# A method of composition orthogonality and new sequences of orthogonal
polynomials and functions for non-classical weights

S. Yakubovich
Department of Mathematics, Fac. Sciences of University of Porto,
Rua do Campo Alegre, 687; 4169-007 Porto (Portugal)

###### Abstract.

A new method of composition orthogonality is introduced. It is applied to
generate new sequences of orthogonal polynomials and functions. In particular,
classical orthogonal polynomials are interpreted in the sense of composition
orthogonality. Finally, new sequences of orthogonal polynomials with respect
to the weight function
$\rho^{2}_{\nu}(x),\rho_{\nu}(x)=2x^{\nu/2}K_{\nu}(2\sqrt{x}),\ x>0,\nu\geq
0$, where $K_{\nu}(z)$ is the modified Bessel function or Macdonald function,
are investigated. Differential properties, recurrence relations, explicit
representations, generating functions and Rodrigues-type formulae are
obtained. The corresponding multiple orthogonal polynomials are exhibited.

###### Key words and phrases:

Classical orthogonal polynomials, Macdonald function, Tricomi function,
generalized hypergeometric function

###### 2000 Mathematics Subject Classification:

33C10, 42C05, 44A15

E-mail<EMAIL_ADDRESS>
The work was partially supported by CMUP, which is financed by national funds
through FCT(Portugal), under the project with reference UIDB/00144/2020.

##  1\. Introduction and preliminary results

In studying Prudnikov’s sequence of orthogonal polynomials (see [10]) with the
weight function $x^{\alpha}\rho_{\nu}(x)$, where
$\rho_{\nu}(x)=2x^{\nu/2}K_{\nu}(2\sqrt{x}),\ x>0,\nu\geq 0,\ \alpha>-1$ and
$K_{\nu}(z)$ is the Macdonald or modified Bessel function [1], Vol. II the
author interpreted this sequence in terms of the Laguerre composition
orthogonality, involving the differential operator $\theta=xDx$, where
$D=d/dx$. The main aim of this paper is to extend the method, employing
classical orthogonal polynomials (Hermite, Laguerre, Jacobi) [1] to find new
sequences of orthogonal polynomials with non-classical weights such as, for
instance, the square of Macdonald function and other hypergeometric functions.
In fact, we will define the composition orthogonality in the following way.

Definition 1. Let $\omega(t),\varphi(t)\ t\in[a,b],\ -\infty\leq
a<b\leq\infty$ be nonnegative functions, $\theta=tDt$. Let $f,g$ be complex-
valued functions such that the product of two operators $f(\theta)g(\theta)$
commutes. Then $f,\ g$ are compositionally orthogonal with respect to the
measure $\omega(t)dt$ relatively to the function $\varphi$ if

$None$ $\int_{a}^{b}f(\theta)g(\theta)\\{\varphi(t)\\}\omega(t)dt=0.$

We will see in the sequel that this definition suits well with the vector
space of polynomials over $\mathbb{R}$, when the composition orthonormality of
the sequence $\\{P_{n}\\}_{n\geq 0}$ is defined by

$None$
$\int_{a}^{b}P_{n}(\theta)P_{m}(\theta)\\{\varphi(t)\\}\omega(t)dt=\delta_{n,m},$

where $\delta_{n,m},\ n,m\in\mathbb{N}_{0}$ is the Kronecker symbol. Moreover,
for some class of functions $\varphi$ it is possible to transform the left-
hand side of the equality (1.2) to the usual orthogonality with respect to a
new weight function. Then since the $n$-th power of the operator $\theta$
satisfies the Viskov-type identity [8]

$None$ $\theta^{n}=\left(xDx\right)^{n}=x^{n}D^{n}x^{n},\quad
n\in\mathbb{N}_{0},$

we can get via integration by parts new properties of the sequence
$\\{P_{n}\\}_{n\geq 0}$ and its relationship, for instance, with classical
orthogonal polynomials. In fact, let us consider a class of functions
$\varphi$ representable by the modified Laplace transform of some nonnegative
function $\psi$

$None$ $\varphi(t)={1\over t}\int_{0}^{\infty}e^{-x/t}\psi(x)dx,\
t\in[a,b]\subset(0,\infty).$

Hence it is easily seen that

$None$
$\theta^{k}\left\\{t^{-1}e^{-x/t}\right\\}=\left(tDt\right)^{k}\left\\{t^{-1}e^{-x/t}\right\\}=x^{k}t^{-1}e^{-x/t},\quad
k\in\mathbb{N}_{0},$

and therefore, differentiating under the integral sign in (1.4), we derive

$None$ $\theta^{k}\left\\{\varphi(t)\right\\}={1\over
t}\int_{0}^{\infty}e^{-x/t}x^{k}\psi(x)dx,\quad k\in\mathbb{N}_{0}.$

It is indeed allowed, for instance, due to the assumed convergence of the
integral

$None$ $\int_{0}^{\infty}e^{-x/b}x^{k}\psi(x)dx<\infty,\quad
k\in\mathbb{N}_{0}.$

Consequently, returning to (1.2) we write its left-hand side in the form

$None$
$\int_{a}^{b}P_{n}(\theta)P_{m}(\theta)\\{\varphi(t)\\}\omega(t)dt=\int_{a}^{b}\int_{0}^{\infty}e^{-x/t}P_{n}(x)P_{m}(x)\psi(x)\omega(t){dxdt\over
t}.$

The interchange of the order of integration on the right-hand side in (1.8) is
permitted by Fubini’s theorem under the imposed condition

$None$ $\int_{a}^{b}\int_{0}^{\infty}e^{-x/t}x^{k}\psi(x)\omega(t){dxdt\over
t}<\infty,\quad k\in\mathbb{N}_{0},$

and, combining with (1.2), we find the equalities

$None$
$\int_{a}^{b}P_{n}(\theta)P_{m}(\theta)\\{\varphi(t)\\}\omega(t)dt=\int_{0}^{\infty}P_{n}(x)P_{m}(x)\psi(x)\Omega(x)dx=\delta_{n,m},$

where

$None$ $\Omega(x)=\int_{a}^{b}e^{-x/t}\omega(t){dt\over t},\quad x>0.$

Thus we see that the sequence $\\{P_{n}\\}_{n\geq 0}$ is orthonormal over
$(0,\infty)$ with respect to the measure $\Omega(x)dx$. Moreover, one can
consider the left-hand side of the first equality in (1.10) as an inner
product on the vector space of polynomials over $\mathbb{R}$ (a pre-Hilbert
space)

$None$ $\langle
p,q\rangle=\int_{a}^{b}p(\theta)q(\theta)\\{\varphi(t)\\}\omega(t)dt,$

inducing the norm by the equality

$None$ $||p||=\sqrt{\langle
p,p\rangle}=\left(\int_{0}^{\infty}p^{2}(x)\psi(x)\Omega(x)dx\right)^{1/2}.$

##  2\. The use of classical orthogonal polynomials

###  2.1. Laguerre polynomials

We begin to consider sequences of orthogonal polynomials which are
compositionally orthogonal with respect to the measure $t^{\nu}e^{-t}dt$ over
$\mathbb{R}_{+}$ related to Laguerre polynomials $\\{L_{n}^{\nu}\\}_{n\geq
0},\ \nu>-1$. In fact, letting $\omega(t)=t^{\nu}e^{-t},\ t>0$ in (1.2), we
get

$None$
$\int_{0}^{\infty}P_{n}(\theta)P_{m}(\theta)\\{\varphi(t)\\}t^{\nu}e^{-t}dt=\delta_{n,m}.$

The corresponding integral (1.11) is calculated in [1], Vol. II, and we obtain

$None$
$\Omega(x)=\int_{0}^{\infty}e^{-x/t-t}t^{\nu-1}dt=2x^{\nu/2}K_{\nu}\left(2\sqrt{x}\right)\equiv\rho_{\nu}(x),\
x>0,$

where $K_{\nu}(z)$ is the modified Bessel function or Macdonald function [9].
The function $\rho_{\nu}$ has the Mellin-Barnes integral representation in the
form (cf. [10])

$None$ $\rho_{\nu}(x)=\frac{1}{2\pi
i}\int_{\gamma-i\infty}^{\gamma+i\infty}\Gamma(\nu+s)\Gamma(s)x^{-s}ds\ ,\quad
x,\gamma\in\mathbb{R}_{+},\ \nu\in\mathbb{R},$

where $\Gamma(z)$ is Euler’s gamma-function [1], Vol. I. The asymptotic
behavior of the modified Bessel function at infinity and near the origin [1],
Vol. II gives the corresponding values for the function $\rho_{\nu},\
\nu\in\mathbb{R}$. Precisely, we have

$None$ $\rho_{\nu}(x)=O\left(x^{(\nu-|\nu|)/2}\right),\ x\to 0,\ \nu\neq
0,\quad\rho_{0}(x)=O(\log x),\ x\to 0,$

$None$ $\rho_{\nu}(x)=O\left(x^{\nu/2-1/4}e^{-2\sqrt{x}}\right),\
x\to+\infty.$

Therefore, if the condition (cf. (1.9))

$None$ $\int_{0}^{\infty}x^{k}\rho_{\nu}(x)\psi(x)dx<\infty,\quad
k\in\mathbb{N}_{0}$

holds valid, we arrive at the following proposition.

Proposition 1. Let $\nu>-1,\varphi,\psi$ be nonnegative functions defined on
$\mathbb{R}_{+}$ which are related by the modified Laplace transform $(1.4)$.
Then under condition $(2.6)$ the finiteness of the integral

$None$ $\int_{0}^{\infty}x^{k}e^{-x/M}\psi(x)dx<\infty,\quad
k\in\mathbb{N}_{0}$

for some $M>0$ the sequence $\\{P_{n}\\}_{n\geq 0}$ of orthogonal polynomials
with respect to the measure $\rho_{\nu}(x)\psi(x)dx$ over $\mathbb{R}_{+}$ is
compositionally orthogonal in the sense of Laguerre relatively to the function
$\varphi$, i.e.

$None$
$\int_{0}^{\infty}P_{n}(\theta)P_{m}(\theta)\\{\varphi(t)\\}t^{\nu}e^{-t}dt=\int_{0}^{\infty}P_{n}(x)P_{m}(x)\rho_{\nu}(x)\psi(x)dx=\delta_{n,m}.$

###### Proof.

Since via (2.7) the integral

${1\over t}\int_{0}^{\infty}e^{-x/t}x^{k}\psi(x)dx,\quad k\in\mathbb{N}_{0}$

converges uniformly with respect to $t\in[1/M,M],M>0$, the consecutive
differentiation under the integral sign is allowed, and we derive, recalling
(1.6)

$P_{n}(\theta)P_{m}(\theta)\\{\varphi(t)\\}=P_{n}(\theta)P_{m}(\theta)\left\\{{1\over
t}\int_{0}^{\infty}e^{-x/t}\psi(x)dx\right\\}$

$={1\over t}\int_{0}^{\infty}e^{-x/t}P_{n}(x)P_{m}(x)\psi(x)dx.$

Then

$\int_{0}^{\infty}P_{n}(\theta)P_{m}(\theta)\\{\varphi(t)\\}t^{\nu}e^{-t}dt=\lim_{M\to\infty}\int_{1/M}^{M}\int_{0}^{\infty}e^{-x/t}P_{n}(x)P_{m}(x)\psi(x)t^{\nu-1}e^{-t}dxdt$

$=\int_{0}^{\infty}P_{n}(x)P_{m}(x)\rho_{\nu}(x)\psi(x)dx=\delta_{n,m},$

where the latter equality is guaranteed by condition (2.6) and Fubini’s
theorem.

∎

Remark 1. Letting $\nu\geq 0,\psi(x)=x^{\alpha},\ \alpha>-1$, we find the
sequence $\\{P^{\nu,\alpha}_{n}\\}_{n\geq 0}$ of Prudnikov’s orthogonal
polynomials studied in [10]

$None$
$\int_{0}^{\infty}P^{\nu,\alpha}_{n}(x)P^{\nu,\alpha}_{m}(x)\rho_{\nu}(x)x^{\alpha}dx=\Gamma(\alpha+1)\int_{0}^{\infty}P_{n}(\theta)P_{m}(\theta)\\{t^{\alpha}\\}t^{\nu}e^{-t}dt=\delta_{n,m}.$

###  2.2. Hermite polynomials

Other interesting case is the Hermite orthogonality with respect to the
measure $e^{-t^{2}}dt$ over $\mathbb{R}$. Our method will work on the
following even extension of the modified Laplace transform (1.4)

$None$ $\varphi(|t|)={1\over|t|}\int_{0}^{\infty}e^{-x/|t|}\psi(x)dx,\
t\in\mathbb{R}\backslash{\\{0\\}}.$

Hence when the Parseval identity for the Mellin transform [6] suggests the
equality

$None$ $\int_{0}^{\infty}e^{-x/t-t^{2}}{dt\over t}={1\over 4\pi
i}\int_{\gamma-i\infty}^{\gamma+i\infty}\Gamma\left({s\over
2}\right)\Gamma(s)\ x^{-s}ds,\quad x,\gamma>0.$

Invoking the duplication formula for the gamma function [1], we get the right-
hand side of (2.11) in the form

${1\over 4\pi i}\int_{\gamma-i\infty}^{\gamma+i\infty}\Gamma\left({s\over
2}\right)\Gamma(s)\ x^{-s}ds={1\over
4\pi^{3/2}i}\int_{\gamma/2-i\infty}^{\gamma/2+i\infty}\Gamma^{2}\left(s\right)\Gamma\left({1\over
2}+s\right)\ \left({x\over 2}\right)^{-2s}ds$

$None$ $={1\over 2\sqrt{\pi}}\ \rho_{1/2,2}\left({x^{2}\over 4}\right),$

where by $\rho_{\nu,k}(x)$ we denote the ultra-exponential weight function
introduced in [10]

$None$ $\rho_{\nu,k}(x)={1\over 2\pi
i}\int_{\gamma-i\infty}^{\gamma+i\infty}\Gamma^{k}\left(s\right)\Gamma\left(\nu+s\right)x^{-s}ds,\
x,\nu,\gamma>0,\ k\in\mathbb{N}_{0}.$

Proposition 2. Let $t>0,\ x\in\mathbb{R},\ \varphi(t),\psi(x)$ be nonnegative
functions which are related by $(2.10)$. If $\psi$ is even then under
conditions $(2.7)$ and

$None$ $\int_{-\infty}^{\infty}|x|^{k}\rho_{1/2,2}\left({x^{2}\over
4}\right)\psi(x)dx<\infty,\quad k\in\mathbb{N}_{0}$

the sequence $\\{P_{n}\\}_{n\geq 0}$ of orthogonal polynomials with respect to
the measure $2^{-1}\pi^{-1/2}\rho_{1/2,2}\left({x^{2}/4}\right)\psi(x)dx$ over
$\mathbb{R}$ is compositionally orthogonal in the sense of Hermite relatively
to the function $\varphi(|t|)$, i.e.

$None$
$\int_{-\infty}^{\infty}P_{n}(\theta)P_{m}(\theta)\\{\varphi(|t|)\\}e^{-t^{2}}dt=\int_{-\infty}^{\infty}P_{n}(x)P_{m}(x)\rho_{1/2,2}\left({x^{2}\over
4}\right)\psi(x){dx\over\sqrt{\pi}}=\delta_{n,m}.$

###### Proof.

Indeed, since

$\int_{-\infty}^{\infty}P_{n}(\theta)P_{m}(\theta)\\{\varphi(|t|)\\}e^{-t^{2}}dt=\int_{0}^{\infty}\left[P_{n}(\theta)P_{m}(\theta)+P_{n}(-\theta)P_{m}(-\theta)\right]\\{\varphi(t)\\}e^{-t^{2}}dt$

we have due to (1.4), (1.6), (2.7)

$None$
$\left[P_{n}(\theta)P_{m}(\theta)+P_{n}(-\theta)P_{m}(-\theta)\right]\\{\varphi(t)\\}={1\over
t}\int_{0}^{\infty}e^{-x/t}\left[P_{n}(x)P_{m}(x)+P_{n}(-x)P_{m}(-x)\right]\psi(x)dx.$

Thus appealing to (2.11), (2.12), (2.14) and Fubini’s theorem, we obtain
finally from (2.16)

$\int_{0}^{\infty}\int_{0}^{\infty}e^{-x/t-t^{2}}\left[P_{n}(x)P_{m}(x)+P_{n}(-x)P_{m}(-x)\right]\psi(x){dxdt\over
t}$

$=\int_{0}^{\infty}\left[P_{n}(x)P_{m}(x)+P_{n}(-x)P_{m}(-x)\right]\rho_{1/2,2}\left({x^{2}\over
4}\right)\psi(x){dx\over\sqrt{\pi}}$

$=\int_{-\infty}^{\infty}P_{n}(x)P_{m}(x)\rho_{1/2,2}\left({x^{2}\over
4}\right)\psi(x){dx\over 2\sqrt{\pi}}=\delta_{n,m}.$

∎

###  2.3. Jacobi polynomials

The modified sequence $\\{P_{n}^{\alpha,\beta}(2t-1)\\}_{n\geq 0}$ of these
classical polynomials is orthogonal with respect to the measure
$(1-t)^{\alpha}t^{\beta}dt,\ \alpha,\beta>-1$ over the interval $[0,1].$
Therefore the kernel $\Omega(x),\ x>0$ (1.11) is calculated accordingly, and
we obtain

$None$
$\Omega(x)=\int_{0}^{1}(1-t)^{\alpha}t^{\beta-1}e^{-x/t}dt=e^{-x}\int_{0}^{\infty}t^{\alpha}\
(1+t)^{-\alpha-\beta-1}e^{-xt}dt=\Gamma(1+\alpha)e^{-x}U(1+\alpha,1-\beta,x),$

where $U(a,b,z)$ is the Tricomi function [4]. Hence we arrive at

Proposition 3. Let $\alpha>-1,\beta>0,\varphi,\psi$ be nonnegative functions
defined on $\mathbb{R}_{+}$ which are related by the modified Laplace
transform $(1.4)$. Then under the condition

$None$ $\int_{0}^{\infty}x^{k}e^{-x}\psi(x)dx<\infty,\quad k\in\mathbb{N}_{0}$

the sequence $\\{P_{n}\\}_{n\geq 0}$ of orthogonal polynomials with respect to
the measure $e^{-x}U(1+\alpha,1-\beta,x)\psi(x)dx$ over $\mathbb{R}_{+}$ is
compositionally orthogonal in the sense of Jacobi relatively to the function
$\varphi$, i.e.

$None$
$\int_{0}^{1}P_{n}(\theta)P_{m}(\theta)\\{\varphi(t)\\}(1-t)^{\alpha}t^{\beta}dt=\Gamma(1+\alpha)\int_{0}^{\infty}P_{n}(x)P_{m}(x)e^{-x}U(1+\alpha,1-\beta,x)\psi(x)dx=\delta_{n,m}.$

###### Proof.

Indeed, since

$P_{n}(\theta)P_{m}(\theta)\\{\varphi(t)\\}=P_{n}(\theta)P_{m}(\theta)\left\\{{1\over
t}\int_{0}^{\infty}e^{-x/t}\psi(x)dx\right\\}={1\over
t}\int_{0}^{\infty}e^{-x/t}P_{n}(x)P_{m}(x)\psi(x)dx,$

where the consecutive differentiation under the integral sign is permitted
owing to the estimate

${1\over
t}\int_{0}^{\infty}e^{-x/t}x^{k}\psi(x)dx\leq{1\over\delta}\int_{0}^{\infty}e^{-x}x^{k}\psi(x)dx,\
0<\delta\leq t\leq 1,\ k\in\mathbb{N}_{0},$

and the latter integral by $x$ is finite via (2.18). Hence, taking into
account (2.17),

$\int_{0}^{1}P_{n}(\theta)P_{m}(\theta)\\{\varphi(t)\\}(1-t)^{\alpha}t^{\beta}dt=\lim_{\delta\to
0+}\int_{\delta}^{1}P_{n}(\theta)P_{m}(\theta)\\{\varphi(t)\\}(1-t)^{\alpha}t^{\beta}dt$

$=\lim_{\delta\to
0+}\int_{\delta}^{1}\int_{0}^{\infty}e^{-x/t}P_{n}(x)P_{m}(x)\psi(x)(1-t)^{\alpha}t^{\beta-1}dxdt$

$=\int_{0}^{1}\int_{0}^{\infty}e^{-x/t}P_{n}(x)P_{m}(x)\psi(x)(1-t)^{\alpha}t^{\beta-1}dxdt$

$=\Gamma(1+\alpha)\int_{0}^{\infty}P_{n}(x)P_{m}(x)e^{-x}U(1+\alpha,1-\beta,x)\psi(x)dx=\delta_{n,m},$

where the interchange of the order of integration is possible by Fubini’s
theorem due to (2.18) and an elementary estimate of the Tricomi function

$\Gamma(1+\alpha)U(1+\alpha,1-\beta,x)=\int_{0}^{\infty}t^{\alpha}\
(1+t)^{-\alpha-\beta-1}e^{-xt}dt\leq\int_{0}^{\infty}t^{\alpha}\
(1+t)^{-\alpha-\beta-1}dt=B(1+\alpha,\beta),$

where $B(a,b)$ is the Euler beta function [1], Vol. I.

∎

##  3\. Properties of Prudnikov’s weight functions and their products

In this section we will exhibit properties of the weight functions
$\rho_{\nu+1}(x)\rho_{\nu}(x),\ \rho_{\nu}^{2}(x),\ x>0$, where
$\rho_{\nu}(x)$ is defined by (2.2), (2.3), in order to involve them in the
sequel to investigate the corresponding orthogonal and multiple orthogonal
polynomials. In particular, we will establish their differential properties,
integral representations and differential equations. Concerning the function
$\rho_{\nu}$, we found in [10] the following integral representation in terms
of Laguerre polynomials

$None$ ${(-1)^{n}x^{n}\over n!}\
\rho_{\nu}(x)=\int_{0}^{\infty}t^{\nu+n-1}e^{-t-x/t}L_{n}^{\nu}(t)dt,\quad
n\in\mathbb{N}_{0}.$

It has a relationship with the Riemann-Liouville fractional integral [9]

$None$
$\left(I_{-}^{\alpha}f\right)(x)\equiv\left(I_{-}^{\alpha}f(x)\right)={1\over\Gamma(\alpha)}\int_{x}^{\infty}(t-x)^{\alpha-1}f(t)dt,\quad{\rm
Re}\alpha>0,$

namely, we get the formula

$None$ $\rho_{\nu}(x)=\left(I_{-}^{\nu}\rho_{0}\right)(x),\ \nu>0.$

Further, the index law for fractional integrals immediately implies

$None$
$\rho_{\nu+\mu}(x)=\left(I_{-}^{\nu}\rho_{\mu}\right)(x)=\left(I_{-}^{\mu}\rho_{\nu}\right)(x).$

The corresponding definition of the fractional derivative presumes the
relation $D^{\mu}_{-}=-DI_{-}^{1-\mu}$. Hence for the ordinary $n$-th
derivative of $\rho_{\nu}$ we find

$None$ $D^{n}\rho_{\nu}(x)=(-1)^{n}\rho_{\nu-n}(x),\quad n\in\mathbb{N}_{0}.$

The function $\rho_{\nu}$ possesses the following recurrence relation (see
[10])

$None$
$\rho_{\nu+1}(x)=\nu\rho_{\nu}(x)+x\rho_{\nu-1}(x),\quad\nu\in\mathbb{R}.$

In the operator form it can be written as follows

$None$ $\rho_{\nu+1}(x)=\left(\nu-xD\right)\rho_{\nu}(x).$

Now we will derive the Mellin-Barnes representations for the product of
functions $\rho_{\nu+1}\rho_{\nu}$ and for the square $\rho_{\nu}^{2}$. In
fact, it can be done, using the related formulas for the product of Macdonald
functions. Thus, appealing to Entries 8.4.23.31, 8.4.23.27 in [5], Vol. III,
we obtain

$None$ $\rho_{\nu+1}(x)\rho_{\nu}(x)={4^{-\nu}\over
2i\sqrt{\pi}}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{\Gamma(s+2\nu+1)\Gamma(s+\nu)\Gamma(s)}{\Gamma(s+\nu+1/2)}(4x)^{-s}ds,\quad
x,\gamma>0,$

$None$ $\rho^{2}_{\nu}(x)={4^{-\nu}\over
i\sqrt{\pi}}\int_{\gamma-i\infty}^{\gamma+i\infty}\frac{\Gamma(s+2\nu)\Gamma(s+\nu)\Gamma(s)}{\Gamma(s+\nu+1/2)}(4x)^{-s}ds,\quad
x,\gamma>0.$

Immediate consequences of these formulas are relationships of the products of
Prudnikov’s weight functions $\rho_{\nu+1}\rho_{\nu},\ \rho_{\nu}^{2}$ with
$\rho_{2\nu+1},\ \rho_{2\nu}$, correspondingly. Indeed, it can be done via
Entry 8.4.2.3 in [5], Vol. III and the Parseval equality for the Mellin
transform. Hence we deduce from (3.8), (3.9), respectively,

$\rho_{\nu+1}(x)\rho_{\nu}(x)=4^{-\nu}\int_{0}^{1}(1-t)^{-1/2}t^{\nu-1}\rho_{2\nu+1}\left({4x\over
t}\right)dt$

$None$
$={x^{\nu}\sqrt{\pi}\over\Gamma(\nu+1/2)}\int_{0}^{1}(1-t)^{\nu-1/2}t^{-\nu-1}\rho_{\nu+1}\left({4x\over
t}\right)dt,$

$\rho^{2}_{\nu}(x)=2^{1-2\nu}\int_{0}^{1}(1-t)^{-1/2}t^{\nu-1}\rho_{2\nu}\left({4x\over
t}\right)dt$

$None$
$={2x^{\nu}\sqrt{\pi}\over\Gamma(\nu+1/2)}\int_{0}^{1}(1-t)^{\nu-1/2}t^{-\nu-1}\rho_{\nu}\left({4x\over
t}\right)dt.$

An ordinary differential equation for the function $\rho_{\nu+1}\rho_{\nu}$ is
given by

Proposition 4. The function $u_{\nu}=\rho_{\nu+1}\rho_{\nu}$ satisfies the
following third order differential equation

$None$ $x^{2}{d^{3}u_{\nu}\over dx^{3}}+x(2-3\nu){d^{2}u_{\nu}\over
dx^{2}}+2(\nu(\nu-1)-2x){du_{\nu}\over dx}+2(2\nu-1)u_{\nu}(x)=0.$

###### Proof.

Differentiating $u_{\nu}$ with the use of (3.5), we have

$None$ ${du_{\nu}\over dx}=-\rho_{\nu}^{2}(x)-\rho_{\nu+1}(x)\rho_{\nu-1}(x).$

Multiplying both sides of (3.13) by $x$ and employing the recurrence relation
(3.6), we get

$x{du_{\nu}\over
dx}=-x\rho_{\nu}^{2}(x)+\nu\rho_{\nu+1}(x)\rho_{\nu}(x)-\rho^{2}_{\nu+1}(x).$

Differentiating both sides of the latter equality, using (3.6), (3.13) and the
notation $u_{\nu}=\rho_{\nu+1}\rho_{\nu}$, we obtain

$None$ ${d\over dx}\left(x{du_{\nu}\over
dx}\right)=-(1+2\nu)\rho_{\nu}^{2}(x)+\nu{du_{\nu}\over dx}+4u_{\nu}(x).$

Differentiating (3.14) and multiplying the result by $x$, we recall (3.6) to
find

$None$ $x{d^{2}\over dx^{2}}\left(x{du_{\nu}\over
dx}\right)=-2\nu(1+2\nu)\rho^{2}_{\nu}(x)+\nu x{d^{2}u_{\nu}\over
dx^{2}}+4x{du_{\nu}\over dx}+2(1+2\nu)u_{\nu}(x).$

Hence, expressing $(1+2\nu)\rho^{2}_{\nu}(x)$ from (3.14) and fulfilling the
differentiation on the left-hand side of (3.15), we get (3.12).

∎

Concerning the differential equation for the function $\rho_{\nu}^{2}$, we
prove

Proposition 5. The function $h_{\nu}=\rho^{2}_{\nu}$ satisfies the following
third order differential equation

$None$ $x^{2}{d^{3}h_{\nu}\over dx^{3}}+3x(1-\nu){d^{2}h_{\nu}\over
dx^{2}}+(2\nu^{2}+1-3\nu-4x){dh_{\nu}\over dx}+2(2\nu-1)h_{\nu}(x)=0.$

###### Proof.

Since
$x\left(\rho^{2}_{\nu}(x)\right)^{\prime}=-2x\rho_{\nu-1}(x)\rho_{\nu}(x)=2\nu\rho^{2}_{\nu}(x)-2\rho_{\nu+1}(x)\rho_{\nu}(x)=2\nu
h_{\nu}(x)-2\rho_{\nu+1}(x)\rho_{\nu}(x)$ we derive, owing to (3.5), (3.6)

$x{d\over dx}\left(x{dh_{\nu}\over dx}-2\nu h_{\nu}(x)\right)=-2x{d\over
dx}\left(\rho_{\nu+1}(x)\rho_{\nu}(x)\right)$

$=2x\
h_{\nu}(x)-2\nu\rho_{\nu+1}(x)\rho_{\nu}(x)+2\rho_{\nu+1}^{2}(x)=2\left(x-\nu^{2}\right)h_{\nu}(x)+\nu
x{dh_{\nu}\over dx}+2\rho_{\nu+1}^{2}(x).$

Hence one more differentiation yields

${d\over dx}\left(x{d\over dx}\left(x{dh_{\nu}\over dx}-2\nu
h_{\nu}(x)\right)\right)={d\over
dx}\left(2\left(x-\nu^{2}\right)h_{\nu}(x)+\nu x{dh_{\nu}\over
dx}\right)-4\rho_{\nu+1}(x)\rho_{\nu}(x).$

But from the beginning of the proof we find

$None$ $2\rho_{\nu+1}(x)\rho_{\nu}(x)=2\nu h_{\nu}(x)-x{dh_{\nu}\over dx}.$

Therefore, substituting this expression into the previous equality, we fulfil
the differentiation to arrive at (3.16) and to complete the proof.

∎

Corollary 1. The following recurrence relations between functions $u_{\nu},\
h_{\nu}$ hold

$None$ $u_{\nu}=\nu h_{\nu}+xu_{\nu-1},$

$None$ $h_{\nu+1}=\nu^{2}h_{\nu}+2x\nu u_{\nu-1}+x^{2}h_{\nu-1}=2\nu
u_{\nu}+x^{2}h_{\nu-1}-\nu^{2}h_{\nu}.$

###### Proof.

Equality (3.18) is a direct consequence of (3.17) and (3.5). Equalities
(3.19), in turn, are obtained, taking squares of both sides of (3.6) and
employing (3.18).

∎

##  4\. Orthogonal polynomials with $\rho^{2}_{\nu}$ weight function

The object of this section is to characterize the sequence of orthogonal
polynomials $\\{P_{n}\\}_{n\geq 0}$, satisfying the orthogonality conditions

$None$
$\int_{0}^{\infty}P_{m}(x)P_{n}(x)\rho^{2}_{\nu}(x)dx=\delta_{n,m},\quad\nu>-{1\over
2}.$

Clearly, up to a normalization factor conditions (4.1) are equivalent to the
equalities

$None$ $\int_{0}^{\infty}P_{n}(x)\rho^{2}_{\nu}(x)x^{m}dx=0,\quad
m=0,1,\dots,n-1,\ \quad n\in\mathbb{N}.$

The moments of the weight $\rho^{2}_{\nu}(x)$ can be obtained immediately from
the Mellin-Barnes representation (3.9), treating it as the inverse Mellin
transform [6]. Hence we get

$None$ $\int_{0}^{\infty}\rho^{2}_{\nu}(x)x^{\mu}dx=\sqrt{\pi}\
\frac{\Gamma(1+\mu+2\nu)\Gamma(1+\mu+\nu)\Gamma(1+\mu)}{2^{1+2(\nu+\mu)}\
\Gamma(\mu+\nu+3/2)},\ \mu>\max\\{-1,-1-\nu,\ 1-2\nu\\}.$

Furthermore, the sequence $\left\\{P_{n}\right\\}_{n\geq 0}$ satisfies the
3-term recurrence relation in the form

$None$ $xP_{n}(x)=A_{n+1}P_{n+1}(x)+B_{n}P_{n}(x)+A_{n}P_{n-1}(x),$

where $P_{-1}^{\nu}(x)\equiv 0,\ P_{n}(x)=\sum_{k=0}^{n}a_{n,k}x^{k},\
a_{n,n}\neq 0$ and

$A_{n+1}={a_{n}\over a_{n+1}},\quad B_{n}={b_{n}\over a_{n}}-{b_{n+1}\over
a_{n+1}},\quad a_{n}\equiv a_{n,n},\ b_{n}\equiv a_{n,n-1}.$

Basing on properties of the functions $\rho^{2}_{\nu},\
\rho_{\nu+1}\rho_{\nu}$ and orthonormality (4.1), we establish

Proposition 6. Let $\nu>-1$. The following formulas hold

$None$ $\int_{0}^{\infty}P^{2}_{n}(x)\rho_{\nu+1}(x)\rho_{\nu}(x)dx={1\over
2}+\nu+n,$

$None$ $\int_{0}^{\infty}P^{2}_{n}(x)\ x\rho_{\nu}(x)\rho_{\nu-1}(x)dx={1\over
2}+n,$

$None$ $\int_{0}^{\infty}P^{2}_{n}(x)\
x^{2}\rho_{\nu}(x)\rho_{\nu-2}(x)dx=B_{n}-(\nu-1)\left({1\over 2}+n\right),$

$None$
$\int_{0}^{\infty}P^{2}_{n}(x)\rho_{\nu+2}(x)\rho_{\nu}(x)dx=B_{n}+(\nu+1)\left({1\over
2}+\nu+n\right).$

###### Proof.

In fact, recurrence relation (3.6) and orthonormality (4.1) implies

$None$
$\int_{0}^{\infty}P^{2}_{n}(x)\rho_{\nu+1}(x)\rho_{\nu}(x)dx=\nu+\int_{0}^{\infty}P^{2}_{n}(x)\
x\rho_{\nu}(x)\rho_{\nu-1}(x)dx.$

Hence, integrating by parts in the latter integral and eliminating integrated
terms by virtue of the asymptotic behavior (2.4), (2.5) , we find

$None$ $\int_{0}^{\infty}P^{2}_{n}(x)\
x\rho_{\nu}(x)\rho_{\nu-1}(x)dx=\int_{0}^{\infty}\left({1\over
2}P^{2}_{n}(x)+xP_{n}(x)P_{n}^{\prime}(x)\right)\rho^{2}_{\nu}(x)dx.$

Therefore, we have from (4.9), (4.10) and orthogonality (4.1)

$\int_{0}^{\infty}P^{2}_{n}(x)\rho_{\nu+1}(x)\rho_{\nu}(x)dx={1\over
2}+\nu+n,$

which proves (4.5). Concerning equality (4.6), it is a direct consequence of
(3.6), (4.1). The same idea is to prove (4.7), (4.8), employing (4.3) as well.

∎

The composition orthogonality which is associated with the sequence
$\\{P_{n}\\}_{n\geq 0}$ is given by

Theorem 1. Let $\nu>-1/2$. The sequence $\\{P_{n}\\}_{n\geq 0}$ is
compositionally orthogonal in the sense of Laguerre relatively to the function
$t^{\nu}e^{t}\Gamma(-\nu,t)$, where $\Gamma(\mu,z)$ is the incomplete gamma
function, i.e.

$None$ $\
\int_{0}^{\infty}t^{\nu}e^{-t}P_{n}\left(\theta\right)P_{m}\left(\theta\right)\left(t^{\nu}e^{t}\Gamma(-\nu,t)\right)dt={\delta_{n,m}\over\Gamma(1+\nu)}.$

###### Proof.

Writing $P_{n}$ explicitly, we appeal to the integral representation (3.1) for
$\rho_{\nu}(x)$ to write the left-hand side of (4.2) in the form

$\int_{0}^{\infty}P_{n}(x)\rho^{2}_{\nu}(x)x^{m}dx=(-1)^{m}m!\sum_{k=0}^{n}a_{n,k}(-1)^{k}k!\int_{0}^{\infty}\int_{0}^{\infty}t^{\nu+k-1}e^{-t-x/t}L_{k}^{\nu}(t)dt$

$None$ $\times\int_{0}^{\infty}y^{\nu+m-1}e^{-y-x/y}L_{m}^{\nu}(y)dydx.$

The Fubini theorem allows to interchange the order of integration in (4.12)
due to the convergence of the following integral for all
$r_{1},r_{2}\in\mathbb{N}_{0}$

$\int_{0}^{\infty}\int_{0}^{\infty}t^{\nu+k+r_{1}-1}e^{-t-x/t}dt\int_{0}^{\infty}y^{\nu+m+r_{2}-1}e^{-y-x/y}dy\
dx$

$=\int_{0}^{\infty}\int_{0}^{\infty}t^{\nu+k+r_{1}}y^{\nu+m+r_{2}}e^{-t-y}{dtdy\over
t+y}\leq{1\over
2}\int_{0}^{\infty}\int_{0}^{\infty}t^{\nu+k+r_{1}-1/2}y^{\nu+m+r_{2}-1/2}e^{-t-y}dtdy$

$={1\over 2}\Gamma(\nu+k+r_{1}+1/2)\Gamma(\nu+m+r_{2}+1/2).$

Therefore after integration with respect to $x$ we find from (4.12)

$None$
$\int_{0}^{\infty}P_{n}(x)\rho^{2}_{\nu}(x)x^{m}dx=(-1)^{m}m!\sum_{k=0}^{n}a_{n,k}(-1)^{k}k!\int_{0}^{\infty}\int_{0}^{\infty}t^{\nu+k}y^{\nu+m}e^{-t-y}L_{k}^{\nu}(t)L_{m}^{\nu}(y){dtdy\over
t+y}.$

However the Rodrigues formula for Laguerre polynomials and Viskov-type
identity (1.3) suggest to write the right-hand side of (4.13) as follows

$(-1)^{m}m!\sum_{k=0}^{n}a_{n,k}(-1)^{k}k!\int_{0}^{\infty}\int_{0}^{\infty}t^{\nu+k}y^{\nu+m}e^{-t-y}L_{k}^{\nu}(t)L_{m}^{\nu}(y){dtdy\over
t+y}$

$None$
$=(-1)^{m}\sum_{k=0}^{n}a_{n,k}(-1)^{k}k!\int_{0}^{\infty}t^{\nu+k}e^{-t}L_{k}^{\nu}(t)\int_{0}^{\infty}\theta^{m}\left(y^{\nu}e^{-y}\right){dydt\over
t+y}.$

The inner integral with respect to $y$ on the right-hand side of the latter
equality can be treated via integration by parts, and we obtain

$None$ $\int_{0}^{\infty}\theta^{m}\left(y^{\nu}e^{-y}\right){dy\over
t+y}=(-1)^{m}\int_{0}^{\infty}y^{\nu}e^{-y}\theta^{m}\left({1\over
t+y}\right)dy.$

Working out the differentiation, we find

$\theta^{m}\left({1\over t+y}\right)=y^{m}{d^{m}\over
dy^{m}}\left(\sum_{k=0}^{m}\binom{m}{k}(-1)^{m-k}t^{m-k}(t+y)^{k-1}\right)$

$=(-1)^{m}t^{m}y^{m}{d^{m}\over dy^{m}}\left({1\over t+y}\right)={m!\
t^{m}y^{m}\over(t+y)^{m+1}}.$

Consequently, combining with (4.14), (4.15), equality (4.13) becomes

$None$
$\int_{0}^{\infty}P_{n}(x)\rho^{2}_{\nu}(x)x^{m}dx=m!\sum_{k=0}^{n}a_{n,k}(-1)^{k}k!\int_{0}^{\infty}t^{\nu+k+m}e^{-t}L_{k}^{\nu}(t)\int_{0}^{\infty}{y^{\nu+m}e^{-y}\over(t+y)^{m+1}}dydt.$

Meanwhile, the integral with respect to $y$ in (4.16) is calculated in (2.17)
in terms of the Tricomi function up to a simple change of variables. Thus we
get

$None$
$\int_{0}^{\infty}P_{n}(x)\rho^{2}_{\nu}(x)x^{m}dx=m!\Gamma(\nu+m+1)\sum_{k=0}^{n}a_{n,k}(-1)^{k}k!\int_{0}^{\infty}t^{2\nu+k+m}e^{-t}L_{k}^{\nu}(t)U(\nu+m+1,1+\nu,t)dt.$

But appealing to the differential formula 13.3.24 in [4] for the Tricomi
function, we have

$m!\
\Gamma(\nu+m+1)t^{\nu+m}U(\nu+m+1,1+\nu,t)=\Gamma(\nu+1)\theta^{m}\left(t^{\nu}U(1+\nu,1+\nu,t)\right)$

$None$ $=\Gamma(\nu+1)\theta^{m}\left(t^{\nu}e^{t}\Gamma(-\nu,t)\right),$

where $\Gamma(a,z)$ is the incomplete gamma function

$None$ $\Gamma(a,z)=\int_{z}^{\infty}e^{-u}u^{a-1}du.$

Hence, plugging this result in (4.17), recalling the Rodrigues formula for
Laguerre polynomials and integrating by parts on its right-hand side, we
combine with (4.2) to end up with equalities

$\int_{0}^{\infty}P_{n}(x)\rho^{2}_{\nu}(x)x^{m}dx=\Gamma(\nu+1)$

$None$
$\times\int_{0}^{\infty}t^{\nu}e^{-t}P_{n}\left(\theta\right)\theta^{m}\left(t^{\nu}e^{t}\Gamma(-\nu,t)\right)dt=0,\quad
m=0,1,\dots,n-1,\ n\in\mathbb{N},$

which yield (4.11) and complete the proof of Theorem 1.

∎

Further, the right-hand side of the first equality in (4.20) can be rewritten
via integration by parts as follows

$None$
$\int_{0}^{\infty}t^{\nu+m}e^{-t}L_{m}^{\nu}(t)P_{n}\left(\theta\right)\left(t^{\nu}e^{t}\Gamma(-\nu,t)\right)dt=0,\quad
m=0,1,\dots,n-1,\ n\in\mathbb{N}.$

Then we expand the function
$F_{n}(t)=\theta^{n}\left(t^{\nu}e^{t}\Gamma(-\nu,t)\right)$ in a series of
Laguerre polynomials

$None$ $F_{n}(t)=\sum_{r=0}^{\infty}c_{n,r}L_{r}^{\nu}(t),$

where the coefficients $c_{n,r}$ are calculated in terms of the
${}_{3}F_{2}$-hypergeometric functions at unity with the aid of Entry
3.31.12.1 in [2]. Precisely, we get via (4.18)

$c_{n,r}={r!\over\Gamma(1+\nu+r)}\int_{0}^{\infty}t^{\nu}e^{-t}L_{r}^{\nu}(t)\theta^{n}\left(t^{\nu}e^{t}\Gamma(-\nu,t)\right)dt$

$={r!n!(1+\nu)_{n}\over\Gamma(1+\nu+r)}\int_{0}^{\infty}t^{2\nu+n}e^{-t}L_{r}^{\nu}(t)U(\nu+n+1,1+\nu,t)dt$

$={(1+\nu)_{n}\over\Gamma(1+\nu)}\left[{n!r!\ \Gamma(\nu)\over(1+\nu)_{r}}\
{}_{3}F_{2}\left(n+1,r+1,\nu+n+1;1-\nu,1;\ 1\right)\right.$

$None$ $\left.+\Gamma(2\nu+n+1)\Gamma(-\nu)\
{}_{3}F_{2}\left(\nu+r+1,\nu+n+1,2\nu+n+1;1+\nu,1+\nu;\ 1\right)\right],$

where $(a)_{z}$ is the Pochhammer symbol [1], Vol. I and it is valid for
nonnegative integers $\nu$ by continuity. Hence after substitution the series
(4.22) into (4.21) these orthogonality conditions take the form

$None$
$\int_{0}^{\infty}t^{\nu+m}e^{-t}L_{m}^{\nu}(t)\sum_{k=0}^{n}a_{n,k}\sum_{r=0}^{2m}c_{k,r}L_{r}^{\nu}(t)dt=0,\quad
m=0,1,\dots,n-1,\ n\in\mathbb{N}.$

Calculating the integral in (4.24) via relation (2.19.14.8) in [5], Vol. II we
obtain

$d_{m,r}=\int_{0}^{\infty}t^{\nu+m}e^{-t}L_{m}^{\nu}(t)L_{r}^{\nu}(t)dt={(-1)^{r}\over
r!}\ (1+\nu)_{r}\ \Gamma(1+\nu+m)$

$None$ $\times{}_{3}F_{2}\left(-r,\nu+m+1,m+1;1+\nu,1;\ 1\right).$

Hence, equalities (4.24) become the linear system of $n$ algebraic equations
with $n+1$ unknowns

$None$ $\sum_{k=0}^{n}a_{n,k}f_{k,m}=0,\quad m=0,1,\dots,n-1,\
n\in\mathbb{N},$

where

$None$ $f_{k,m}=\sum_{r=0}^{2m}c_{k,r}d_{m,r}.$

Consequently, explicit values of the coefficients $a_{n,k},\ k=1,2,\dots,n$
can be expressed via Cramer’s rule in terms of the free coefficient $a_{n,0}$
as follows

$None$ $a_{n,k}=-a_{n,0}\ {D_{n,k}\over D_{n}},\quad k=1,\dots,n,$

where

$None$ $D_{n}=\begin{vmatrix}f_{1,0}&f_{2,0}&\dots&\dots&f_{n,0}\\\
f_{1,1}&\dots&\dots&\dots&f_{n,1}\\\ \dots&\dots&\dots&\dots&\dots\\\
\vdots&\ddots&\ddots&\ddots&\vdots\\\ f_{1,n-1}&\dots&\dots&\dots&f_{n,n-1}\\\
\end{vmatrix},$

$None$
$D_{n,k}=\begin{vmatrix}f_{1,0}&\dots&f_{k-1,0}&f_{0,0}&f_{k+1,0}&\dots&f_{n,0}\\\
f_{1,1}&\dots&f_{k-1,1}&f_{0,1}&f_{k+1,1}&\dots&f_{n,1}\\\
\dots&\dots&\dots&\dots&\dots&\dots&\dots\\\
\vdots&\ddots&\ddots&\vdots&\ddots&\ddots&\vdots\\\
\vdots&\ddots&\ddots&\vdots&\ddots&\ddots&\vdots\\\
\vdots&\ddots&\ddots&\vdots&\ddots&\ddots&\vdots\\\
f_{1,n-1}&\dots&f_{k-1,n-1}&f_{0,n-1}&f_{k+1,1}&\dots&f_{n,n-1}\\\
\end{vmatrix}.$

The free coefficient can be determined, in turn, from the orthogonality
conditions (4.1), (4.2), which imply the formula

$None$ $\int_{0}^{\infty}P_{n}(x)\rho^{2}_{\nu}(x)\ x^{n}dx={1\over a_{n,n}}.$

Therefore from (4.28) and (4.3) we derive

$None$ ${1\over a_{n,n}}=-{a_{n,0}\sqrt{\pi}\over D_{n}}\
\frac{\Gamma(1+2\nu)\Gamma(1+\nu)}{2^{1+2\nu}\ \Gamma(\nu+3/2)}\
\sum_{k=0}^{n}D_{n,k}\ (n+k)!\ \frac{(1+2\nu)_{n+k}(1+\nu)_{n+k}}{4^{n+k}\
(\nu+3/2)_{n+k}},$

where $D_{n,0}\equiv-D_{n}.$ Hence

$None$ $a_{n,0}=\pm\ {D_{n}\over[D_{n,n}]^{1/2}}\left[\sqrt{\pi}\
\frac{\Gamma(1+2\nu)\Gamma(1+\nu)}{2^{1+2\nu}\ \Gamma(\nu+3/2)}\
\sum_{k=0}^{n}D_{n,k}\ (n+k)!\ \frac{(1+2\nu)_{n+k}(1+\nu)_{n+k}}{4^{n+k}\
(\nu+3/2)_{n+k}}\right]^{-1/2},$

where the sign can be chosen accordingly, making positive expressions under
the square roots. Assuming also the positivity of the leading coefficient
$a_{n,n}$ we have its value, correspondingly,

$None$ $a_{n,n}=\mp\ [D_{n,n}]^{1/2}\left[\sqrt{\pi}\
\frac{\Gamma(1+2\nu)\Gamma(1+\nu)}{2^{1+2\nu}\ \Gamma(\nu+3/2)}\
\sum_{k=0}^{n}D_{n,k}\ (n+k)!\ \frac{(1+2\nu)_{n+k}(1+\nu)_{n+k}}{4^{n+k}\
(\nu+3/2)_{n+k}}\right]^{-1/2}.$

Theorem 2. Let $\nu>-1/2$. The sequence of orthogonal polynomials
$\left\\{P_{n}\right\\}_{n\geq 0}$ can be expressed explicitly, where the
coefficients $a_{n,k},\ k=1,2,\dots,n$ are calculated by relations $(4.28)$
and the free term $a_{n,0}$ is defined by the equality $(4.33)$. Besides, it
satisfies the 3-term recurrence relation $(4.4)$, where

$None$ $A_{n+1}={a_{n,0}\ D_{n+1}\ D_{n,n}\over a_{n+1,0}\ D_{n}\
D_{n+1,n+1}},\quad B_{n}={D_{n,n-1}\over D_{n,n}}-{D_{n+1,n}\over
D_{n+1,n+1}}.$

An analog of the Rodrigues formula for the sequence
$\left\\{P_{n}\right\\}_{n\in\mathbb{N}_{0}}$ can be established, appealing to
the integral representation of an arbitrary polynomial in terms of the
associated polynomial of degree $2n$ (see [10]). Hence, employing (2.2), we
deduce

$P_{n}(x)={1\over\rho_{\nu}(x)}\int_{0}^{\infty}t^{\nu-1}e^{-t-x/t}q_{2n}(t)dt={1\over\rho^{2}_{\nu}(x)}\int_{0}^{\infty}u^{\nu-1}e^{-u-x/u}du\int_{0}^{\infty}t^{\nu-1}e^{-t-x/t}q_{2n}(t)dt$

$None$ $={(-1)^{n}\over\rho^{2}_{\nu}(x)}{d^{n}\over
dx^{n}}\int_{0}^{\infty}\int_{0}^{\infty}(ut)^{\nu+n-1}e^{-u-t-x(1/u+1/t)}q_{2n}(t){dtdu\over(u+t)^{n}},$

where

$None$ $q_{2n}(x)=\sum_{k=0}^{n}a_{n,k}(-1)^{k}k!x^{k}L_{k}^{\nu}(x)$

and the $k$-th differentiation under the integral sign is permitted due to the
estimate

$\int_{0}^{\infty}\int_{0}^{\infty}(ut)^{\nu+k-1}e^{-u-t-x(1/u+1/t)}\left|q_{2n}(t)\right|{dtdu\over(u+t)^{k}}$

$\leq
2^{-k}\int_{0}^{\infty}u^{\nu+k/2-1}e^{-u}du\int_{0}^{\infty}t^{\nu+k/2-1}e^{-t}\left|q_{2n}(t)\right|dt$

$=2^{-k}\Gamma\left(\nu+{k\over
2}\right)\int_{0}^{\infty}t^{\nu+k/2-1}e^{-t}\left|q_{2n}(t)\right|dt<\infty,\quad
k=1,\dots,n.$

Then writing

${(ut)^{n}\over(u+t)^{n}}={1\over(n-1)!}\int_{0}^{\infty}e^{-(1/u+1/t)y}y^{n-1}dy,$

we get from (4.36)

$P_{n}(x)={(-1)^{n}\over(n-1)!\ \rho^{2}_{\nu}(x)}{d^{n}\over
dx^{n}}\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\infty}y^{n-1}(ut)^{\nu-1}e^{-u-t-(x+y)/u-(x+y)/t}q_{2n}(t)dtdudy$

$None$ $={(-1)^{n}\over(n-1)!\ \rho^{2}_{\nu}(x)}{d^{n}\over
dx^{n}}\int_{0}^{\infty}\int_{0}^{\infty}y^{n-1}t^{\nu-1}e^{-t-(x+y)/t}q_{2n}(t)\rho_{\nu}(x+y)dtdy.$

Then, expressing $q_{2n}$ in terms of the Laguerre polynomials

$None$ $q_{2n}(x)=\sum_{k=0}^{2n}h_{2n,k}L_{k}^{\nu}(x),$

where the coefficients $h_{2n,k}$ are calculated by virtue of (4.28), (4.37)
and relation (2.19.14.15) in [5], Vol. II, namely,

$h_{2n,k}={k!\over\Gamma(1+\nu+k)}\int_{0}^{\infty}t^{\nu}e^{-t}L_{k}^{\nu}(t)q_{2n}(t)dt$

$=-{k!\ a_{n,0}\over D_{n}\
\Gamma(1+\nu+k)}\sum_{r=0}^{n}D_{n,r}(-1)^{r}r!\int_{0}^{\infty}t^{\nu+r}e^{-t}L_{k}^{\nu}(t)L_{r}^{\nu}(t)dt$

$=-{a_{n,0}\over D_{n}}\sum_{r=0}^{n}D_{n,r}\ r!\ (1+\nu)_{r}\
{}_{3}F_{2}\left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right).$

So,

$None$ $h_{2n,k}=-{a_{n,0}\over D_{n}}\sum_{r=0}^{n}D_{n,r}\ r!\ (1+\nu)_{r}\
{}_{3}F_{2}\left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right),$

and the values of the generalized hypergeometric function can be simplified
via relations (7.4.4; 90,91,92,93) in [5], Vol. III. Precisely, we get for
$k=0,1\dots,n$

$None$ ${}_{3}F_{2}\left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right)=0,\quad
k>2r,$

$(1+\nu)_{r}\ {}_{3}F_{2}\left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\
1\right)={(2r)!\over r!},\quad k=2r,$

$(1+\nu)_{r}\ {}_{3}F_{2}\left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\
1\right)=-{(2r-1)!(\nu+3r)\over(r-1)!},\quad k=2r-1,$

$(1+\nu)_{r}\ {}_{3}F_{2}\left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\
1\right)={(2(r-1))!\over 2\ r!}\left(2r^{2}(2r+\nu-1)(2r-1)\right.$

$\left.+r(r-1)(r+\nu-1)(r+\nu)\right),\quad k=2(r-1).$

Therefore, returning to (4.38) and minding (4.39), (4.40), (4.41), (3.1),
(3.2) we find

$P_{n}(x)={(-1)^{n+1}a_{n,0}\over D_{n}\ (n-1)!\ \rho^{2}_{\nu}(x)}{d^{n}\over
dx^{n}}\sum_{k=0}^{2n}\sum_{r=0}^{n}D_{n,r}\ r!\ (1+\nu)_{r}\
{}_{3}F_{2}\left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right)$

$\times\int_{0}^{\infty}\int_{0}^{\infty}y^{n-1}t^{\nu-1}e^{-t-(x+y)/t}L_{k}^{\nu}(t)\rho_{\nu}(x+y)dtdy$

$={(-1)^{n+1}a_{n,0}\over D_{n}\ (n-1)!\ \rho^{2}_{\nu}(x)}{d^{n}\over
dx^{n}}\sum_{k=0}^{2n}{1\over k!}\sum_{r=0}^{n}D_{n,r}\ r!\ (1+\nu)_{r}\
{}_{3}F_{2}\left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right)$

$\times\int_{0}^{\infty}y^{n-1}{d^{k}\over
dx^{k}}\left((x+y)^{k}\rho_{\nu}(x+y)\right)\rho_{\nu}(x+y)dy$

$={(-1)^{n+1}a_{n,0}\over D_{n}\ (n-1)!\ \rho^{2}_{\nu}(x)}{d^{n}\over
dx^{n}}\sum_{k=0}^{2n}{1\over k!}\sum_{r=0}^{n}D_{n,r}\ r!\ (1+\nu)_{r}\
{}_{3}F_{2}\left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right)$

$\times\int_{x}^{\infty}(y-x)^{n-1}{d^{k}\over
dy^{k}}\left(y^{k}\rho_{\nu}(y)\right)\rho_{\nu}(y)dy=-{a_{n,0}\over D_{n}\
\rho_{\nu}(x)}\sum_{k=0}^{2n}{d^{k}\over
dx^{k}}\left(x^{k}\rho_{\nu}(x)\right)\sum_{r=0}^{n}D_{n,r}\ r!$

$\times\ {(1+\nu)_{r}\over k!}\ {}_{3}F_{2}\left(-k,\ 1+\nu+r,\ 1+r;\ 1+\nu,\
1;\ 1\right)$

$=-{a_{n,0}\over D_{n}\ \rho_{\nu}(x)}\sum_{r=0}^{n}D_{n,r}\ r!\
(1+\nu)_{r}\sum_{k=0}^{2r}{d^{k}\over
dx^{k}}\left(x^{k}\rho_{\nu}(x)\right){1\over k!}\ {}_{3}F_{2}\left(-k,\
1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right).$

Thus it proves

Theorem 3. Let $\nu>-1/2,\ n\in\mathbb{N}_{0}$. Orthogonal polynomials $P_{n}$
satisfy the Rodrigues-type formula

$None$ $P_{n}(x)=-{a_{n,0}\over D_{n}\
\rho_{\nu}(x)}\sum_{r=0}^{n}D_{n,r}r!(1+\nu)_{r}\sum_{k=0}^{2r}{1\over
k!}{d^{k}\over dx^{k}}\left(x^{k}\rho_{\nu}(x)\right){}_{3}F_{2}\left(-k,\
1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right),$

where $a_{n,0}$ is defined by $(4.33)$ and $D_{n},\ D_{n,r}$ by
$(4.29),(4.30)$, respectively.

Corollary 2. Orthogonal polynomials $P_{n}$ have the following representation

$P_{n}(x)=-{a_{n,0}\over D_{n}\
}\sum_{r=0}^{n}D_{n,r}r!(1+\nu)_{r}\left[\sum_{k=0}^{r}{A_{k,k-1}(x)\over(2k)!}{}_{3}F_{2}\left(-2k,\
1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right)\right.$

$None$
$\left.+\sum_{k=0}^{r-1}{A_{k,k}(x)\over(2k+1)!}{}_{3}F_{2}\left(-2k-1,\
1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right)\right],$

where $A_{k,k-1},A_{k,k}$ are the type $1$ multiple orthogonal polynomials of
degree $k$, associated with the vector of weight functions
$(\rho_{\nu},\rho_{\nu+1})$ .

###### Proof.

In fact, we write (4.42) in the form

$P_{n}(x)=-{a_{n,0}\over D_{n}\
\rho_{\nu}(x)}\sum_{r=0}^{n}D_{n,r}r!(1+\nu)_{r}\left[\sum_{k=0}^{r}{1\over(2k)!}{d^{2k}\over
dx^{2k}}\left(x^{2k}\rho_{\nu}(x)\right){}_{3}F_{2}\left(-2k,\ 1+\nu+r,\ 1+r;\
1+\nu,\ 1;\ 1\right)\right.$

$None$ $\left.+\sum_{k=0}^{r-1}{1\over(2k+1)!}{d^{2k+1}\over
dx^{2k+1}}\left(x^{2k+1}\rho_{\nu}(x)\right){}_{3}F_{2}\left(-2k-1,\ 1+\nu+r,\
1+r;\ 1+\nu,\ 1;\ 1\right)\right].$

Meanwhile, appealing to the Rodrigues formulas for the type $1$ multiple
orthogonal polynomials associated with the vector of weight functions
$(\rho_{\nu},\rho_{\nu+1})$ (see in [7]), it gives

${d^{2k}\over
dx^{2k}}\left(x^{2k}\rho_{\nu}(x)\right)=A_{k,k-1}(x)\rho_{\nu}(x)+B_{k,k-1}(x)\rho_{\nu+1}(x),$

${d^{2k+1}\over
dx^{2k+1}}\left(x^{2k+1}\rho_{\nu}(x)\right)=A_{k,k}(x)\rho_{\nu}(x)+B_{k,k}(x)\rho_{\nu+1}(x),$

where $A_{k,k-1},B_{k,k-1}$ are polynomials of degree $k,k-1$, respectively,
and $A_{k,k},B_{k,k}$ are polynomials of degree $k$. Therefore, substituting
these expressions into (4.44), we obtain

$P_{n}(x)=-{a_{n,0}\over D_{n}\
}\sum_{r=0}^{n}D_{n,r}r!(1+\nu)_{r}\left[\sum_{k=0}^{r}{A_{k,k-1}(x)\over(2k)!}{}_{3}F_{2}\left(-2k,\
1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right)\right.$

$\left.+\sum_{k=0}^{r-1}{A_{k,k}(x)\over(2k+1)!}{}_{3}F_{2}\left(-2k-1,\
1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right)\right]$

$-{a_{n,0}\ \rho_{\nu+1}(x)\over D_{n}\
\rho_{\nu}(x)}\sum_{r=0}^{n}D_{n,r}r!(1+\nu)_{r}\left[\sum_{k=0}^{r}{B_{k,k-1}(x)\over(2k)!}{}_{3}F_{2}\left(-2k,\
1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right)\right.$

$\left.+\sum_{k=0}^{r-1}{B_{k,k}(x)\over(2k+1)!}{}_{3}F_{2}\left(-2k-1,\
1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right)\right].$

But the existence of a multiple orthogonal polynomial sequence with respect to
the vector of weight functions $(\rho_{\nu},\rho_{\nu+1})$ implies the
identity

$\sum_{r=0}^{n}D_{n,r}r!(1+\nu)_{r}\left[\sum_{k=0}^{r}{B_{k,k-1}(x)\over(2k)!}{}_{3}F_{2}\left(-2k,\
1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right)\right.$

$\bigg{.}+\sum_{k=0}^{r-1}{B_{k,k}(x)\over(2k+1)!}{}_{3}F_{2}\left(-2k-1,\
1+\nu+r,\ 1+r;\ 1+\nu,\ 1;\ 1\right)\bigg{]}\equiv 0,$

which drives to (4.43) and completes the proof.

∎

Remark 2. In a similar manner orthogonal polynomials with the weight function
$\rho_{\nu+1}(x)\rho_{\nu}(x)$ can be investigated. We leave this topic to the
interested reader.

Finally, in this section we establish the generating function for polynomials
$P_{n}$, which is defined as usually by the equality

$None$ $G(x,z)=\sum_{n=0}^{\infty}P_{n}(x){z^{n}\over n!},\quad x>0,\
z\in\mathbb{C},$

where $|z|<h_{x}$ and $h_{x}>0$ is a convergence radius of the power series.
To do this, we employ (3.1), (4.36) and (4.39), having the following equality
from (4.45)

$G(x,z)={1\over\rho_{\nu}(x)}\sum_{n=0}^{\infty}{z^{n}\over
n!}\sum_{k=0}^{2n}{h_{2n,k}\over k!}{d^{k}\over
dx^{k}}\left[x^{k}\rho_{\nu}(x)\right]={1\over\rho_{\nu}(x)}\sum_{n=0}^{\infty}{z^{n}\over
n!}\sum_{k=0}^{2n}h_{2n,k}\sum_{j=0}^{k}\binom{k}{j}{(-1)^{j}\over(k-j)!}\
x^{j}\rho_{\nu-j}(x).$

Meanwhile, the product $x^{j}\rho_{\nu-j}(x)$ is expressed in [3] as follows

$x^{j}\rho_{\nu-j}(x)=x^{j/2}r_{j}(2\sqrt{x};\nu)\rho_{\nu}(x)+x^{(j-1)/2}r_{j-1}(2\sqrt{x};\nu-1)\rho_{\nu+1}(x),\quad
j\in{\mathbb{N}}_{0},$

where $r_{-1}(z;\nu)=0$,

$None$
$x^{j/2}r_{j}(2\sqrt{x};\nu)=(-1)^{j}\sum_{i=0}^{[j/2]}(\nu+i-j+1)_{j-2i}(j-2i+1)_{i}{x^{i}\over
i!}.$

Therefore this leads to the final expression of the generating function for
the sequence $\left(P_{n}\right)_{n\in\mathbb{N}_{0}}$, namely,

$G(x,z)=\sum_{n=0}^{\infty}\sum_{k=0}^{2n}\sum_{j=0}^{k}\binom{k}{j}{(-1)^{j}\
h_{2n,k}\over n!\ (k-j)!}x^{j/2}r_{j}(2\sqrt{x};\nu)z^{n}$

$None$
$+{\rho_{\nu+1}(x)\over\rho_{\nu}(x)}\sum_{n=0}^{\infty}\sum_{k=0}^{2n}\sum_{j=0}^{k}\binom{k}{j}{(-1)^{j}\
h_{2n,k}\over n!\ (k-j)!}x^{(j-1)/2}r_{j-1}(2\sqrt{x};\nu-1)z^{n},$

where coefficients $h_{2n,k}$ are defined by (4.40).

##  5\. Note on the multiple orthogonal polynomials

In this section we will exhibit two types of multiple orthogonal polynomials
for the vector of weight functions $(\rho^{2}_{\nu},\ \rho^{2}_{\nu+1},\
\rho_{\nu}\rho_{\nu+1}),\ \nu>-1/2$ over $\mathbb{R}_{+}$ with an additional
factor $x^{\alpha},\ \alpha>-1$. Precisely, we consider the type $1$
polynomials $(A^{\alpha}_{n},\ B^{\alpha}_{n-1},\ C^{\alpha}_{n-1}),\
n\in\mathbb{N}$ of degree $n,\ n-1$, respectively, satisfying the
orthogonality conditions

$None$
$\int_{0}^{\infty}\left[A^{\alpha}_{n}(x)\rho^{2}_{\nu}(x)+B^{\alpha}_{n-1}(x)\rho^{2}_{\nu+1}(x)+C^{\alpha}_{n-1}(x)\rho_{\nu}(x)\rho_{\nu+1}(x)\right]x^{\alpha+m}dx=0,\
m=0,1,\dots,3n-1.$

So, we have $3n$ linear homogeneous equations with $3n+1$ unknown coefficients
of polynomials $A^{\alpha}_{n},\ B^{\alpha}_{n-1},\ C^{\alpha}_{n-1}$.
Therefore we can find type $1$ polynomials up to a multiplicative factor. Let
us denote the function $q^{\alpha}_{n,n-1,n-1}$ for the convenience

$None$
$q^{\alpha}_{n,n-1,n-1}(x)=A^{\alpha}_{n}(x)\rho^{2}_{\nu}(x)+B^{\alpha}_{n-1}(x)\rho^{2}_{\nu+1}(x)+C^{\alpha}_{n-1}(x)\rho_{\nu}(x)\rho_{\nu+1}(x).$

Type $2$ polynomials $p^{\alpha}_{n,n-1,n}$ are monic polynomials of degree
$3n-1$ which satisfy the multiple orthogonality conditions

$None$
$\int_{0}^{\infty}p^{\alpha}_{n,n-1,n}(x)\rho^{2}_{\nu}(x)x^{\alpha+m}dx=0,\quad
m=0,1,\dots,n-1,$

$None$
$\int_{0}^{\infty}p^{\alpha}_{n,n-1,n}(x)\rho^{2}_{\nu+1}(x)x^{\alpha+m}dx=0,\quad
m=0,1,\dots,n-2,$

$None$
$\int_{0}^{\infty}p^{\alpha}_{n,n-1,n}(x)\rho_{\nu}(x)\rho_{\nu+1}(x)x^{\alpha+m}dx=0,\quad
m=0,1,\dots,n-1.$

This gives $3n-1$ linear equations with $3n-1$ unknown coefficients since the
leading coefficient is equal to $1$. Hence it can be uniquely determined.

When $|\nu|<1/2$ the uniqueness of the representation (5.2) is validated by
the following

Theorem 4. Let $n,m,l\in\mathbb{N}_{0},|\nu|<1/2,\ f_{n},\ g_{m},\ h_{l}$ be
polynomials of degree at most $n,\ m,\ l$, respectively. Let

$None$
$f_{n}(x)\rho^{2}_{\nu}(x)+g_{m}(x)\rho^{2}_{\nu+1}(x)+h_{l}(x)\rho_{\nu}(x)\rho_{\nu+1}(x)=0$

for all $x>0$. Then $f_{n}\equiv 0,\ g_{m}\equiv 0,\ h_{l}\equiv 0.$

###### Proof.

As is known (cf. [10]) the quotient $\rho_{\nu}/\rho_{\nu+1}$ is represented
by the Ismail integral

$None$
${\rho_{\nu}(x)\over\rho_{\nu+1}(x)}={1\over\pi^{2}}\int_{0}^{\infty}{s^{-1}ds\over(x+s)(J_{\nu+1}^{2}(2\sqrt{s})+Y_{\nu+1}^{2}(2\sqrt{s}))},$

where $J_{\nu},Y_{\nu}$ are Bessel functions of the first and second kind,
respectively [5], Vol. II. In fact, let $r\geq\max\\{n,m+1,l+1\\}.$ Hence,
dividing (5.6) by $\rho_{\nu}(x)\rho_{\nu+1}(x)$ and using (3.6), we find

$None$
$f_{n}(x){\rho_{\nu}(x)\over\rho_{\nu+1}(x)}+xg_{m}(x){\rho_{\nu-1}(x)\over\rho_{\nu}(x)}+\nu
g_{m}(x)+h_{l}(x)=0.$

Then, differentiating $r$ times, it gives

$None$ ${d^{r}\over
dx^{r}}\left[f_{n}(x){\rho_{\nu}(x)\over\rho_{\nu+1}(x)}\right]+{d^{r}\over
dx^{r}}\left[xg_{m}(x){\rho_{\nu-1}(x)\over\rho_{\nu}(x)}\right]=0.$

Assuming $f_{n}(x)=\sum_{k=0}^{n}f_{n,k}\ x^{k}$ and employing (5.7), the
first term on the left-hand side of (5.9) can be treated as follows

${d^{r}\over
dx^{r}}\left[f_{n}(x){\rho_{\nu}(x)\over\rho_{\nu+1}(x)}\right]={1\over\pi^{2}}{d^{r}\over
dx^{r}}\sum_{k=0}^{n}f_{n,k}x^{k}\int_{0}^{\infty}e^{-xy}dy\int_{0}^{\infty}{e^{-sy}\
s^{-1}ds\over J_{\nu+1}^{2}(2\sqrt{s})+Y_{\nu+1}^{2}(2\sqrt{s})}$

$={1\over\pi^{2}}\sum_{k=0}^{n}f_{n,k}(-1)^{k}{d^{r}\over
dx^{r}}\int_{0}^{\infty}{d^{k}\over
dy^{k}}\left[e^{-xy}\right]dy\int_{0}^{\infty}{e^{-sy}\ s^{-1}ds\over
J_{\nu+1}^{2}(2\sqrt{s})+Y_{\nu+1}^{2}(2\sqrt{s})}$

$={1\over\pi^{2}}\sum_{k=0}^{n}f_{n,k}(-1)^{k}\int_{0}^{\infty}{\partial^{k+r}\over\partial
y^{k}\partial x^{r}}\left[e^{-xy}\right]dy\int_{0}^{\infty}{e^{-sy}\
s^{-1}ds\over J_{\nu+1}^{2}(2\sqrt{s})+Y_{\nu+1}^{2}(2\sqrt{s})}$

$={1\over\pi^{2}}\sum_{k=0}^{n}f_{n,k}(-1)^{k+r}\int_{0}^{\infty}{d^{k}\over
dy^{k}}\left[y^{r}e^{-xy}\right]dy\int_{0}^{\infty}{e^{-sy}\ s^{-1}ds\over
J_{\nu+1}^{2}(2\sqrt{s})+Y_{\nu+1}^{2}(2\sqrt{s})},$

where the differentiation under the integral sign is possible via the absolute
and uniform convergence. Now, we integrate $k$ times by parts in the outer
integral with respect to $y$ on the right-hand side of the latter equality,
eliminating the integrated terms due to the choice of $r$, and then
differentiate under the integral sign in the inner integral with respect to
$s$ owing to the same arguments, to obtain

$None$ ${d^{r}\over
dx^{r}}\left[f_{n}(x){\rho_{\nu}(x)\over\rho_{\nu+1}(x)}\right]={1\over\pi^{2}}\int_{0}^{\infty}y^{r}e^{-xy}\int_{0}^{\infty}{e^{-sy}\
s^{-1}\over
J_{\nu+1}^{2}(2\sqrt{s})+Y_{\nu+1}^{2}(2\sqrt{s})}\left(\sum_{k=0}^{n}f_{n,k}(-1)^{k+r}s^{k}\right)ds.$

In the same fashion the second term in (5.9) is worked out to find
$(g_{m}(x)=\sum_{k=0}^{m}g_{m,k}\ x^{k}$ )

$None$ ${d^{r}\over
dx^{r}}\left[xg_{m}(x){\rho_{\nu-1}(x)\over\rho_{\nu}(x)}\right]={1\over\pi^{2}}\int_{0}^{\infty}y^{r}e^{-xy}\int_{0}^{\infty}{e^{-sy}\over
J_{\nu}^{2}(2\sqrt{s})+Y_{\nu}^{2}(2\sqrt{s})}\left(\sum_{k=0}^{m}g_{m,k}(-1)^{k+r+1}s^{k}\right)ds.$

Substituting (5.10), (5.11) into (5.9) and cancelling twice the Laplace
transform via its injectivity for integrable and continuous functions [6], we
arrive at the equality

$None$
$xg_{m}(-x)\left[J_{\nu+1}^{2}(2\sqrt{x})+Y_{\nu+1}^{2}(2\sqrt{x})\right]+f_{n}(-x)\left[J_{\nu}^{2}(2\sqrt{x})+Y_{\nu}^{2}(2\sqrt{x})\right]=0,\
x>0.$

The sum of squares of Bessel functions in brackets is called the Nicholson
kernel, which has the Mellin-Barnes representation by virtue of Entry
8.4.20.35 in [5], Vol. III

$x^{k}\left[J_{\nu}^{2}(2\sqrt{x})+Y_{\nu}^{2}(2\sqrt{x})\right]={2^{1-2k}\cos(\pi\nu)\over
2\pi^{7/2}i}\int_{\gamma-i\infty}^{\gamma+i\infty}\Gamma(s+k)\Gamma(s+k+\nu)\Gamma(s+k-\nu)$

$None$ $\times\Gamma\left({1\over
2}-s-k\right)(4x)^{-s}ds,\quad|\nu|-k<\gamma<{1\over 2}-k.$

Then, using on the right-hand side of (5.13) the reflection formula for gamma
function [1], Vol. I, it can be written as follows

${2^{1-2k}\cos(\pi\nu)\over
2\pi^{7/2}i}\int_{\gamma-i\infty}^{\gamma+i\infty}\Gamma(s+k)\Gamma(s+k+\nu)\Gamma(s+k-\nu)\Gamma\left({1\over
2}-s-k\right)(4x)^{-s}ds$

$None$ $={2^{1-2k}(-1)^{k}\cos(\pi\nu)\over
2\pi^{7/2}i}\int_{\gamma-i\infty}^{\gamma+i\infty}\Gamma(s+k)\Gamma(s+k+\nu)\Gamma(s+k-\nu){\Gamma(1/2-s)\over(1/2+s)_{k}}(4x)^{-s}ds.$

Our goal now is to shift the contour to the right to make integration along
the straight line with $|\nu|<{\rm Re}s<1/2$. To do this we should take into
account the residues at $k$ simple poles $s_{m}=-1/2-m,\ m=0,1,\dots,k-1$
which have the values

$\hbox{Res}_{s=s_{m}}\left(\frac{\Gamma(s+k)\Gamma(s+k+\nu)\Gamma(s+k-\nu)\Gamma(1/2-s)}{(1/2+s)(3/2+s)\dots(s+k-1/2)}(4x)^{-s}\right)$

$=\frac{\Gamma(s_{m}+k)\Gamma(s_{m}+k+\nu)\Gamma(s_{m}+k-\nu)\Gamma(1/2-s_{m})}{(1/2+s_{m})(1/2+s_{m}+1)\dots(1/2+s_{m}+m-1)(1/2+s_{m}+m+1)\dots(1/2+s_{m}+k-1)}(4x)^{-s_{m}}$

$={(-1)^{m}(4x)^{1/2+m}\over(k-m-1)!}\
\Gamma(k-m-1/2)\Gamma(k-m-1/2+\nu)\Gamma(k-m-1/2-\nu).$

Therefore we get from (5.14)

${2^{1-2k}\cos(\pi\nu)\over
2\pi^{7/2}i}\int_{\gamma-i\infty}^{\gamma+i\infty}\Gamma(s+k)\Gamma(s+k+\nu)\Gamma(s+k-\nu)\Gamma\left({1\over
2}-s-k\right)(4x)^{-s}ds$

$={2^{1-2k}(-1)^{k}\cos(\pi\nu)\over
2\pi^{7/2}i}\int_{\mu-i\infty}^{\mu+i\infty}\Gamma(s+k)\Gamma(s+k+\nu)\Gamma(s+k-\nu){\Gamma(1/2-s)\over(1/2+s)_{k}}(4x)^{-s}ds$

$None$
$-{4^{1-k}\sqrt{x}\cos(\pi\nu)\over\pi^{5/2}}\sum_{m=0}^{k-1}{(-1)^{k+m}(4x)^{m}\over(k-m-1)!}\
\Gamma(k-m-1/2)\Gamma(k-m-1/2+\nu)\Gamma(k-m-1/2-\nu),$

where $|\nu|<\mu<1/2$. Now, recalling Parseval’s equality for the Mellin
transform and Entries 8.4.2.5, 8.4.23.27 in [5], Vol. III, we derive from
(5.13), (5.15)

$x^{k}\left[J_{\nu}^{2}(2\sqrt{x})+Y_{\nu}^{2}(2\sqrt{x})\right]={2^{3-2k}(-1)^{k}\sqrt{x}\over\pi^{3}}\cos(\pi\nu)\int_{0}^{\infty}K^{2}_{\nu}(\sqrt{t})\
{t^{k-1/2}\over 4x+t}dt$

$None$ $-\
{4^{1-k}\sqrt{x}\cos(\pi\nu)\over\pi^{5/2}}\sum_{m=0}^{k-1}{(-1)^{k+m}(4x)^{m}\over(k-m-1)!}\
\Gamma(k-m-1/2)\Gamma(k-m-1/2+\nu)\Gamma(k-m-1/2-\nu).$

Analogously, we find

$x^{k+1}\left[J_{\nu+1}^{2}(2\sqrt{x})+Y_{\nu+1}^{2}(2\sqrt{x})\right]={2^{1-2k}(-1)^{k}\sqrt{x}\over\pi^{3}}\cos(\pi\nu)\int_{0}^{\infty}K^{2}_{\nu+1}(\sqrt{t})\
{t^{k+1/2}\over 4x+t}dt$

$None$ $-\
{4^{-k}\sqrt{x}\cos(\pi\nu)\over\pi^{5/2}}\sum_{m=0}^{k}{(-1)^{k+m+1}(4x)^{m}\over(k-m)!}\
\Gamma(k-m+1/2)\Gamma(k-m+1/2+\nu)\Gamma(k-m+1/2-\nu).$

Substituting expressions (5.16), (5.17) in (5.12), we derive after
straightforward simplifications

$\int_{0}^{\infty}K^{2}_{\nu+1}(2\sqrt{t})\ g_{m}(t){\sqrt{t}\over
x+t}dt+\int_{0}^{\infty}K^{2}_{\nu}(2\sqrt{t})\
f_{n}(t){dt\over\sqrt{t}(x+t)}$

$+\sqrt{\pi}\sum_{k=0}^{m}\sum_{j=0}^{k}{(-x)^{j}\over 4^{k-j}(k-j)!}\
\Gamma(k-j+1/2)\Gamma(k-j+1/2+\nu)\Gamma(k-j+1/2-\nu)$

$None$ $-\sqrt{\pi}\sum_{k=0}^{n-1}\sum_{j=0}^{k}{(-x)^{j}\over
4^{k-j}(k-j)!}\ \Gamma(k-j+1/2)\Gamma(k-j+1/2+\nu)\Gamma(k-j+1/2-\nu)=0,\
x>0.$

Last two terms in (5.18) are polynomials of degree $m,n-1$, respectively.
Hence, differentiating through $r_{1}\geq\max\\{n,m+1\\}$ times by $x$, we
obtain

$None$ ${d^{r_{1}}\over dx^{r_{1}}}\int_{0}^{\infty}K^{2}_{\nu+1}(2\sqrt{t})\
g_{m}(t){\sqrt{t}\over x+t}dt+{d^{r_{1}}\over
dx^{r_{1}}}\int_{0}^{\infty}K^{2}_{\nu}(2\sqrt{t})\
f_{n}(t){dt\over\sqrt{t}(x+t)}=0.$

The left-hand side of (5.19) is the $r_{1}$-th derivative of the sum of two
Stieltjes transforms which are, in turn, two fold Laplace transforms.
Consequently, fulfilling the differentiation under the integral sign in (5.19)
as above owing to the absolute and uniform convergence by $x\geq x_{0}>0$, we
cancel Laplace transforms of integrable functions via the injectivity. Then
with (2.2) it yields the equality

$None$ $f_{n}(x)\rho^{2}_{\nu}(x)+g_{m}(x)\rho^{2}_{\nu+1}(x)=0,\quad x>0.$

Comparing with (5.6), we see that $h_{l}\equiv 0.$ Further, identity (5.20)
implies that $f_{n},\ g_{m}$ have the same positive roots, if any. Let $x>0$
be not a root of $g_{m}$. Then dividing (5.20) by $g_{m}$ and making a
differentiation, we get

$-2\rho_{\nu}(x)\rho_{\nu-1}(x){f_{n}(x)\over
g_{m}(x)}+\rho^{2}_{\nu}(x){f^{\prime}_{n}(x)g_{m}(x)-f_{n}(x)g_{m}^{\prime}(x)\over
g_{m}^{2}(x)}-2\rho_{\nu+1}(x)\rho_{\nu}(x)=0.$

Since $\rho_{\nu}(x)>0$, we divide the previous equation by $\rho_{\nu}$,
multiply by $x,g_{m}^{2}$ and employ (3.6) to find

$None$
$-2g_{m}(x)\rho_{\nu+1}(x)\left(f_{n}(x)+xg_{m}(x)\right)+\rho_{\nu}(x)\left(2\nu
f_{n}(x)g_{m}(x)+x\left(f^{\prime}_{n}(x)g_{m}(x)-f_{n}(x)g_{m}^{\prime}(x)\right)\right)=0.$

However, the existence of the type 1 multiple orthogonal polynomials with
respect to the vector of weight functions $(\rho_{\nu},\rho_{\nu+1})$ suggests
the equalities

$None$ $g_{m}(x)\left(f_{n}(x)+xg_{m}(x)\right)\equiv 0,\quad\quad 2\nu
f_{n}(x)g_{m}(x)+x\left(f^{\prime}_{n}(x)g_{m}(x)-f_{n}(x)g_{m}^{\prime}(x)\right)\equiv
0.$

So, if $g_{m}\equiv 0$, it proves the theorem. Otherwise
$f_{n}(x)+xg_{m}(x)\equiv 0$, and with the second equation in (5.22) we have
$x(2\nu+1)g^{2}_{m}(x)\equiv 0.$ Thus $g_{m}\equiv 0$ and $f_{n}\equiv 0$ from
(5.20). Theorem 4 is proved.

∎

Remark 3. The choice of $\nu$ is important. For instance, for $\nu=-1/2$ the
theorem fails. This can be seen, taking $f_{n}(x)\equiv-x,\ g_{m}(x)\equiv 1,\
h_{l}(x)\equiv 0$.

Theorem 5. Let $\nu\in[0,1/2)$. For every $\alpha>0$

$None$ ${d\over
dx}\left[x^{\alpha}q^{\alpha}_{n,n-1,n-1}(x)\right]=x^{\alpha-1}q^{\alpha-1}_{n,n-1,n}(x)$

and the following differential recurrence relations hold

$None$
$A^{\alpha-1}_{n}(x)=(\alpha+2\nu)A^{\alpha}_{n}(x)+x[A^{\alpha}_{n}(x)]^{\prime}-xC^{\alpha}_{n-1}(x),$

$None$ $B^{\alpha-1}_{n-1}(x)=\alpha
B^{\alpha}_{n-1}(x)+x[B^{\alpha}_{n-1}(x)]^{\prime}-C^{\alpha}_{n-1}(x),$

$None$
$C^{\alpha-1}_{n}(x)=(\alpha+\nu)C^{\alpha}_{n-1}(x)+x[C^{\alpha}_{n-1}(x)]^{\prime}-2A^{\alpha}_{n}(x)-2xB^{\alpha}_{n-1}(x).$

###### Proof.

From (5.1), (5.2) and integration by parts we get

$\int_{0}^{\infty}q^{\alpha}_{n,n-1,n-1}(x)x^{\alpha+m}dx=-{1\over
m+1}\int_{0}^{\infty}{d\over
dx}\left[x^{\alpha}q^{\alpha}_{n,n-1,n-1}(x)\right]x^{m+1}dx=0,$

where the integrated terms vanish for every $\alpha>-1$ via asymptotic
behavior (2.4), (2.5). Hence (5.1) suggests the equality

$\int_{0}^{\infty}{d\over
dx}\left[x^{\alpha}q^{\alpha}_{n,n-1,n-1}(x)\right]x^{m}dx=0,\quad
m=1,\dots,3n.$

But, evidently,

$\int_{0}^{\infty}{d\over
dx}\left[x^{\alpha}q^{\alpha}_{n,n-1,n-1}(x)\right]dx=0,\quad\alpha>0.$

Therefore

$None$ $\int_{0}^{\infty}{d\over
dx}\left[x^{\alpha}q^{\alpha}_{n,n-1,n-1}(x)\right]x^{m}dx=0,\quad
m=0,\dots,3n.$

Now, working out the differentiation in (5.27), involving (3.5), (3.6), we
find

${d\over dx}\left[x^{\alpha}q^{\alpha}_{n,n-1,n-1}(x)\right]=\alpha\
x^{\alpha-1}q^{\alpha}_{n,n-1,n-1}(x)+x^{\alpha-1}\left(x[A^{\alpha}_{n}(x)]^{\prime}\rho^{2}_{\nu}(x)+x[B^{\alpha}_{n-1}(x)]^{\prime}\rho^{2}_{\nu+1}(x)+x[C^{\alpha}_{n-1}(x)]^{\prime}\rho_{\nu}(x)\rho_{\nu+1}(x)\right.$

$\left.+2A^{\alpha}_{n}(x)\rho_{\nu}(x)(\nu\rho_{\nu}(x)-\rho_{\nu+1}(x))-2xB^{\alpha}_{n-1}(x)\rho_{\nu+1}(x)\rho_{\nu}(x)\right.$

$\left.-xC^{\alpha}_{n-1}(x)\rho^{2}_{\nu}(x)+C^{\alpha}_{n-1}(x)\rho_{\nu+1}(x)(\nu\rho_{\nu}(x)-\rho_{\nu+1}(x))\right).$

Thus

$None$ ${d\over
dx}\left[x^{\alpha}q^{\alpha}_{n,n-1,n-1}(x)\right]=x^{\alpha-1}\left[A^{\alpha-1}_{n}(x)\rho^{2}_{\nu}(x)+B^{\alpha-1}_{n-1}(x)\rho^{2}_{\nu+1}(x)+C^{\alpha-1}_{n}(x)\rho_{\nu}(x)\rho_{\nu+1}(x)\right],$

where $A^{\alpha-1}_{n}(x),\ B^{\alpha-1}_{n-1}(x),\ C^{\alpha-1}_{n}(x)$ are
polynomials of degree at most $n,n-1,n$, respectively, being defined by
formulas (5.24), (5.25), (5.26). The linear homogeneous system (5.27) of
$3n+1$ equations contains $3n+2$ unknowns. Therefore up to a constant,
choosing to be one, the left-hand side of (5.28) is equal to
$x^{\alpha-1}q^{\alpha-1}_{n,n-1,n}(x)$, and the representation (5.28) is
unique by virtue of Theorem 4. This proves (5.23) and completes the proof of
Theorem 5.

∎

Remark 4. The same analysis for the function $q^{\alpha}_{n,n-1,n}(x)$ does
not work. In fact, in this case we derive analogously

$\int_{0}^{\infty}{d\over
dx}\left[x^{\alpha}q^{\alpha}_{n,n-1,n}(x)\right]x^{m}dx=0,\quad
m=0,\dots,3n+1.$

However, working out the differentiation, we will get polynomials of degree at
most $n+1,n,n$, respectively, which implies $3n+4$ unknowns for $3n+2$
equations (the so-called quasi multiple orthogonal case.)

Finally, we establish the differentiation property for the type $2$ multiple
orthogonal polynomials $p^{\alpha}_{n,n-1,n}$ (or $3$-orthogonal polynomials).

Theorem 6. For every $\nu\geq 0,\ \alpha>-1$

$None$ ${d\over
dx}\left[p^{\alpha}_{n,n-1,n}(x)\right]=(3n-1)p^{\alpha+1}_{n,n-1,n-1}(x).$

###### Proof.

Recalling (3.5), (3.6) and asymptotic behavior of the weight functions (2.4),
(2.5), we integrate by parts in (5.3), eliminating the integrated terms, to
deduce

$None$ $\int_{0}^{\infty}{d\over
dx}\left[p^{\alpha}_{n,n-1,n}(x)\right]\rho^{2}_{\nu}(x)x^{\alpha+1+m}dx=0,\quad
m=0,1,\dots,n-1.$

Concerning equalities (5.4), (5.5), it corresponds the following ones

$None$ $\int_{0}^{\infty}{d\over
dx}\left[p^{\alpha}_{n,n-1,n}(x)\right]\rho^{2}_{\nu+1}(x)x^{\alpha+1+m}dx=0,\quad
m=0,1,\dots,n-2,$

$None$ $\int_{0}^{\infty}{d\over
dx}\left[p^{\alpha}_{n,n-1,n}(x)\right]\rho_{\nu}(x)\rho_{\nu+1}(x)x^{\alpha+1+m}dx=0,\quad
m=0,1,\dots,n-2.$

Now $\left[p^{\alpha}_{n,n-1,n}(x)\right]^{\prime}$ is a polynomial of degree
$3n-2$ with leading coefficient $3n-1$ and by equalities (5.30), (5.31),
(5.32) it satisfies orthogonality conditions (5.3), (5.4), (5.5) for the type
$2$ multiple orthogonal polynomial $p^{\alpha+1}_{n,n-1,n-1}(x)$. Hence we get
(5.29) by unicity. Theorem 6 is proved.

∎

## References

  * [1] A. Erdélyi,W. Magnus, F. Oberhettinger, and F.G. Tricomi, Higher Transcendental Functions,Vols. I and II, McGraw-Hill, NewYork, London, Toronto, 1953.
  * [2] Yu.A. Brychkov, O.I. Marichev, N.V. Savischenko, Handbook of Mellin Transforms. Advances in Applied Mathematics, CRC Press, Boca Raton, 2018.
  * [3] E. Coussement, W. Van Assche, Some properties of multiple orthogonal polynomials associated with Macdonald functions, J. Comput. Appl. Math. 133 (2001), 253-261.
  * [4] NIST Digital Library of Mathematical Functions. http://dlmf.nist.gov/, Release 1.0.17 of 2017-12-22. F. W. J. Olver, A. B. Olde Daalhuis, D. W. Lozier, B. I. Schneider, R. F. Boisvert, C. W. Clark, B. R. Miller and B. V. Saunders, eds.
  * [5] A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, Vol. I: Elementary Functions, Gordon and Breach, New York, London, 1986; Vol. II: Special Functions, Gordon and Breach, New York, London, 1986; Vol. III: More Special Functions, Gordon and Breach, New York, London, 1990.
  * [6] E.C. Titchmarsh, An Introduction to the Theory of Fourier Integrals, Clarendon Press, Oxford, 1937.
  * [7] W. Van Assche and S. Yakubovich, Multiple orthogonal polynomials associated with Macdonald functions, Integral Transforms and Special Functions, 9 (2000), N 3, 229- 244.
  * [8] O.V. Viskov, H.M. Srivastava, New approaches to certain identities involving differential operators, J. Math. Anal. Appl. 186 (1994), 1-10.
  * [9] S. Yakubovich and Yu. Luchko, The Hypergeometric Approach to Integral Transforms and Convolutions, Kluwer Academic Publishers, Mathematics and Applications. Vol.287, 1994.
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##  6\. Addendum

Corrigendum to ”A method of composition orthogonality and new sequences of
orthogonal polynomials and functions for non-classical weights” [J. Math.
Anal. Appl. 499(2021) 125032]. DOI of original article:
https://doi.org/10.1016/j.jmaa.2021.125032

In the last section of the article one should consider the type $2$ multiple
monic orthogonal polynomials $p^{\alpha}_{n+1,n,n+1}$ of degree $3n+2$ which
satisfy the orthogonality conditions

$None$
$\int_{0}^{\infty}p^{\alpha}_{n+1,n,n+1}(x)\rho^{2}_{\nu}(x)x^{\alpha+m}dx=0,\quad
m=0,1,\dots,n,$

$None$
$\int_{0}^{\infty}p^{\alpha}_{n+1,n,n+1}(x)\rho^{2}_{\nu+1}(x)x^{\alpha+m}dx=0,\quad
m=0,1,\dots,n-1,$

$None$
$\int_{0}^{\infty}p^{\alpha}_{n+1,n,n+1}(x)\rho_{\nu}(x)\rho_{\nu+1}(x)x^{\alpha+m}dx=0,\quad
m=0,1,\dots,n.$

This gives $3n+2$ linear equations with $3n+2$ unknown coefficients since the
leading coefficient is equal to $1$. Hence it can be uniquely determined.
Theorem 6 reads as follows.

Theorem 6. For every $\nu\geq 0,\ \alpha>-1$

$None$ ${d\over
dx}\left[p^{\alpha}_{n+1,n,n+1}(x)\right]=(3n+2)p^{\alpha+1}_{n+1,n,n}(x).$

###### Proof.

In fact, recalling (3.5), (3.6) and asymptotic behavior of the weight
functions (2.4), (2.5), we integrate by parts in (1), eliminating the
integrated terms, to deduce

$\int_{0}^{\infty}{d\over
dx}\left[p^{\alpha}_{n+1,n,n+1}(x)\right]\rho^{2}_{\nu}(x)x^{\alpha+1+m}dx=0,\quad
m=0,1,\dots,n.$

Concerning equalities (2), (3), it corresponds the following ones

$\int_{0}^{\infty}{d\over
dx}\left[p^{\alpha}_{n+1,n,n+1}(x)\right]\rho^{2}_{\nu+1}(x)x^{\alpha+1+m}dx=0,\quad
m=0,1,\dots,n-1,$

$\int_{0}^{\infty}{d\over
dx}\left[p^{\alpha}_{n+1,n,n+1}(x)\right]\rho_{\nu}(x)\rho_{\nu+1}(x)x^{\alpha+1+m}dx=0,\quad
m=0,1,\dots,n-1.$

Now $\left[p^{\alpha}_{n+1,n,n+1}\right]^{\prime}$ is a polynomial of degree
$3n+1$ with leading coefficient $3n+2$ and by the latter equalities it
satisfies orthogonality conditions (1), (2), (3) for the type $2$ multiple
orthogonal polynomial $p^{\alpha+1}_{n+1,n,n}$. Hence we get (4) by unicity
and complete the proof.

∎

The author would like to apologise for any inconvenience caused.


Furthermore, we adapted our algorithm to allow rotation on the items in the resource augmentation schemes, in this case, in all dimensions.

At last, we were able to extend these results for a more generalized version of the knapsack problem where some additional constraints regarding the items need to be satisfied, such as pairwise conflicts and capacity constraints.
This is mainly due to the versatility provided by the linear program that we use for attending such modifications.

Given the flexibility of our framework, we believe the techniques presented here may be useful for other problems as well.

Missing 'biblatex' package
The bibliography requires the 'biblatex' package.

Elsevier BV

titleA beam search algorithm for the circular packing problem
journaltitleComputers & Operations Research

AIP Publishing

titleCircle packing in regular polygons
journaltitlePhysics of Fluids

titleBin Packing in Multiple Dimensions: Inapproximability Results
and Approximation Schemes
journaltitleMathematics of Operations Research

Society for Industrial
Applied Mathematics

booktitleProceedings of the Twenty-Fifth Annual ACM-SIAM Symposium
on Discrete Algorithms (SODA)
m for Two-Dimensional Bin
Portland, Oregon

* Can't use 'location' + 'address'

titleOptimal Tile Salvage.

titlePacking circles within ellipses
journaltitleInternational Transactions in Operational Research

Elsevier BV

titleA matheuristic approach with nonlinear subproblems for
large-scale packing of ellipsoids
journaltitleEuropean Journal of Operational Research

Springer Science
Business Media LLC

titlePacking ellipsoids by nonlinear optimization
journaltitleJournal of Global Optimization

Elsevier BV

titleMinimizing the object dimensions in circle and sphere packing
journaltitleComputers & Operations Research

Springer Nature Switzerland

booktitleApproximation and Online Algorithms
titleApproximation Schemes Under Resource Augmentation
for Knapsack and Packing Problems of Hyperspheres and Other Shapes

titleApproximation and online algorithms for multidimensional bin
packing: A survey
journaltitleComputer Science Review

Bin packing, Cutting stock, Approximation algorithm
Cutting stock problems and bin packing problems are basically the same
problems. They differ essentially on the variability of the input items. In
the first, we have a set of items, each item with a given multiplicity; in
the second, we have simply a list of items (each of which we may assume to
have multiplicity 1). Many approximation algorithms have been designed for
packing problems; a natural question is whether some of these algorithms can
be extended to cutting stock problems. We define the notion of
“well-behaved” algorithms and show that well-behaved approximation
algorithms for one, two and higher dimensional bin packing problems can be
translated to approximation algorithms for cutting stock problems with the
same approximation ratios.

titleA note on the approximability of cutting stock problems
journaltitleEuropean Journal of Operational Research

Springer New York

booktitleHandbook of Combinatorial Optimization
titleBin Packing Approximation Algorithms: Survey and

titlePerformance Bounds for Level-Oriented Two-Dimensional Packing
journaltitleSIAM Journal on Computing

Annals of Mathematics

titleThe sphere packing problem in dimension 24
journaltitleAnnals of Mathematics

A K Peters

booktitleOrigami$^5$: Proceedings of the 5th International
Conference on Origami in Science, Mathematics and Education (OSME 2010)
titleCircle Packing for Origami Design Is Hard

NP-complete, packing, covering, intersection graph, geometric
titleOptimal packing and covering in the plane are NP-complete
journaltitleInformation Processing Letters

Royal Society of Chemistry (RSC)

titleHard sphere packings within cylinders
journaltitleSoft Matter

Schloss Dagstuhl – Leibniz-Zentrum für Informatik

booktitle37th International Symposium on Computational Geometry
(SoCG 2021)
seriesLeibniz International Proceedings in Informatics (LIPIcs)
titleImproved Approximation Algorithms for 2-Dimensional
Knapsack: Packing into Multiple L-Shapes, Spirals, and More
Dagstuhl, Germany

Keywords: Approximation algorithms, two-dimensional knapsack, geometric

titleA Formulation of the Kepler Conjecture
journaltitleDiscrete & Computational Geometry

Orthogonal Knapsack problem, Square packing, Hypercube packing
noteAutomata, Languages and Programming (ICALP 2006)
titleApproximation algorithms for orthogonal packing problems for
journaltitleTheoretical Computer Science

Hindawi Publishing Corporation

titleA Literature Review on Circle and Sphere Packing Problems:
Models and Methodologies
journaltitleAdvances in Operations Research

Elsevier BV

titleA local search-based method for sphere packing problems
journaltitleEuropean Journal of Operational Research

Schloss Dagstuhl - Leibniz-Zentrum für Informatik

booktitle49th International Colloquium on Automata, Languages, and
Programming, ICALP 2022, July 4-8, 2022, Paris, France
titleA PTAS for Packing Hypercubes into a Knapsack

Institute for Operations Research
the Management Sciences (INFORMS)

titleA Near-Optimal Solution to a Two-Dimensional Cutting Stock
journaltitleMathematics of Operations Research

titleStrena seu de Nive Sexangula (The six-cornered snowflake)

booktitleEuropean Conference on Computational Geometry (EuroCG
titlePacking Segments in a Simple Polygon is APX-hard

titleBin packing can be solved within 1+ $\varepsilon$ in linear

It is shown that the integer linear programming problem with a fixed number
of variables is polynomially solvable. The proof depends on methods from
geometry of numbers.

issn0364765X, 15265471
titleInteger Programming with a Fixed Number of Variables
journaltitleMathematics of Operations Research

titlePacking squares into a square
journaltitleJournal of Parallel and Distributed Computing

Springer International Publishing

booktitleLATIN 2018: Theoretical Informatics
titleTwo-Dimensional Knapsack for Circles

titleOn packing of squares and cubes
journaltitleJournal of combinatorial theory

Schloss Dagstuhl–Leibniz-Zentrum für Informatik

booktitle47th International Colloquium on Automata, Languages, and
Programming (ICALP 2020)
seriesLeibniz International Proceedings in Informatics (LIPIcs)
titleOn the Two-Dimensional Knapsack Problem for Convex
Dagstuhl, Germany

Keywords: Approximation algorithms, geometric knapsack problem, polygons,

titlePolynomial-Time Approximation Schemes for Circle and Other
Packing Problems

titleTechniques and results on approximation algorithms for
packing circles
journaltitleSão Paulo Journal of Mathematical Sciences

Springer Science
Business Media LLC

titleProportional Packing of Circles in a Circular Container
journaltitleCybernetics and Systems Analysis

Annals of Mathematics

titleThe sphere packing problem in dimension 8
journaltitleAnnals of Mathematics

Elsevier BV

titleIterated Tabu Search and Variable Neighborhood Descent for
packing unequal circles into a circular container
journaltitleEuropean Journal of Operational Research

get arXiv to do 4 passes: Label(s) may have changed. Rerun