votes
Accepted
Evaluation of Hankel determinants for the reverse Bessel polynomials
There is a combinatorial formula for the Hankel determinant $H_n= \det\left([\varphi_{i+j}]_{i,j=0}^{n}\right)$ in terms of weighted sums of disjoint collections of Schröder paths (and also for ...
- 1,271
1
vote
Number of bounded Dyck paths with negative length as Hankel determinants
As stated in the previous question, $C_n^{(2k+1)}$ satisfies
$$\sum_{j=0}^{k+1} (-1)^j \binom{2k+2-j}{j} C_{n-j}^{(2k+1)}=0.$$
The formula
$$C_{ - n}^{(2k + 1)} = \det \left( {C_{n + 1 + i + j}^{(2k + ...
- 34.3k
1
vote
The Golay-Rudin-Shapiro sequence as “Hankel transform”
Here is an attempt at analyzing this problem. It seems better to adjust the index of the sequence, so let $\;A_n = b_k \mbox{ if } n=2^k,\; 0\;$ otherwise. Let $\;c_n\; := \det \left( {{A_{i + j}}} \...
- 2,784
1
vote
Some curious Hankel determinants
Let
$$
D_{n}(q)=\left[\prod_{k=1}^{i+j}a(q^k)\right]_{i,j=0}^{n},
$$
$$c_n(q)=a(q)...a(q^{2n})$$
and
$$ {\bf b}_{n}(q)=\left[a(q)...a(q^{n}), a(q)...a(q^{n+1}), \cdots , a(q)...a(q^{2n-1})\right]$$
...
- 977
1
vote
Some Hankel Determinants
IMHO Theorem 1 is a lucky coincidence. These Hankel determinants do not so much depend on the values of their entries but rather on the linear functionals which they generate.
More precisely: Let $...
- 5,612
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