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\begin{center}
\vskip 1cm{\Large\bf
On a Family of Sequences Related to \\
\vskip .1in
Chebyshev Polynomials
}
\vskip 1cm
Andrew N. W. Hone\footnote{Corresponding author. Currently on leave in the
School of Mathematics and Statistics,
University of New South Wales, Sydney NSW 2052, Australia. } \\
School of Mathematics, Statistics \& Actuarial Science \\
Sibson Building \\
University of Kent \\
Canterbury CT2 7FS \\
UK \\
\href{mailto:anwh@kent.ac.uk}{\tt anwh@kent.ac.uk} \\
\ \\
L. Edson Jeffery \\
11918 Vance Jackson Road \\
San Antonio, TX 78230 \\
USA \\
\href{mailto:lejeffery2@gmail.com}{\tt lejeffery2@gmail.com} \\
\ \\
Robert G. Selcoe \\
16214 Madewood Street \\
Cypress, TX 77429 \\
USA \\
\href{mailto:rselcoe@gmail.com}{\tt rselcoe@gmail.com}
\end{center}
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\begin{abstract}
We consider the appearance of primes in
a family of linear recurrence sequences labelled by a positive integer $n$.
The terms of each sequence correspond to a particular class of Lehmer
numbers, or (viewing them as polynomials in $n$) dilated versions of
the so-called Chebyshev polynomials of the fourth kind, also known as
airfoil polynomials. We prove that when the value of $n$ is given
by a dilated Chebyshev polynomial of the first kind evaluated at a
suitable integer, either the sequence contains a single prime, or no
term is prime. For all other values of $n$, we conjecture that the
sequence contains infinitely many primes, whose distribution has
analogous properties to the distribution of Mersenne primes among the
Mersenne numbers. Similar results are obtained for the sequences
associated with negative integers $n$, which correspond to Chebyshev
polynomials of the third kind, and to another family of Lehmer
numbers.
\end{abstract}
\section{Introduction}
Consider the linear recurrence of second order given by
\beq \label{srec}
s_{k+2} - n\, s_{k+1} + s_k =0,
\eeq
together with the initial conditions
\beq\label{inits}
s_0 = 1, \qquad s_1 = n+1.
\eeq
For each integer $n$, this generates an integer sequence that begins
\beq\label{sseq}
\begin{array}{l}
1,n+1,n^2+n-1, n^3+n^2-2n-1, n^4+n^3-3n^2-2n+1,
\\ n^5+n^4-4n^3-3n^2+3n+1, \ldots .
\end{array}
\eeq
The sequence can also be extended backwards to negative indices $k$, so that in particular
$s_{-1}=-1=-s_0$, which implies
that it has the
symmetry
\beq\label{sym}
s_k(n)=-s_{-k-1}(n)
\eeq
for all $k$.
In this way we obtain a sequence that we denote
by $(\,s_k(n)\,)_{k\in\Z}$, where the argument denotes the dependence
on $n$.
We can also interpret this as a sequence of polynomials in
the variable $n$, with the integer sequences being obtained by
substituting particular values for the argument. From this point of view, it is apparent from the recursive definition that,
for each $k\geq 0$, $s_k(n)$ is a monic polynomial of degree $k$ in $n$ with integer coefficients. In fact, these are
rescaled (or dilated) versions of polynomials that are used to determine the pressure distribution in linear airfoil theory, being given
by
\beq\label{airfoil}
s_k(n) = W_k({n}/{2}), \qquad W_k (\cos\theta ) = \frac{ \sin\Big( (2k+1)\theta /2 \Big) } {\sin(\theta /2)},
\eeq
where $W_k$ are known as the Chebyshev polynomials of the fourth kind \cite{mh}, or the airfoil polynomials of the second kind (see \cite{nasa}, where the notation $u_k$ is used in place of $W_k$).
As a function of $\theta$, the expression on the far right-hand side of (\ref{airfoil}) is known as the Dirichlet kernel in Fourier
analysis, where it is usually denoted $D_k(\theta)$ \cite{dym}.
Compared with those of the third and fourth kinds,
the properties of Chebyshev polynomials of the first and second kinds are much
better known,
and in what follows we will make extensive use of connections with the latter two sets of polynomials.
The
primary goal of
this article is to describe the case where $n$ is a positive integer, but before proceeding, we
consider the sequences obtained for some particular small values of $|n|\leq 2$, which will mostly be excluded from subsequent analysis, but are relevant nevertheless. In the case $n=0$, the sequence
$(\,s_k(0)\,)$ begins
\beq\label{modn}
1,1,-1,-1,\ldots,
\eeq
and repeats with period 4; we mention this case because it is equivalent to the sequence $(\,s_k(n) \,\bmod n\,)$.
When $n=1$ the sequence has period 6, being specified by the six initial terms
\beq\label{n1}
1,2,1,-1,-2,-1, \ldots,
\eeq
and for $n=-1$ the sequence repeats the values
\beq\label{nm1}1,0,-1
\eeq
with period 3.
For $n=2$ the sequence grows linearly with $k$, beginning with
\beq\label{n2}
1,3,5,7,9,11,\ldots,
\eeq
and consists of the odd integers, that is
\beq\label{lin}
s_k(2) = 2k+1,
\eeq
while for $n=-2$ the sequence has period 2, being given by
\beq\label{nm2}
s_k(-2) =(-1)^k.
\eeq
For each integer $n\geq 3$ the sequence increases monotonically for $k\geq 0$ and grows exponentially with $k$ (see below
for details).
Sequence \seqnum{A269254} in the Online Encyclopedia of Integer Sequences (OEIS) \cite{oeis} records the first appearance of a prime
term in $(\,s_k(n)\,)$.
\begin{definition} (Sequence \seqnum{A269254}.) \label{adef}
For each integer $n\geq 1$, if the sequence of terms $(\,s_k(n)\,)_{k\geq 0}$ with non-negative indices contains
a prime, then let $a_n$ be the smallest value of $k\geq 1$ such that $s_k(n)$ is prime; or otherwise, if there is no such
term, let $a_n=-1$.
\end{definition}
There is also sequence \seqnum{A269253}, whose $n$th term is given by the first prime to
appear in $(\,s_k(n)\,)_{k\geq 0}$, or by $-1$ if no prime appears.
To illustrate the above definition,
let us start with $n=1$: since the first prime term in
the sequence (\ref{n1}) is $s_1(1)=2$, it follows that $a_1=1$. Similarly, for $n=2$, the first prime in (\ref{n2}) is $s_1(2)=3$, so $a_2=1$; but for $n=3$, the sequence $(\,s_k(3)\,)$ begins $1,4,11,\ldots$, so $a_3=2$. In cases where a prime term has appeared in the sequence $(\,s_k(n)\,)$, the value of $a_n$ is immediately determined. The sequence
$(a_n)_{n\geq 1}$ begins with the following terms for $1\leq n\leq 34:$
\beq\label{anseq}
\begin{array}{l}
1, 1, 2, 1, 2, 1, -1, 2, 2, 1, 2, 1, 2, -1, 2, 1, 3, 1, 2, 2, 2, 1, -1, 2, 6,
2, 3, 1, \\ 3, 1,
2, 9, 9, -1,\ldots .
\end{array}
\eeq
All of the positive values above can be checked very rapidly, and it turns out that all values of $a_n>0$ are of the form $(p-1)/2$, where $p$ is an odd prime: this is a direct consequence of Lemma \ref{2kp1} below. What is less easy to verify is the negative values $a_{7}=a_{14}=a_{23}=a_{34}=-1$ displayed above, indicating no primes. For instance, when $n=7$, the sequence
$(\,s_k(7)\,)$ begins with
\beq\label{n7}
1,8,55, 377,2584,17711,121393,832040,5702887,\ldots,
\eeq
and it can be verified that none of these first few terms are prime;
but to show that $a_7=-1$ it is necessary to prove that
$s_k(7)$ is composite for all $k>0$: a proof of this fact can be found in section \ref{fact}, while another proof appears in section \ref{chebv} in a broader setting.
In fact, in order to understand the family of sequences $(\,s_k(n)\,)$ with positive $n$, it will be natural
to consider negative integer values of $n$ as well. In that case, it is helpful to define the family of sequences $(\,r_k(n)\,)$
given by
\beq\label{rsrel}
r_k(n)=(-1)^k\, s_k(-n).
\eeq
It is straightforward to show by induction that, for fixed $n$, the sequence
$(\,r_k(n)\,)$ satisfies the same recurrence (\ref{srec}) but with different initial conditions,
namely
$$
r_{k+2} - n\, r_{k+1} + r_k =0,
$$
together with
\beq\label{rinits}
r_0 = 1, \qquad r_1 = n-1.
\eeq
For integer $n$,
this generates an integer sequence that begins
\beq\label{rseq} \begin{array}{l}
1,n-1,n^2-n-1, n^3-n^2-2n+1, n^4-n^3-3n^2+2n+1, \\ n^5-n^4-4n^3+3n^2+3n-1, \ldots .
\end{array}
\eeq
Up to rescaling $n$ by a factor of 2, this sequence of polynomials arises in describing the downwash distribution
in linear airfoil theory,
and in this context they are referred to as
the airfoil polynomials of the first kind \cite{nasa}, denoted $t_k$; with the alternative notation
$V_k$ they are also referred to as the Chebyshev polynomials of the third kind \cite{mh}, so that
\beq\label{1stairfoil}
r_k(n) = V_k({n}/{2}), \qquad V_k (\cos\theta ) = \frac{ \cos\Big( (2k+1)\theta /2 \Big) } {\cos(\theta /2)}.
\eeq
Since $n=2\cos\theta$, the identity (\ref{rsrel}) can also be obtained immediately by taking $\theta\to\theta+\pi$ in (\ref{airfoil}), and comparing with (\ref{1stairfoil}).
There is another OEIS sequence that is relevant here, corresponding to the first appearance of a prime in the
sequence defined by (\ref{rsrel}) for each positive integer $n$.
\begin{definition} (Sequence \seqnum{A269252}.) \label{atdef}
For each integer $n\geq 1$, if the sequence of terms $(\,r_k(n)\,)_{k\geq 0}$ with non-negative indices contains
a prime, then let $\tilde{a}_n$ be the smallest value of $k\geq 1$ such that $r_k(n)$ is prime; or otherwise, if there is no such
term, let $\tilde{a}_n=-1$.
\end{definition}
There is also sequence \seqnum{A269251}, whose $n$th term is given by the first prime to
appear in $(\,r_k(n)\,)_{k\geq 0}$, or by $-1$ if no prime appears.
For comparison with
\seqnum{A269254}, note that the first few terms of \seqnum{A269252} for $1\leq n\leq 34$ are given by
\beq\label{atnseq} \begin{array}{l}
-1, -1, 1, 1, 2,
1, 2, 1, 2, 2,
2, 1, 3, 1, 3,
2, 2, 1, 14, 1,
2, 2, 3, 1, 2,
5, 2, 36, \\ 2, 1,
2, 1, 15, -1,
\ldots .
\end{array}
\eeq
The two initial $-1$ values that appear above for $n=1,2$ clearly correspond to
(\ref{nm1}) and (\ref{nm2}), respectively, while the first non-trivial case
to consider is the value $-1$ that appears for $n=34$, corresponding to the sequence
$(\, r_k(34)\,)$, which begins with
\beq\label{nm34}
1,
33,
1121,
38081,
1293633,
43945441,
1492851361,
50713000833
,\ldots;
\eeq
again it can be verified that none of these first few terms are prime, while a proof
that all terms with $k>0$ are composite for this and certain other values of $n$
is given in section 5.
Aside from the connection with Chebyshev polynomials, the numbers
$s_k(n)$ and $r_k(n)$ also correspond to particular instances of Lehmer numbers with
odd index, which are closely related to sequences of Lucas numbers.
Prime divisors
in sequences of Lucas and
Lehmer numbers have been studied for some time; see
e.g., \cite{bilu, rotk, schinzel, stewart} for some
general results, or see
\cite{estw} for a more
elementary introduction to primitive divisors. However, to the best of our
knowledge, the question of when such sequences are
without prime terms, or of where the first prime appears in such
sequences, has not been considered in detail before,
except in the case $n=6$.
The case $n=6$ corresponds to the so-called NSW numbers,
named after \cite{nsw} (sequence \seqnum{A002315}). An NSW number $q$ can be characterized
by
there being some $r$ such that the pair of positive integers $(q,r)$
satisfies the Diophantine equation
$$
q^2+1=2r^2.
$$
The sequence of NSW numbers is given by $q=s_k(6)$ for $k\geq 0$,
with the corresponding solution to the above equation
being $(q,r)=( s_k(6),r_k(6))$. The subsequence of prime NSW numbers
is of particular interest in relation to finite simple groups of square order:
the symplectic group of dimension 4 over the finite field $\F_q$ has a
square order if and only if $q$ is a prime NSW number, with the order
being $(q^2(q^2-1)r)^2$. The first prime NSW number is $s_1(6)=7$,
and the symplectic group of dimension 4 over $\F_7$ is of order
$11760^2$.
For an arbitrary linear recurrence relation of second order, that is
$$
x_{k+2}=a\, x_{k+1}+b\, x_k, \qquad (a,b)\in\Z^2,
$$
the general question of whether it generates a sequence
without prime terms has been considered for some time. If either the coefficients
$a,b$ or the two initial values $x_0,x_1$ have a common factor then it is obvious
that all terms $x_k$ for $k\geq2$ have the same common factor, so
the main case of interest is where $\gcd(a,b)=1=\gcd(x_0,x_1)$.
In the case of the Fibonacci recurrence with $a=b=1$,
the groundbreaking result was due to
Graham, who found a sequence
whose first two terms are relatively prime and which consists only of composite integers \cite{graham}.
This result was generalized to arbitrary
second-order recurrences by Somer \cite{somer} and Dubickas et al.\ \cite{composite}.
An outline of the paper is as follows. The next section serves to set up notation
and provide a rapid introduction
to the properties of dilated Chebyshev polynomials of the first and second kinds,
which will be used extensively in the sequel, and also contains the
required definitions of the corresponding
sequences of Lucas and Lehmer numbers that appear subsequently. Section 3 provides a
very brief review of some standard facts about linear recurrence sequences and products
of such sequences, before a presentation of examples and preliminary results about values of $n$ for which the terms
$s_k(n)$ factor into a product of two linear recurrence sequences; this serves
to illustrate and motivate
the results which appear in section 5. As preparation for the latter,
section 4 contains a collection of various general properties
of the sequences
$(\,s_k(n)\,)$. The main results of the paper,
on the factorization of $(\,s_k(n)\,)$ and $(\,r_k(n)\,)$ when
$n$ is a dilated Chebyshev polynomial of the first kind evaluated at
integer argument (Chebyshev values),
are presented in section 5. Section 6 considers the appearance
of primes in these sequences in the case that $n$ is not one of the Chebyshev values,
and gives heuristic arguments and numerical evidence to support a
conjecture to the effect that the
behaviour is analogous to that of the sequence of Mersenne primes.
Some conclusions are made in the final section, and there are two appendices:
the first is a collection of data on prime appearances, and the second
is a brief catalogue of related
sequences in the OEIS.
This paper arose out of a series of posts to the {\tt SeqFan} mailing list, with contributions from many people,
both professional and recreational mathematicians.
Our aim throughout has been to make the presentation as explicit as possible, and for the sake of completeness
we have stated several standard facts and definitions, as well as providing direct, elementary proofs of almost every statement (even when some of them are particular cases of more general results
in the literature). We hope that in this form it will be possible for our work to be appreciated by sequence
enthusiasts of every persuasion.
\section{Dilated Chebyshev polynomials and Lehmer numbers}
The families of Chebyshev polynomials arise in the theory of orthogonal polynomials, and have diverse applications in numerical analysis \cite{mh}. There are
four such families,
and while the Chebyshev polynomials of the first and second kinds are well studied in the literature, those of the third and fourth kinds are not so well known, and some of their
connections to arithmetical problems have only been considered quite recently \cite{jonny}.
In order to define scaled versions of the standard Chebsyhev polynomials in terms of trigonometric functions,
let
$$
n = 2\cos\theta = \la + \la^{-1},
$$
so that we may write
\beq\label{ladef}
\la = \frac{n+\sqrt{n^2 - 4}}{2} = e^{\ri\theta},
\eeq
where $\ri=\sqrt{-1}$.
Then the formulae
\beq\label{1st}
\Tc_k(2\cos\theta) = 2\, \cos (k\theta), \qquad
\Uc_k(2\cos\theta) = \frac{\sin\left((k+1)\theta\right)}{\sin\theta}
\eeq
define $\Tc_k$, $\Uc_k$ as polynomials in $n$, for all $k\in\Z$.
In \cite[Chapter 18]{dlmf} these polynomials are referred to as the dilated Chebyshev polynomials of the first and second kinds, and they are denoted by $C_k$, $S_k$ respectively. In standard notation, the classical Chebyshev polynomials of the first and second kinds are written as $T_k$ and $U_k$,
and their precise
relationship with the dilated polynomials used here is as follows:
$$
\Tc_k(n) = 2\,T_k\left(\frac{n}{2}\right), \qquad \Uc_k(n) = U_k\left(\frac{n}{2}\right).
$$
It is straightforward to show from the definitions (\ref{1st}) that the dilated Chebyshev polynomials of the first and second kinds satisfy the same recurrence (\ref{srec}) as the sequence $(\,s_k(n) \,)$, but with different initial values. For example, to verify that the sequence $(\,\Uc_k(n) \,)$ satisfies the recurrence, it is sufficient to note that
\beq \label{urec} \begin{array}{rcl}
\Uc_{k}(n) - n \, \Uc_{k-1}(n) +\Uc_{k-2}(n) &
= & \frac{\sin\left((k+1)\theta\right) -2\cos\theta \,{\sin(k\theta)}+ {\sin\left((k-1)\theta\right)} }{\sin\theta},
\end{array}
\eeq
and then observe that the right-hand side above vanishes as a consequence of
the addition formula for sine in the form
\beq\label{sine}
\sin (\theta+\phi)+\sin(\theta -\phi) = 2\sin\theta \cos\phi.
\eeq
For comparison with other texts, we note that the sequence of dilated first kind polynomials begins thus:
\beq\label{1stk}
(\, \Tc_k(n)\,): \qquad 2,n, n^2 -2, n^3-3n, n^4 - 4n^2+2, n^5-5n^3+5n, \ldots .
\eeq
In contrast, the sequence of dilated second kind polynomials begins as
\beq\label{2ndk}
(\, \Uc_k(n)\,): \qquad 1,n, n^2 -1, n^3-2n, n^4 - 3n^2+1, n^5-4n^3+3n, \ldots .
\eeq
For future reference, we note the standard identities
\beq\label{tid}
\Tc_{ab}(n) = \Tc_a(\Tc_b(n))
\eeq
and
\beq\label{uid}
\Uc_{ab-1}(n) = \Uc_{a-1}(\Tc_b(n)) \,\Uc_{b-1}(n),
\eeq
which follow from the trigonometric definitions above.
In order to get a formula for the coefficients of the polynomials $s_k(n)$, we present an explicit expansion for
the dilated Chebyshev polynomials of the second kind. Although this can be found elsewhere in the literature (cf.\ equations (5.74) and (6.129) in \cite{conc}), for
completeness we present an elementary proof.
\begin{proposition}\label{uprop}
The dilated Chebyshev polynomials of the second kind are given by
\beq\label{uexp}
\Uc_k(n) = \sum_{i=0}^{\left \lfloor{\frac{k}{2}}\right \rfloor}(-1)^i { k-i \choose i}
n^{k-2i}. \eeq
\end{proposition}
\begin{proof} First note that for $k=0,1$ the sum on the right-hand side of (\ref{uexp})
agrees with the initial terms $\Uc_0=1$, $\Uc_1=n$. Then, upon substituting the sum formula into the recurrence (\ref{urec})
and comparing powers of $n$, after dividing by $(-1)^i$ we see that the coefficient of $n^{k-2i}$ yields the identity
$$
{ k-i \choose i} - { k-i -1 \choose i} - { k-i -1 \choose i-1 } = 0
$$
for binomial coefficients.
Thus the sequences defined by the left-hand and right-hand sides of (\ref{uexp}) satisfy the same recurrence with the
same initial conditions, so they must coincide.
\end{proof}
For use in what follows, we also define Lehmer numbers. Given a quadratic polynomial in $X$
with roots $\al,\be$, that is
\beq\label{lquad}
X^2 -\sqrt{R}\, X + Q=(X-\al)(X-\be), \qquad Q,R\in \Z,
\eeq
where it is assumed that $Q,R$ are coprime and $\al/\be$ is not a root of unity,
there are two associated sequences of Lehmer numbers,
which (adapting the
notation of
\cite{epsw}) we denote by $L^-_k(\sqrt{R},Q)$ and
$L^+_k(\sqrt{R},Q)$, where
\beq\label{lm}
L^-_k(\sqrt{R},Q) =
\begin{cases}
\frac{\al^k-\be^k}{\al-\be}, & \text{if } k\,\,\text{odd}; \\
\frac{\al^k-\be^k}{\al^2-\be^2}, & \text{if } k\,\,\text{even},
\end{cases}
\eeq
and
\beq\label{lp}
L^+_k(\sqrt{R},Q) = \begin{cases}
\frac{\al^k+\be^k}{\al+\be}, & \text{if } k\,\,\text{odd}; \\
{\al^k+\be^k}, & \text{if } k\,\,\text{even}.
\end{cases}
\eeq
The sequences of Lehmer numbers can be viewed as generalizations of Lucas sequences. Assuming that
$R$ is a perfect square, so $P=\sqrt{R}\in\Z$, the two types of Lucas sequences associated to the
quadratic $X^2-PX+Q=(X-\al)(X-\be)$
are given
by
\beq\label{luc}
\ell^-_k (P,Q) = \frac{\al^k-\be^k}{\al - \be}, \qquad \ell^+_k (P,Q)= \al^k +\be^k;
\eeq
the corresponding Lehmer numbers $L^{\pm}_k(P,Q)$ are obtained from the Lucas numbers
$\ell^{\pm}_k(P,Q)$ by removing trivial factors.
From the above definitions, there is a
clear link between Chebyshev polynomials and Lucas/Lehmer numbers, which can be summarized in the following
\begin{proposition}\label{lehch}
For integer values $n$, the sequences of dilated Chebyshev polynomials of the first and second kinds coincide with
particular Lucas sequences, that is
\beq\label{lucas}
\Tc_k(n) = \ell_k^+(n,1), \qquad \Uc_{k-1}(n) = \ell_k^-(n,1),
\eeq
while the sequences
generated by (\ref{srec}) with initial values (\ref{rinits}) and (\ref{inits}) consist of Lehmer numbers
with odd index, namely
\beq\label{skleh}
r_k(n) = L^+_{2k+1}\left(\sqrt{n+2},1\right), \quad
s_k(n) = L^-_{2k+1}\left(\sqrt{n+2},1\right)
\eeq
respectively,
for all $k$.
\end{proposition}
\begin{proof}
The formulae for $\Tc_k$ and $\Uc_k$ follow immediately from comparison of (\ref{1st}) with (\ref{luc}), requiring from (\ref{ladef}) that $\al =\la =e^{\ri\theta} =\be^{-1}$ in (\ref{lquad}). For the proof of the second part of the statement, note that
taking $k=0,1$ gives $L_1^-(\sqrt{n+2},1)=1$
and
$L_3^-(\sqrt{n+2},1)=n+1$, while a short calculation shows that
$L^-_{2k+1}(\sqrt{n+2},1)$ satisfies the same recurrence (\ref{srec}) as $s_k(n)$,
and similarly for the other sequence given by
$r_k(n)=(-1)^k s_k(-n)$; so for each equation in (\ref{skleh}), the sequences
given by their left/right-hand sides coincide.
\end{proof}
\begin{remark}
There are also expressions for
$r_k(n)$ and $s_k(n)$
in terms of
dilated Chebyshev polynomials of the first/second kinds, respectively, with
argument $\sqrt{n+2}$: see (\ref{p2f}) and (\ref{mainf}) below.
\end{remark}
By writing the roots of the polynomial $X^2-\sqrt{n+2}\,X +1$ as
\beq\label{aldef}
\al^{\pm 1} = \frac{\sqrt{n+2}\pm\sqrt{n-2}}{2},
\eeq
we have an alternative way to identify the terms in sequence \seqnum{A269254}.
\begin{corollary} (Alternative characterization of sequence \seqnum{A269254}.) \label{altdef}
For each $n\geq 3$, if $\al$ is defined by (\ref{aldef}), then $a_n$ is that positive integer $k$ yielding the smallest prime of
the form
\beq\label{alform}
\frac{\al^{2k+1} - \al^{-(2k+1)}}{\al-\al^{-1}},
\eeq
or $a_n=-1$
if no such $k$ exists.
\end{corollary}
The sequence \seqnum{A269252} can be identified in terms of the characteristic roots of
(\ref{aldef}) in a similar way.
\begin{corollary} (Alternative characterization of sequence \seqnum{A269252}.) \label{altrdef}
For each $n\geq 3$, if $\al$ is defined by (\ref{aldef}), then $\tilde{a}_n$ is that positive integer $k$ yielding the smallest prime of
the form
\beq\label{ralform}
\frac{\al^{2k+1} + \al^{-(2k+1)}}{\al+\al^{-1}},
\eeq
or $\tilde{a}_n=-1$
if no such $k$ exists.
\end{corollary}
\section{Some surprising factorizations} \label{fact}
In this section we briefly recall some basic facts about sequences generated by linear recurrences, before looking at
some special properties of the family of sequences $(\,s_k(n)\,)$. We assume that all recurrences are defined over the field $\C$ of complex numbers. (In the next
section we will also consider recurrences in finite fields or residue rings.)
For a broad review of linear recurrences in a more general setting,
the reader is referred to \cite{epsw}.
For what follows, it is convenient to make use of the forward shift, denoted $\rS$, which
is a linear operator that acts on any
sequence $(f_k)$ with index $k$ according to
$$\rS \, f_k = f_{k+1}.$$
With this notation, the fact that a sequence $(x_k)$ satisfies a linear recurrence relation of
order $N$ with constant coefficients can be expressed in the form
\beq\label{linrel}
F(\rS) \, x_k =0,
\eeq
where $F$ (of degree $N$) is the characteristic polynomial of the recurrence.
\begin{definition} A {\it decimation} of a sequence $(x_k)_{k\in\Z}$ is any subsequence of the form
$(x_{i+dk})_{k\in\Z}$, for some fixed integers $i,d$, with $d\geq 2$. A particular name for the case $d=2$ is a
{\it bisection}, $d=3$ is a
{\it trisection}, and in general this is a decimation
of order $d$.
\end{definition}
\begin{remark} The case of decimations of linear recurrences defined over finite fields is considered in \cite{dm}.
\end{remark}
Since, at least in the case that all the roots $\la_1,\la_2,\ldots, \la_N$ of $F$ are distinct,
the general solution of (\ref{linrel}) can be written as a linear combination of $k$th powers of the $\la_j$,
it is apparent that
the terms of a decimation
of order $d$ are given by
$
x_{i+dk} = \sum_{j=1}^NA_j\,\la_j^{dk}
$,
for some coefficients $A_j$.
Hence
the decimation
satisfies the linear recurrence
\beq \label{diss}
\prod_{j=1}^N (\rS -\la_j^d) \, x_{i+dk} = 0.
\eeq
(The recurrence for the decimation
has the same form in the case of repeated roots.)
Decimations of the sequence $(\,s_k(n)\,)$ will be considered in Proposition \ref{linrels} in the next section.
Given two sequences $(x_k)$, $(y_k)$ that satisfy linear recurrences of order $N,M$ respectively, the product
sequence $$(z_k)=(x_ky_k)$$ also satisfies a linear recurrence. The following result is well known.
\begin{theorem}\label{prodt}
The product $(z_k)=(x_ky_k)$ of two sequences that satisfy linear recurrences of order $N,M$ satisfies a
linear recurrence of order at most $NM$.
\end{theorem}
To prove the theorem in the generic situation where the recurrences for $(x_k)$, $(y_k)$ both have
distinct characteristic roots, given by
$\la_i$, $1\leq i\leq N$ and
$\mu_j$, $1\leq j\leq M$ respectively, observe that each product $\nu_{i,j}= \la_i\mu_j$ is a characteristic root for the linear recurrence satisfied by $(z_k) $,
that is
\beq\label{nusum}
\prod_{i,j}(\rS - \nu_{i,j}) \, z_k =0,
\eeq
where the sum is over a maximum set of $i,j$ that give distinct $\nu_{i,j}$; so
if the $\nu_{i,j}$ are all different from each other
then the order of the recurrence is exactly $MN$, but the order could be smaller if
some of the $\nu_{i,j}$ coincide. For the general situation with repeated roots, see \cite{zm}.
We now consider an observation concerning the sequences $(\,s_k(n)\,)$ for the special values
$n=j^2-2$ where $j\in \Z$, which includes the cases $n=7,14,23,34$ that have $a_n=-1$ in (\ref{anseq}).
The fact is that for all these values, there is a
factorization of the form
\beq\label{t2fac}
s_k(j^2-2) =r_k(j) s_k(j) ,
\eeq
where both factors on the right-hand side above satisfy a
linear recurrence of second order. This is surprising, because in the light of Theorem \ref{prodt} one would naively
expect such a product to satisfy a recurrence of order 4.
\begin{theorem}\label{t2}
For the values $n=j^2-2$, the terms of the sequence $(\,s_k(n)\,)$ admit the factorization
(\ref{t2fac}),
where $r_k(j)$ satisfies the same recurrence as $s_k(j)$, that is
\beq\label{p2}
r_{k+2}(j) -j\, r_{k+1}(j) +r_k(j)=0,
\eeq
with the initial values
\beq\label{rjinits}
r_0(j)=1, \qquad r_1(j)=j-1.
\eeq
Thus for all $j\in\Z$ the formula (\ref{t2fac}) expresses $s_k(j^2-2)$ as a product of two integers.
\end{theorem}
\begin{proof}
In the case $n=j^2-2$, the formula (\ref{aldef}) fixes the characteristic roots of the recurrence (\ref{srec})
as $\la=\al^2$, $\la^{-1}=\al^{-2}$, where $\al = (j+\sqrt{j^2-4})/2$; so $\al+\al^{-1}=j$, and the square root of
$\al$ can be fixed so that $\al^{1/2}+\al^{-1/2}=\sqrt{j+2}$. Then, by applying the difference of two squares to the numerator and denominator of (\ref{alform}), it follows that
\beq\label{dos}
s_k(j^2-2)
= \left(\frac{\al^{(2k+1)/2}+\al^{-(2k+1)/2}}{\al^{1/2}+\al^{-1/2}}\right)\,
\left(\frac{\al^{(2k+1)/2}-\al^{-(2k+1)/2}}{\al^{1/2}-\al^{-1/2}}\right)
\eeq
which is the factorization (\ref{t2fac}) with
$r_k(j)$
and $s_k(j)$
given by making the replacement $n \to j$ in
(\ref{skleh}).
(For an alternative expression for these factors, see (\ref{rsu}) in
Remark \ref{altf} below.)
Each of the factors above is a linear combination of $k$th powers of the characteristic roots $\al, \al^{-1}$, and $r_k(j)=(-1)^k s_k(-j)$ as in (\ref{rsrel}), so they each
satisfy the same recurrence
(\ref{p2}) with an appropriate set of initial values.
\end{proof}
\begin{remark}\label{klee}
Generically, the product of any two solutions of the recurrence (\ref{p2}) would have
three characteristic roots, namely
$\al^2,\al^{-2},1$, giving a recurrence of order 3 in (\ref{nusum}), but the potential root 1 cancels from the product
(\ref{dos}), giving the second-order recurrence (\ref{srec}) with $n=j^2-2$.
An inductive proof of the preceding result was given by Klee in a post to the Seqfan mailing list: see \cite{klee} for
details. However, the factorization (\ref{dos}) in the form
$ L^-_{2k+1}(j,1) = L^+_{2k+1}(\sqrt{j+2},1)\, L^-_{2k+1}(\sqrt{j+2},1)$ appears to be well known in the literature on Lehmer numbers; see e.g., \cite{caldwell}
and references.\footnote{In particular, see \url{http://primes.utm.edu/top20/page.php?id=47 }
for a sketch of a proof of Theorem \ref{t2}.}
\end{remark}
\begin{example}
In the case $n=7$, there is the factorization
$$
s_k(7)=r_k(3)\,s_k(3),
$$
where the first terms of the factor sequences are
$$ \begin{array}{l}
(\,r_k(3)\,): 1,2,5,13,34,89,233,610,1597,\ldots , \\
(\,s_k(3)\,): 1,4,11,29,76,199,521,1364,3571,\ldots,
\end{array}
$$
which multiply together to give the terms in (\ref{n7}). Since both $(\,r_k(3)\,)$
and $(\,s_k(3)\,)$ are strictly increasing sequences, it follows that $s_k(7)$ is composite for
all $k\geq 1$, and hence $a_7=-1$, as asserted previously.
\end{example}
As a consequence of the factorization (\ref{t2fac}), one can show similarly that for all integers $j\geq 3$, the terms $s_k(j^2-2)$
are composite for $k\geq 1$, and thus $a_{j^2-2}=-1$ for all $j\geq 3$ (for full details, see the proof of
Theorem \ref{maint} below). In particular, Theorem \ref{t2} accounts for all the values $n=7,14,23,34$ with $a_n=-1$ that are shown in the list (\ref{anseq}).
The question is now whether there are other cases with $a_n=-1$, for which $n\neq j^2 -2$ for
some $j$.
It turns out that the answer to this question is affirmative, and the first case with $a_n=-1$ that does not fit into the above pattern is $n=110$ \cite{setal}.
\begin{example}\label{n110}
The sequence $(\,s_k(110)\,)_{k\geq 0}$, beginning with
\beq\label{n110s} \begin{array}{l}
1, 111, 12209, 1342879, 147704481, 16246150031, 1786928798929, \\
196545921732159, \ldots ,
\end{array}
\eeq
appears as number \seqnum{A298677} in the OEIS. To see that none of the terms are prime, first of all note that the sequence
$(\,s_k(110)\, \bmod 111\,)$ is periodic with period 3: it is equivalent to the sequence (\ref{nm1}); this observation is
a special case of Lemma \ref{resper} below. Thus it is helpful to consider the three
trisections $(\,s_{3k+i}(110)\,)$ for $i=0,1,2$, each of which satisfy the
second-order recurrence
\beq\label{tris}
s_{3(k+2)+i}(110) - 1330670\,s_{3(k+1)+i}(110)+ s_{3k+i}(110)=0,
\eeq
as follows by applying the formula (\ref{diss}).
The easiest case is $i=1$, since $s_{3k+1}\equiv 0 \pmod{111}$ for all
$k$;
so in this subsequence, the first term $111=3\times 37$ is composite, and subsequent terms
$ 147704481=111\times 1330671$, $196545921732159=111\times 1770683979569$, etc. are all
multiples of 111. The trisection $(\,s_{3k}(110)\,)$ is the subsequence beginning with
$s_0(110)=1$, and then $s_3(110)=1342879=9661\times 139$,
$s_6(110)=1786928798929=116876761\times 15289$, and by
induction it can be shown that each of these terms is divisible by the
corresponding one for the sequence $(\,s_{3k}(5)\,)=1,139,15289,\ldots$, so
that
\beq\label{trifac}
s_{3k}(110) = R_{3k}(5) \, s_{3k}(5),
\eeq
where the integer sequence of prefactors satisfies the third order recurrence
\beq\label{3rdo}
R_{3(k+3)}(5) -12099\, \Big( R_{3(k+2)}(5) - R_{3(k+1)}(5)\Big) -R_{3k}(5) = 0.
\eeq
Similarly, for the remaining trisection, namely
$(\,s_{3k+2}(110)\,)$, one has
\beq\label{trifac2}
s_{3k+2}(110) = R_{3k+2}(5)\, s_{3k+2}(5),
\eeq
where the prefactor sequence $(\, R_{3k+2}(5)\,)$ consists of integers and satisfies the same recurrence
(\ref{3rdo}). In fact, it is not necessary to consider this trisection separately, since its properties follow
immediately from
extending $(\,s_{3k}(110)\,)$ to $k<0$ and using the symmetry (\ref{sym}).
These observations show that all the terms in (\ref{n110s}) are composite
for $k>0$, confirming that $a_{110}=-1$ as claimed. Moreover,
for all $k$ there is a factorization
\beq\label{trifac3}
s_{k}(110) = R_{k}(5) \,s_{k}(5),
\eeq
where
\beq\label{3rdo2}
R_{k+3}(5) -24\, \Big( R_{k+2}(5) - R_{k+1}(5)\Big) -R_{k}(5) = 0,
\eeq
but the prefactors making up the full sequence $(\,R_k(5)\,)_{k\geq 0}$, that is
$$1,\frac{37}{2},421,9661,\frac{443557}{2},5091241,116876761,
\frac{5366148517}{2}, \ldots,$$
are only integers in the cases (\ref{trifac}) and (\ref{trifac2}), and not when $k\equiv 1 \pmod{3}$.
\end{example}
The values of $n$ with $a_n=-1$ mentioned so far all have one thing in common: they
correspond to values of dilated Chebyshev polynomials of the first kind. Indeed, the
four -1 terms displayed in (\ref{anseq}) appear at the index values
$$7=\Tc_2(3), \quad 14=\Tc_2(4), \quad 23=\Tc_2(5), \quad
34=\Tc_2(6), $$
and Theorem (\ref{t2}) implies that
$s_k(n)$ is composite for all $k\geq 1$ when $n=\Tc_2(j)$, $j\geq 3$, while
$$
110 = \Tc_3(5).
$$
It turns out that for any Chebyshev value $n=\Tc_p(j)$ with $p>1$, there is a factorization analogous to (\ref{t2fac}) or (\ref{trifac3}):
see Theorem \ref{factors} below.
Due to the identity (\ref{tid}), it is sufficient to consider the case of prime $p$ only.
The curious reader might wonder why the values $n=18=\Tc_3(3)$ and
$n=52=\Tc_3(4)$ are missing from the discussion. The reason is that,
although there is a factorization
analogous to (\ref{trifac3}) for these values of $n$, there are the prime terms $s_1(18)=19$ and
$s_1(52)=53$, which imply that $a_{18}=1=a_{52}$;
but it turns out that there are no
other primes in the sequences $(\, s_k(n)\, )_{k\geq 0}$ for $n=18$ or $52$.
See Theorem \ref{maint} for a more general statement which includes all these
Chebyshev values.
\section{General properties of the defining sequences}
By writing the general solution of (\ref{srec}) in terms of the roots of
its characteristic quadratic, and using various expressions for the
dilated
Chebyshev polynomials, as in
section 2,
we immediately obtain a number of equivalent explicit formulae for the
sequence $(\,s_k(n)\,)$.
\begin{proposition}
The terms of the sequence generated by (\ref{srec}) with the initial values (\ref{inits})
are given explicitly by
\beq\label{expl}
s_k(n) = \frac{\la^{k+1} - \la^{-k}}{\la -{1}}
= \Uc_{k-1}(n) + \Uc_k(n),
\eeq
where $\la$ is given in terms of $n$ according to (\ref{ladef}), and by
\beq\label{mainf}
s_k(n) = \Uc_{2k}(\sqrt{n+2})
= \frac{ \sin\Big( (2k+1)\theta /2 \Big) }
{ \sin ( \theta /2 ) } ,
\eeq
and they have the generating function
\beq\label{gf}
G(X,n): = \sum_{j=0}^\infty s_j(n)\, X^j =\frac{1+X}{1-nX+X^2}.
\eeq
\end{proposition}
\begin{proof}
The first formula in (\ref{expl}) is equivalent to (\ref{alform}), with $\la=\al^2$, and the other one
follows by rewriting the Chebyshev polynomials as linear combinations of $\la^k$ and $\la^{-k}$, which
generically provide two independent
solutions of (\ref{srec}).\footnote{The first equality
is invalid when $n=\pm 2$, due to
repeated
roots $\la=\la^{-1}=\pm 1$,
cf.\ (\ref{lin}) and (\ref{nm2}).
}
For the latter set of identities,
let $m=\sqrt{n+2}$, and note that $\Tc_2(m)=n$, so $\theta$ can always be chosen such that
$m=2\cos(\theta/2)$.
The expression on the far right-hand side of (\ref{mainf}) is obtained by
by applying (\ref{sine}) to
the last equality in (\ref{expl}), or by setting
$\al=e^{\ri\theta/2}$ in (\ref{alform}),
and this expression equals $\Uc_{2k}(2\cos(\theta/2))=\Uc_{2k}(m)$.
The generating function (\ref{gf}) follows from using the
first formula in (\ref{expl}) and summing a pair of geometric series.
\end{proof}
\begin{remark} \label{altf}
The last formula in (\ref{expl}) together with (\ref{rsrel}) shows that
the terms on the right-hand side of the factorization (\ref{t2fac})
in the case $n=\Tc_2(j)=j^2-2$ can also be written as
\beq\label{rsu}
r_k(j) = \Uc_k(j)- \Uc_{k-1}(j), \qquad
s_k(j) = \Uc_k(j)+ \Uc_{k-1}(j).
\eeq
\end{remark}
If $5/22$, so $\la^{-k}/(\la -1)<1$ for all $k\geq 0$, and so we have
\begin{corollary} \label{floorf}
For all real $n>5/2$, the terms $s_k(n)$ for $k\geq 0$ are given by
$$
s_k(n)= \left \lfloor{\frac{\la^{k+1}}{\la-1}}\right \rfloor .
$$
\end{corollary}
The recurrence (\ref{srec}) can also be rewritten in matrix form, as
\beq\label{matf}
{\bf v}_{j} = {\bf A}\, {\bf v}_{j-1} ,
\eeq
where
$$
{\bf A} =
\left(\begin{array}{cc} 0 & 1 \\ -1 & n \end{array}\right), \qquad
{\bf v}_{j} = \left(\begin{array}{c} s_{j}(n) \\ s_{j+1} (n) \end{array}\right) ,
$$
hence for all $j$ the terms of the sequence are given in terms of the powers of $\bf A$ by
$$
{\bf v}_{j} = {\bf A}^{j}\, {\bf v}_{0} .
$$
By a standard method of repeated squaring, this allows
rapid calculation of the terms of the sequence.
\begin{proposition}\label{mata}
The $j$th power of the matrix $\bf A$ is given explicitly by
\beq\label{pow}
{\bf A}^{j} = \left(\begin{array}{cc} -\Uc_{j-2}(n) & \Uc_{j-1}(n) \\ -\Uc_{j-1}(n) & \Uc_{j}(n) \end{array}\right),
\eeq
and this can be calculated in $O(\log j )$ steps.
\end{proposition}
\begin{proof}
The formula (\ref{pow}) follows by induction, noting that the columns of the matrix on the right-hand side
satisfy the same recurrence (\ref{matf}) as the vector ${\bf v}_j$, and it is trivially true for $j=0$.
To calculate the powers of $\bf A$ quickly, compute the binary expansion $j=\sum_{i=0}^{d-1}b_i\,2^i$,
where $b_{d-1}=1$ and $d=\log_2 j+1$ is the number of bits, then use repeated squaring to obtain the
sequence $ \tilde{{\bf A}}_i={\bf A}^{2^i}$ for $i=0,1,\ldots, d-1$, and finally evaluate
${\bf A}^j =\prod_{i=0}^{d-1} \tilde{{\bf A}}_i^{b_i}$.
\end{proof}
There are other useful
representations for the terms $s_k(n)$, two of which we record in the following
\begin{proposition}\label{expa}
For $k\geq 0$, the terms of the sequence $(\,s_k(n)\,)$ admit the expansion
\beq\label{nexp}
s_k(n) =
\sum_{i=0}^{\left \lfloor{\frac{k}{2}}\right \rfloor}(-1)^i { k-i \choose i}
n^{k-2i} + \sum_{i=0}^{\left \lfloor{\frac{k-1}{2}}\right \rfloor}(-1)^i
{ k-i -1\choose i}
n^{k-2i-1}
\eeq
in powers of $n$, and the expansion
\beq\label{texp}
s_k(n) = \frac{1}{2}\Tc_0(n) + \sum_{i=1}^k \Tc_i(n)
\eeq
in terms of dilated Chebyshev polynomials of the first kind.
\end{proposition}
\begin{proof} The first expansion (\ref{nexp}) follows from the expression on the far right-hand side of (\ref{expl}),
together with equation (\ref{uexp}). The second expansion (\ref{texp}) corresponds to a standard identity
for the Dirichlet kernel; it can be proved
by noting that dilated first/second kind Chebyshev
polynomials are related via
the identity
$
\Tc_k(n) = 2\Uc_k(n) - n\Uc_{k-1}(n)$, which is easily verified.
Taken together with the recurrence (\ref{urec}),
as well as the last expression in
(\ref{expl}), this gives
\beq\label{diff}
s_k(n) - s_{k-1}(n) = \Tc_k(n).
\eeq
Thus the expansion (\ref{texp}) is obtained by starting from $s_0=1=\frac{1}{2}\Tc_0$ and then taking the
telescopic sum of the first difference formula (\ref{diff}).
\end{proof}
\begin{remark} A different form of series expansion for the airfoil polynomials of the second kind is given in \cite{nasa}.
\end{remark}
\begin{proposition}\label{pshift}
For any odd integer $p$,
\beq\label{psh}
s_{k+p}(n) - s_k(n) = \Tc_{k+\frac{p+1}{2}}(n) \, s_{(p-1)/2}(n).
\eeq
\end{proposition}
\begin{proof} This follows from the trigonometric expression on the far right-hand side of (\ref{mainf}), by applying the addition formula
(\ref{sine}).
\end{proof}
\begin{remark} The formula (\ref{diff}) is the particular case $p=1$ of the above identity.
\end{remark}
\begin{proposition}\label{linrels}
Any decimation
$(\, s_{i+dk}(n)\,)$ of the sequence of order $d$, satisfies the linear recurrence
\beq\label{lind}
s_{i+d(k+1)}(n) - \Tc_d(n)\, s_{i+dk}(n) + s_{i+d(k-1)}(n) = 0.
\eeq
\end{proposition}
\begin{proof} By the first formula for $s_k(n)$ in (\ref{expl}), the terms of the decimated sequence can be written
as linear combinations of $k$th powers of $\la^d$ and $\la^{-d}$, so from the formula (\ref{diss}) we find
$$
\Big(\rS^2 - (\la^d+\la^{-d})\, \rS +1\Big) \,s_{i+dk}(n) =0,
$$
and by using (\ref{ladef}) we see that $\la^d+\la^{-d} = 2\cos(d\theta) = \Tc_d(n)$, which verifies (\ref{lind}).
\end{proof}
It is worth highlighting some particular cases of the preceding two results, namely
the formulae
\beq\label{qpm1a}
s_{2j}(n) +1= s_j(n)\, \Tc_j(n),
\qquad s_{2j+1}(n) -1= s_j(n)\, \Tc_{j+1}(n),
\eeq
of which the first arises by setting $i=0$, $k= 1$, $d= j$ in (\ref{lind}),
while the second comes from taking $k= 0$, $p=2j+1$ in (\ref{psh}). For primality testing of a number $q$, it is often useful to have a factorization, or a
partial factorization, of either $q-1$ or $q+1$ \cite{bls, pom}, and each of the
identities in (\ref{qpm1a})
also has an analogue where the sign of the $\pm 1$ term on the
left-hand side is reversed.
\begin{proposition} \label{qpm} For any integer $j$,
\beq\label{qpm1b}
s_{2j}(n) -1=(n+2)\, r_j(n)\, \Uc_{j-1}(n),
\qquad s_{2j+1}(n) + 1= (n+2)\, r_j(n)\, \Uc_{j}(n).
\eeq
\end{proposition}
\begin{proof}
For the first identity in (\ref{qpm1b}),
using $\la =\al^2 $ with $\al=e^{\ri\theta/2}$ and $n =\la+\la^{-1}$ yields
$n+2=(\al+\al^{-1})^2$, and then from (\ref{ralform}) and the definition
of the dilated Chebyshev polynomials of the second kind it follows
that
$(n+2) r_j(n)\, \Uc_{j-1}(n)$ is equal to
$$
(\al+\al^{-1})^2\, \left(\frac{\al^{2j+1}+\al^{-(2j+1)}}{\al+\al^{-1}}\right)\,
\left(\frac{\al^{2j}-\al^{-2j}}{\al^2-\al^{-2}}\right)
=\left(\frac{\al^{4j+1}-\al^{-(4j+1)}}{\al-\al^{-1}}\right) - 1
$$
which is precisely $s_{2j}(n) -1$, by (\ref{alform}). The proof of the
second identity is similar.
\end{proof}
Another basic fact we shall use is that, with a suitable restriction on $n$, $s_k(n)$ is
monotone increasing with $k$.
\begin{proposition} \label{inc}
For each real $n\geq 2$, the sequence $(\,s_k(n)\,)$ is strictly increasing, and grows exponentially with
leading order asymptotics
$$
s_k(n)\sim \frac{1}{2}\left( 1+ \sqrt{\frac{n+2}{n-2}}\right) \,\left(\frac{n+\sqrt{n^2 - 4}}{2} \right)^k
\qquad
as \quad k\to\infty ,
$$
for all $n>2$.
\end{proposition}
\begin{proof} For real $n\geq 2$, from (\ref{ladef}) we can set
$$\tau=\ri\theta = \log\left(\frac{n+\sqrt{n^2 - 4}}{2} \right), $$
which defines a bijection
from the interval $n\in [2,\infty )$ to $\tau\in [0,\infty)$.
The inverse is
$$
n = 2\,\mathrm{cosh} \tau \implies \frac{\rd n}{\rd\tau}=2\,\mathrm{sinh} \tau >0,
$$
and we have
$$
\Tc_k(n) = 2\, \mathrm{cosh} (k\tau ) \implies \frac{\rd }{\rd\tau}\,\Tc_k(n) =2k\,\mathrm{sinh} (k\tau) >0
$$
for $\tau>0$; hence, for all fixed $k$, $\Tc_k(n)$ is a strictly increasing function of $n$ for $n\geq 2$. Similarly,
$\frac{\rd }{\rd k}\,\Tc_k(n) =2\tau\,\mathrm{sinh} (k\tau)$ so for all fixed $n>2$, the sequence $(\,\Tc_k(n)\,)_{k\geq 0}$ is also strictly increasing
with $k$.
Then since $\Tc_k(2) =2$ for all $k$, it follows that, for all $k$,
\beq\label{tb}
\Tc_k(n) \geq 2\qquad \forall n \geq 2,
\eeq
so from (\ref{diff})
we have $$s_k(n) - s_{k-1}(n)\geq 2.$$
Upon taking the leading term of the explicit expression in terms of $\la$ in (\ref{expl}) and rewriting it as a
function of $n$, the
asymptotic formula results.
\end{proof}
We can now use the explicit formulae above to derive various arithmetical properties of the integer sequences defined by
$s_k(n)$ for positive integers $n$.
This will culminate in Lemma \ref{gcd}
below, which describes coprimality conditions on the terms, as well as Lemma
\ref{pdiv} and its corollaries, which constrain where particular prime factors can appear.
To begin with we describe where primes can appear in the sequence.
\begin{lemma}\label{2kp1}
For all integers $n\geq 2$, if $s_k(n)$ is prime then $k=(p-1)/2$ for $p$ an odd prime.
\end{lemma}
\begin{proof}
If $2k+1=ab$ is composite, for some odd integers $a,b\geq 3$, then the identity (\ref{uid}) can be applied to the middle expression in
(\ref{mainf}), to write
$s_k(n)$ as the product
$$
\begin{array}{rcl}
s_k(n) & = &
\Uc_{a-1}\left(\Tc_b(\sqrt{n+2})\right) \,\Uc_{b-1}(\sqrt{n+2}) \\
& = &
s_{(a-1)/2}\left(\Tc_b (\sqrt{n+2})^2 -2\right) \, s_{(b-1)/2}(n).
\end{array}
$$
Then, since $\Tc_2(j)=j^2-2$, by using (\ref{tid}) we have
\beq\label{facid} \begin{array}{rcl}
s_k(n)
& = & s_{(a-1)/2}\left(\Tc_{2b} (\sqrt{n+2})\right) \, s_{(b-1)/2}(n) \\
& = & s_{(a-1)/2}\left(\Tc_{b} (n)\right) \, s_{(b-1)/2}(n) ,
\end{array}
\eeq
and each factor above is an integer greater than 1.
\end{proof}
Henceforth we will consider only integer values of $n$. It is well known that all linear recurrence sequences defined over $\Z$ are
eventually periodic $\bmod \,m$ for any modulus $m$ \cite{ward}; and for the recurrence (\ref{srec}) we can say further that it is strictly periodic $\bmod \,m$, because the linear map $(s_k,s_{k+1})\mapsto (s_{k+1},s_{k+2})$ defined by the matrix ${\bf A}$ in (\ref{matf}) is always invertible $\bmod \,m$ (since $\det{\bf A} =1$).
However, in order to obtain coprimality conditions, we need a lemma that explicitly describes the periodicity of the terms $s_k(n)\,\bmod s_{j}(n)$ for fixed $j$.
\begin{lemma}\label{resper}
For all integers $n\geq 2$ and any odd number $p\geq 3$,
the sequence of residues $s_k(n)\,\bmod s_{(p-1)/2}(n)$ is periodic with period $p$, and
$s_k(n) \equiv 0 \pmod{s_{(p-1)/2}(n)}$ if and only if $k \equiv (p-1)/2 \pmod{p}$.
\end{lemma}
\begin{proof} The identity (\ref{psh}) implies that, for all $k$,
$$s_{k+p}(n)\equiv s_k(n) \pmod{s_{(p-1)/2}(n)},$$
so the residues repeat with period $p$.
By the monotonicity result in Proposition \ref{inc},
$$
1=s_0(n) 0$, then
by writing $2j+1=(2m+1)(2j'+1)$, $2k+1=(2m+1)(2k'+1)$ with $\gcd(2j'+1, 2k'+1)=1$, and applying (\ref{facid}), we have
$$
\gcd\Big(s_j(n),s_k(n)\Big)=s_m(n) \, \gcd\Big(s_{j'}(\Tc_{2m+1}(n)),s_{k'}(\Tc_{2m+1}(n))\Big) = s_m(n),
$$
by applying Lemma \ref{star} once again.
\end{proof}
\begin{remark} The preceding result is a special case of a result on the greatest common divisor of a pair of Lehmer numbers: see Lemma 3 in \cite{stewart}.
\end{remark}
\begin{remark} Since the argument $n$ plays a passive role in most of the above, it is clear that, mutatis mutandis,
Lemmas \ref{resper}, \ref{star}
and \ref{gcd}
also apply to the sequence of polynomials $(\,s_k(n)\,)$
in $\Z[n]$. Analogous divisibility properties for the Chebyshev polynomials of the first kind are described in \cite{rtw}.
\end{remark}
The preceding results allow the periodicity of the sequence modulo any prime to be described quite precisely.
The notation $\legendre{\,\cdot\,}{\,\cdot\,}$ is used below to denote the Legendre
symbol.
\begin{lemma}\label{pdiv} Let $n\geq 2$ be fixed, and for any prime $q$ let $\pi (q)$ denote the period of the sequence $(s_k(n) \bmod \,q)$.
Then $\pi (2)=3$ if and only if $n$ is odd, in which case $s_k(n)$ is even$\iff k\equiv 1\pmod{3}$,
while $\pi(2)=1$ and all $s_k(n)$ are odd when $n$ is even. Moreover, for
$q$ an odd prime, one of three possibilities can occur: (i) $\legendre{n^2-4}{q}=\pm1$ and $\pi(q)|q\mp 1$;
(ii) $n\equiv 2\pmod{q}$ and $\pi(q)=q$ with $s_k(n)\equiv 0\pmod{q}\iff q|2k+1$;
(iii) $n\equiv -2\pmod{q}$ and $\pi(q)=2$ with $s_k(n)\equiv (-1)^k\pmod{q}$.
\end{lemma}
\begin{proof} When $n$ is even, then since $s_0(n)=1$ and $s_1(n)=n+1$ are both odd, it follows from (\ref{srec})
that $s_k(n)$ is odd for all $k$, so $\pi(2)=1$. For $n$ odd, $s_1(n)$ is even, so by
Lemma \ref{star},
$s_k(n)$ is even if and only if $k\equiv 1\pmod{3}$, and $\pi(2)=3$.
Now let $q$ be an odd prime. For case (i) it is most convenient to consider the behaviour of $(s_k \bmod \, q)$ in terms of the equivalent sequence
defined by the recurrence (\ref{srec}) in the finite field $\F_q$. In that case we have $n>2$,
and when $\legendre{n^2-4}{q}=1$ it follows
that $n^2-4$ is a quadratic residue mod $q$, so the
first formula in (\ref{expl}), which can be rewritten as
\beq\label{forml}
s_k(n)= \frac{\la^{-k}(\la^{2k+1}-1)}{\la -1},
\eeq
remains valid in terms of $\la\in\F_q$, with $\la\neq \pm 1$, and $\la^{q-1}=1$ in $\F_q$ by Fermat's little theorem. The terms
of the sequence repeat with period $\pi(q)=\mathrm{ord}(\la)>2$, the multiplicative order of $\la$ in the group
$\F_q^*$, and this divides $q-1$ by Lagrange's theorem. The case $\legendre{n^2-4}{q}=-1$ is similar, but now
$n^2-4$ is a quadratic nonresidue mod $q$, so $\la$ is not defined in $\F_q$ and the formula (\ref{forml})
should be interpreted in the field extension $ \F_q[\sqrt{n^2-4}] \simeq\F_{q^2}$. The Frobenius automorphism
$\la\to\la^q$ exchanges the roots of the quadratic $X^2-n\,X+1=(X-\la)(X-\la^{-1})$, hence $\la^q=\la^{-1}$.
Thus $\la^{q+1}=1$, and now the sequence given by (\ref{forml})
repeats with period $\pi(q)=\mathrm{ord}(\la)$, the order of $\la$ in
$\F_{q^2}^*$, which divides $q+1$. In case (ii), the sequence $s_k(n)\bmod \,q$ is the same as the
sequence (\ref{lin}) mod $q$, which first vanishes when $k=(q-1)/2$ and repeats with period $q$, and in case
(iii) the sequence is equivalent to (\ref{nm2}), which is never zero mod $q$.
\end{proof}
At this stage it is convenient to introduce the notion of a primitive prime divisor (sometimes just referred to as a primitive divisor), which is a
prime factor $q$ that divides $s_k(n)$ but does not divide any of the previous terms in the sequence
\cite{estw}, and by convention does not divide the discriminant $n^2-4$ either \cite{bilu, schinzel}.
\begin{definition} Let the product of the discriminant and the first $k$ terms be denoted by
\beq\label{pik}
\Pi_{k}(n)=(n^2-4)\, s_1(n)s_2(n) \cdots s_{k}(n).
\eeq
A {\it primitive prime divisor} of $s_k(n)$ is a prime $q|s_k(n)$ such that
$q\not | \,\Pi_{k-1}(n)$.
\end{definition}
Case (i) of Lemma \ref{pdiv} is the most interesting one.
In that case it is clear from (\ref{forml}) that a prime $q|s_k(n)$ for some $k$ whenever
$\la^{2k+1}=1$ in $\F_{q^2}\supset\F_q$, and then $\pi(q)=\mathrm{ord}(\la)=2k^*+1$ must be odd, where
$k^*=(\pi(q)-1)/2>0$ is the smallest
$k$ for which this happens; and if $\pi(q)$ is even then this cannot happen.
If we include $q=2$, then
we can rephrase the latter by saying that the prime factors $q$ appearing in the sequence
$(\,s_k(n)\,)$ are precisely those $q$ which have an odd period $\pi(q)>1$, and this
consequence of Lemma \ref{pdiv} can be restated in terms of primitive prime divisors.
\begin{corollary} \label{papp}
A prime $q$ is a primitive divisor of $s_k(n)$ if and only if $k=(\pi(q)-1)/2$
where $\pi(q)$ is odd.
Moreover, if $q$ is odd and $\legendre{n^2-4}{q}=\pm1$ then
$q=2a\pi(q)\pm 1$ for some positive integer $a$.
\end{corollary}
The latter statement just says that an odd primitive divisor of $s_k(n)$ has the form
$q=2a(2k+1)\pm 1$ for some $a\geq 1$, so the minus sign with $a=1$ gives the lower bound $q\geq 4k+1$. Hence
the primes that do not appear as factors in the sequence can also be characterized.
\begin{corollary} \label{rp}
If a prime $q<4k+1$ is not a factor of $\Pi_{k-1}(n)$,
then it never
appears as a factor of $s_j(n)$ for $j\geq k$, and $\pi(q)$ is even.
\end{corollary}
So far we have concentrated on properties of $s_k(n)$ for fixed $n$ and allowed $k$ to vary. However,
if one is interested in finding factors of $s_k(n)$ for large $n$, then it may also be worthwhile
to consider other values of $n$, as the following result shows.
\begin{proposition}
Suppose that an integer $f|s_k(n)$ for some $k,n$. Then $f|s_k(m)$
whenever $m\equiv n \pmod{f}$.
\end{proposition}
\begin{proof}
If $m\equiv n \pmod{f}$ then $m^j\equiv n^j \pmod{f}$ for any exponent $j\geq 0$, and
since $s_k(m)$ is a polynomial in $m$ with integer coefficients, it follows that
$s_k(m)\equiv s_k( n)\equiv 0 \pmod{f}$, as required.
\end{proof}
\section{Generic factorization for Chebyshev values}
\label{chebv}
The sequence $(\,s_k(n)\,)$ has special properties when $n$ is given by a dilated Chebyshev polynomial of the
first kind evaluated at an integer value of the argument.
\begin{theorem} \label{factors}
For all integers $p\geq 2$, when $n = \Tc_p(j)$ for some integer $j$
the terms of the sequence $(\,s_k(n)\,)$
can be factorized as a product of
rational numbers, that is
\beq\label{fac}
s_k (\Tc_p(j)) = R_k(j) \, s_k(j) ,
\eeq
where the prefactors $R_k(j)\in\Q$
are given by
\beq\label{rform}
R_k(j) = \frac{\Uc_{p-1}(\Tc_{2k+1}(\sqrt{j+2}))}{\Uc_{p-1}(\sqrt{j+2})}
\eeq
and satisfy a linear recurrence of order $p$.
In particular, for $p=2$ the prefactor is
$R_k(j)=r_k(j)\in \Z$,
as given in Theorem \ref{t2},
while for all odd $p$ the prefactor can be written as
\beq \label{poddf}
R_k(j) = \frac{s_{(p-1)/2}(\Tc_{2k+1}(j))}{s_{(p-1)/2}(j)} \in\Q,
\eeq
and satisfies the recurrence
\beq\label{podd}
(\rS-1)\prod_{i=1}^{(p-1)/2}(\rS^2 -\Tc_{2i}(j)\,\rS +1) \, R_k(j) =0.
\eeq
\end{theorem}
\begin{proof} Upon introducing $\phi$ such that $\ell = \sqrt{j+2}=2\,\cos(\phi /2)$,
the formula (\ref{mainf}) gives
$$
R_k(j) = \frac{s_k (\Tc_p(j)) }{s_k(j)} = \frac{\sin\Big((2k+1)p\phi/2\Big)\sin(\phi/2)}
{\sin(p\phi/2)\sin\Big((2k+1)\phi/2\Big)},
$$
and the definition of the dilated Chebyshev polynomials
of the second kind in (\ref{1st}) produces (\ref{rform}).
For integer $j$, $R_k(j)$ is a ratio of integers, so it is a rational number
(positive for $j\geq 2$).
In the case $p=2$, $R_k(j)=r_k(j)$, which can be written in the form
\beq\label{p2f} r_k(j) = \frac{\Tc_{2k+1}(\sqrt{j+2})}{\sqrt{j+2}}, \eeq
which is an integer, as follows from the fact that $\Uc_1(\ell) =\ell$,
and this ratio is an even polynomial of degree $2k$ in $\ell$ with integer coefficients, hence it is a polynomial of degree $k$ in $j$; and by (\ref{tb}) it is positive for real $j\geq 2$, and takes positive integer values for integers $j$ in this range.
In the case that $p$ is odd, the expression (\ref{poddf}) is found by applying the formula (\ref{mainf}) to the numerator and denominator of (\ref{rform}).
To see that $R_k(j)$ satisfies a linear recurrence of order $p$, note that, upon setting $\mu = \exp(\ri\phi )$ and applying the first formula in (\ref{expl}) with $\la=\mu^p$, the factorization (\ref{fac}) can be seen as a consequence of the elementary algebraic identity
\beq\label{algi}
\frac{\mu^{p(k+1)}-\mu^{-pk}}{\mu^p -1} =
\left(
\frac{ \sum_{j=0}^{p-1}\mu^{(k+1)(p-1)-(2k+1)j}}{\mu^{p-1}+\mu^{p-2}+\cdots + 1}
\right)
\, \left(\frac{\mu^{k+1}-\mu^{-k}}{\mu -1}\right) ,
\eeq
where the first factor on the right-hand side above is just $R_k(j)$.
Thus the denominator of the expression for $R_k(j)$ in (\ref{algi}) is $\sum_{j=0}^{p-1}\mu^j$,
which is independent of $k$, while the numerator is
a linear combination of $k$th powers of the characteristic
roots $\mu^{(p-1)}, \mu^{(p-3)}, \ldots, \mu^{-(p-3)}, \mu^{-(p-1)}$, giving a total of $p$ distinct roots. When $p$ is
even, the roots come in $p/2$ pairs, namely $\mu^{\pm(2i-1)}$ for $i=1,\ldots,p/2$, which gives the characteristic
polynomial
$$
F(\la) = \prod_{i=1}^{p/2}(\la^2-\Tc_{2i-1}(j)\la+1),
$$
so that, in particular, for $p=2$ the recurrence satisfied by $R_k(j)$ is (\ref{p2}),
while for $p$ odd there are the pairs $\mu^{\pm 2i}$ for $i=1,\ldots,(p-1)/2$ together with the root 1,
which yields (\ref{podd}).
\end{proof}
\begin{theorem}\label{maint}
Let $(a_n)_{n\geq 1}$ be the sequence specified by Definition \ref{adef}. If
$n = \Tc_2(j)$ for some $j\geq 3$, then $a_n = -1$. Furthermore, if
$n = \Tc_p(j)$ for some $j\geq 3$ with $p$ an odd prime, then either
$s_{(p-1)/2}(n)$ is not prime and $a_n = -1$, or $s_{(p-1)/2}(n)$ is
the only prime in the sequence $(\,s_k(n)\,)_{k\geq 0}$ and
$a_n = (p-1)/2$.
\end{theorem}
\begin{proof} First of all, consider the factorization (\ref{fac}) when $p=2$, with prefactor $r_k(j)$ as in Theorem \ref{t2}, given by (\ref{p2f}).
When $j=2$ this is not interesting, because it gives $r_k(j)=1$ for all $j$. However, note the property (mentioned in passing in the proof of Proposition \ref{inc}),
that for real $n>2$, the sequence $(\,\Tc_k(n)\,)_{k\geq 0}$ is strictly
increasing with the index. Hence, for all $k>0$, $\Tc_{2k+1}(\sqrt{j+2})>\sqrt{j+2} = \Tc_{1}(\sqrt{j+2})$. Thus for all
$j\geq 3$ and $k\geq 1$,
both factors $r_k(j)$, $s_k(j)$ are greater than 1, so $s_k(T_2(j))$ can never be prime, and $a_n=-1$.
Now for any odd prime $p$, note that, a priori, the prefactor $R_k(j)$ in (\ref{fac}) is a positive rational number,
and the formula (\ref{poddf}) gives
\beq\label{sikid}
s_k (\Tc_p(j)) = \frac{ s_k(j) \, s_{(p-1)/2}(\Tc_{2k+1}(j))}{s_{(p-1)/2}(j)}.
\eeq
However, according to Lemma \ref{resper}, $s_{(p-1)/2}(j)|s_k(j)$ whenever $k\equiv (p-1)/2 \pmod{p}$.
On the other hand, for all values of $k\not\equiv (p-1)/2 \pmod{p}$,
Lemma \ref{star} gives
$\gcd(s_k(j) , s_{(p-1)/2}(j))=1$.
Therefore $s_{(p-1)/2}(j)$ divides $s_{(p-1)/2}(\Tc_{2k+1}(j))$ and $R_k(j)\in\Z$.
Thus, for all $k$, the terms $s_k (\Tc_p(j))$ can be written as a product of two integers, that is
\beq\label{cases}
s_k (\Tc_p(j)) = \begin{cases}
\hat{R}_k(j)\, s_{(p-1)/2}(\Tc_{2k+1}(j)), &\text{if} \,\,k\equiv (p-1)/2 \pmod{p}; \\
R_k(j)\,s_k(j), & \text{otherwise},
\end{cases}
\eeq
where $R_k(j)$ is given by (\ref{poddf}) as above, and
\beq\label{rt}
\hat{R}_k(j) = \frac{ s_k(j) }{s_{(p-1)/2}(j)} = s_i(\Tc_p(j)) \quad \mathrm{for}\,\, k=(p-1)/2+ip,
\eeq
with the latter formula being obtained from (\ref{facid}).
In the first case of (\ref{cases}) above, for $k=(p-1)/2$ the prefactor is $\hat{R}_{(p-1)/2}(j)=1$, while
$\hat{R}_{k}(j)>1$ for all $k=(p-1)/2+i p$, $i\geq 1$,
by Lemma \ref{inc}, and the other factor is $s_{(p-1)/2}(\Tc_{2k+1}(j))>1$ for all these values of $k$.
In the second case, for $k>0$,
we can use (\ref{texp}) together with (\ref{tid}) to write
$$
\begin{array}{rcl}
s_{(p-1)/2}(\Tc_{2k+1}(j)) & = & \frac{1}{2}\Tc_0 + \sum_{i=1}^{(p-1)/2}\Tc_i(\Tc_{2k+1}(j)) \\
& = & \frac{1}{2}\Tc_0 + \sum_{i=1}^{(p-1)/2}\Tc_{(2k+1)i}(j) \\
& > & \frac{1}{2}\Tc_0 + \sum_{i=1}^{(p-1)/2}\Tc_{i}(j) = s_{(p-1)/2}(j) ,
\end{array}
$$
so $s_{(p-1)/2}(\Tc_{2k+1}(j)) / s_{(p-1)/2}(j)>1$.
Hence both factors $R_k(j)$, $s_k(j)$ are greater than 1 in the second case of (\ref{cases}). Thus
the only term that can be prime is $s_{(p-1)/2} (\Tc_p(j))$,
and the result is proved.
\end{proof}
\begin{remark}
For any odd $p=2i+1$, the identity (\ref{sikid}) can
be rewritten in the symmetric form
\beq\label{symid}
s_{i}(j)\,
s_k (\Tc_{2i+1}(j)) = s_k(j) \, s_{i}(\Tc_{2k+1}(j)) .
\eeq
\end{remark}
\begin{remark} \label{polj}
Similarly to the remark after Lemma \ref{gcd}, the formula (\ref{cases}) also corresponds to factorizations of
the corresponding polynomials in $\Z[j]$, according to whether $k\equiv (p-1)/2 \pmod{p}$ or not.
\end{remark}
It is clear from the factorizations (\ref{cases}) that in the second case, $s_k(\Tc_p(j)) \equiv 0 \pmod{s_k(j)}$ whenever $k$ is not
congruent to $(p-1)/2 \,\bmod \,p$. It turns out that an explicit expression for
$s_k(\Tc_p(j)) \,\bmod \,s_k(j)$ can be given in the first case as well. Before doing so, it is convenient to define
some more polynomials, which are shifted versions of the airfoil polynomials.
\begin{definition} \label{ppol}
Polynomials $\Pc_k(z)$ are defined as elements of $\Z[z]$ by
$$
\Pc_k(z) = s_k(2-z),
$$
or equivalently by
\beq\label{sinf}
\Pc_k(4\sin^2\theta) = \frac{\sin\Big((2k+1)\theta\Big)}{\sin\theta}.
\eeq
They satisfy the linear recurrence
\beq\label{prec}
\Pc_{k+1}(z) +(z-2) \Pc_k(z) +\Pc_{k-1}(z) =0,
\eeq
and
for $k\geq 0$ their expansion in powers of $z$ takes the form
\beq\label{expan}
\Pc_k(z) = 2k+1 - c_k^{(1)}\, z +c_k^{(2)}\, z^2+\cdots+ (2k+1)(-z)^{k-1} + (-z)^k.
\eeq
with
$$
c_k^{(1)}=\frac{k(k+1)(2k+1)}{3!} ,\qquad c_k^{(2)}= \frac{k(k-1)(k+1)(2k^2+5k+2)}{5!}.
$$
\end{definition}
\begin{theorem} \label{rema}
For all odd integers $p$,
\beq\label{pform}
s_k(\Tc_p(j)) = s_i(\Tc_p(j))\,\Pc_{(p-1)/2}\Big((2-j)\, s_k(j)^2\Big) \quad \mathrm{for} \quad k=(p-1)/2+ip,
\eeq
and in particular,
$$
s_k(\Tc_p(j)) \equiv p\, s_i(\Tc_p(j)) \pmod{(j-2)\,s_k(j)^2}
$$
holds in that case.
\end{theorem}
\begin{proof}Upon setting $p=2q+1$,
by using (\ref{rt}) together with the first formula in (\ref{cases}),
we
have
$$
\begin{array}{rcl}
{s_k(\Tc_p(j))}/{s_i(\Tc_p(j))} & = &
{\sin\Big((2q+1)(2k+1)\phi/2\Big)}\bigg/{\sin\Big((2k+1)\phi/2\Big)} \\
& = & \Pc_q(z), \qquad
\mathrm{where} \qquad z= 4\sin^2\Big((2k+1)\phi/2\Big),
\end{array}
$$
and (with the same notation as in the proof of Theorem \ref{factors}) we also have
$j=2\,\cos\phi$. Comparing the expressions for $z$ and $j$ gives
$
z =4\sin^2(\phi/2) \, s_k(j)^2 = (2-j)\, s_k(j)^2$,
which yields the identity (\ref{pform}) in terms of the shifted airfoil polynomial $\Pc_{(p-1)/2}$. The
terms displayed in the
expansion (\ref{expan}) are easily obtained from the recurrence (\ref{prec}),
or by substituting $n=2-z$ in (\ref{nexp}),
and the leading term
gives the reduction of (\ref{pform}) $\bmod \, s_k(j)^2$.
\end{proof}
We now turn to the sequences $(\,r_k(n)\,)$ for $n>0$, which are associated with negative values of $n$ via (\ref{rsrel}).
It turns out that these sequences also admit factorizations for certain Chebyshev values of $n$. The case of
$n=\Tc_p(j)$ for odd index $p$ can be inferred immediately from Theorem \ref{factors} together with
(\ref{rsrel}), since $\Tc_p$ is an odd function
of its argument in that case. However, the even case $n=\Tc_2(j)$ does not translate directly to
the sequences $(\,r_k(n)\,)$, and
requires a separate treatment. That there should be a significant difference for values of even Chebyshev polynomials
is also
apparent from comparison of the fact that
$a_7=a_{14}=a_{23}=a_{34}=-1$ in
(\ref{anseq}),
but $\tilde{a}_1=\tilde{a}_2=\tilde{a}_{34}=-1$ in (\ref{atnseq}), while
$\tilde{a}_{7}$, $\tilde{a}_{14}$ and $\tilde{a}_{23}$ are all positive.
The following analogue of Theorem \ref{t2} for the sequences $(\,r_k(n)\,)$ only provides
a factorization of the terms for a particular subset of the values $n=\Tc_2(j)$ .
\begin{theorem}\label{rt2} When $n=\Tc_2(j)=j^2-2$ with $j=2(\ell^2-1)$ for integer $\ell\geq 2$,
the terms of the sequence $(\,r_k(n)\,)$ admit the factorization
\beq\label{rt2fac}
r_k(j^2-2) = f_k^+(j)\, f_k^-(j),
\eeq
where
\beq\label{fform}
f_k^{\pm}(j) = \frac{\ell r_k(j) \pm \delta_k}{\ell\pm 1}\in\Z,
\qquad \delta_k = (-1)^{\left \lfloor{\frac{k+1}{2}}\right \rfloor}.
\eeq
\end{theorem}
\begin{proof} Since $\delta_k^2=1$ and
$r_k(j) =(j+2)^{-1/2} \left(\al^{(2k+1)/2}+ \al^{-(2k+1)/2}\right)$, using
$\al^{1/2}+\al^{-1/2}=\sqrt{j+2}$ as in (\ref{dos}),
it follows that
$$
\begin{array}{rcl}
f_k^+(j)\, f_k^-(j) &= &
{(\ell^2 r_k(j)^2 -1)}/{(\ell^2-1)} = j^{-1}\Big((j+2)r_k(j)^2-2\Big) \\
& = &
{\Big( \left(\al^{(2k+1)/2}+ \al^{-(2k+1)/2}\right)^2-2\Big)}/{(\al+\al^{-1})},
\end{array}
$$
which coincides with the formula (\ref{ralform}) for $r_k(n)$ with $n=j^2-2=4\ell^4-8\ell^2+2$ in this case.
To see that each factor $f_k^\pm (j)$ is an integer for all $k$, note that
$(\rS^2-j\rS+1)r_k(j)=0$, and checking the sequence of signs $+1,-1,-1,+1$ for
$k=0,1,2,3$ shows that $(\rS^2-j\rS+1)\delta_k =j \delta_{k-1}$. Hence
the factors in (\ref{t2fac}) each satisfy an inhomogeneous linear
recurrence of second order, that is
\beq\label{finh}
f_{k+2}^\pm(j) - j \, f_{k+1}^\pm(j) +f_k^\pm(j) = 2(\pm\ell - 1)\delta_{k-1}.
\eeq
From (\ref{fform}) it can be seen that
$$
f_{-1}^{\pm}(j) = f_{0}^{\pm}(j) =1
$$
provide integer initial values for (\ref{finh}) in each case,
so these two sequences consist entirely of integers.
\end{proof}
\begin{example}
For $\ell=2$, the above result gives $j=\Tc_2(2\sqrt{2})=6$ and $n=\Tc_2(6)=34$ ,
with the sequence $(\,r_k(34)\,)=(\,f_k^+(6)\,f_k^-(6)\,)$ beginning
\beq\label{r34}
1,33,1121,38081,1293633,\ldots,
\eeq
where the factors are
\beq\label{fpm6}
(\,f_k^+(6)\,): \,\, 1,3,19,113,657,\ldots, \quad
(\,f_k^-(6)\,): \,\, 1,11,59,337,1969,\ldots,
\eeq
and these satisfy the inhomogeneous recurrences
$$ \begin{array}{rcl}
f_{k+2}^+(6) - 6 \, f_{k+1}^+(6) +f_k^+(6)
& = & 2(-1)^{\left \lfloor{\frac{k}{2}}\right \rfloor} , \\
f_{k+2}^-(6) - 6 \, f_{k+1}^-(6) +f_k^-(6)
& = & 6(-1)^{\left \lfloor{\frac{k}{2}}\right \rfloor+1}
.
\end{array}
$$
\end{example}
For the case where $n=\Tc_p(j)$ for $p$ odd, the formula (\ref{poddf}) can be applied, together with
(\ref{rsrel}), to yield the factorization
\beq\label{rfactor}
r_k (\Tc_p(j)) = \tilde{R}_k(j) \, r_k(j),
\eeq
where
\beq\label{Rform}
\tilde{R}_k(j) =R_{k}(-j) = \frac{r_{(p-1)/2}(\Tc_{2k+1}(j))}{r_{(p-1)/2}(j)}
\eeq
satisfies the same recurrence (\ref{podd}) as $R_k(j)$.
\begin{example}
When $n=\Tc_3(3)=18$, the sequence $(\,r_k(18)\,)$ begins
\beq\label{r18}
1,17,305,5473,98209,1762289,31622993,567451585,\ldots,
\eeq
so that $\tilde{a}_{18}=1$ since $r_1(18)=17$ is prime.
By adapting Theorem \ref{factors},
the terms can be factored as
$$
r_k(18)=\tilde{R}_k(3)\,r_k(3),
$$
where the sequence $(\,r_k(3)\,)$ begins with
\beq\label{r3}
1,2,5,13,34,89,233,610,\ldots,
\eeq
and $(\,\tilde{R}_k(3)\,)$ is a sequence
of rational numbers, starting with
$$
1,\frac{17}{2},61,421, \frac{5777}{2}, 19801, 135721, \frac{1860497}{2}, \ldots,
$$
which satisfies the third order recurrence
$$
\tilde{R}_{k+3}(3)- 8 \, \Big(\tilde{R}_{k+2}(3) - \tilde{R}_{k+1}(3)\Big) - \tilde{R}_{k}(3)=0.
$$
The prefactors $\tilde{R}_k(3)$ are integers whenever $k\equiv 0$ or $2\pmod{3}$,
so $r_k(3)$ divides $r_k(18)$ for all such $k$, while
the terms of the trisection $(\, r_{3k+1}(18)\,)$ are all divisible by $r_1(18)$, which is the only
prime in the sequence $(\,r_k(18)\,)_{k\geq 0}$.
\end{example}
Having described the situation for even Chebyshev values, and given
the above example of an odd Chebyshev value, the analogue of
Theorem \ref{maint} for $(\,r_k(n)\,)$ can now be stated.
\begin{theorem}\label{tmaint}
Let $(\tilde{a}_n)_{n\geq 1}$ be the sequence specified by Definition \ref{atdef}. If
$n = \Tc_2(j)$ where $j=2(\ell^2-1)$ for some $\ell\geq 2$, then $\tilde{a}_n = -1$. Furthermore, if
$n = \Tc_p(j)$ for some $j\geq 3$ with $p$ an odd prime, then either
$r_{(p-1)/2}(n)$ is not prime and $\tilde{a}_n = -1$, or $r_{(p-1)/2}(n)$ is
the only prime in the sequence $(\,r_k(n)\,)_{k\geq 0}$ and
$\tilde{a}_n = (p-1)/2$.
\end{theorem}
\begin{proof}
For the case $n = \Tc_2(j)$, $j=2(\ell^2-1)$ with $\ell\geq 2$, note that each factor in
(\ref{rt2fac}) is an integer, and $r_0(\Tc_2(j))=f_0^\pm (j)=1$. Now as noted in the proof of Lemma \ref{inc},
for fixed argument the sequence of Chebyshev polynomials of the first kind is strictly increasing
with the index $k\geq 0$, so from (\ref{p2f}) it follows that $(\,r_k(j)\,)_{k\geq 0}$ is strictly increasing.
Thus, from their explicit expressions in (\ref{fform}), both sequences $(\,f_k^\pm (j)\,)$ are
strictly increasing as well. Hence, for these values of $j$, $r_k(\Tc_2(j))$ is composite for $k\geq 1$.
Hence there are no primes in the sequence $(\,r_k(n)\,)_{k\geq 0}$ for any of these even Chebyshev values of $n$.
When $n = \Tc_p(j)$, $j\geq 2$ with $p$ an odd prime, there is the factorization (\ref{rfactor}), with
$\tilde{R}_k(j) \in\Q$ given by (\ref{Rform}). Just as for the sequences $(\,s_k(n)\,)$, it is necessary
to consider whether $k\equiv (p-1)/2 \pmod{p}$ or not. One can show
that the analogue of Lemma \ref{resper} holds for the sequences $(\,r_k(n)\,)$, $n\geq 3$,
so that when $k\equiv (p-1)/2 \pmod{p}$ the denominator of $\tilde{R}_k(j)$ divides $s_k(j)$; or,
by using (\ref{rsrel}) together with (\ref{facid}), one can write the explicit factorization
\beq\label{rifac}
r_k(\Tc_p(j))= r_i(\Tc_p(j))\, r_{(p-1)/2}(\Tc_{2k+1}(j)) \qquad \mathrm{for} \quad k=(p-1)/2+ip ,
\eeq
where both integer factors above are greater than 1 for $i>0$.
On the other hand, for the case $k\not\equiv (p-1)/2 \pmod{p}$, one can show that
the analogue of Lemma \ref{star} also holds for the sequences $(\,r_k(n)\,)$, so
in that case the denominator in (\ref{Rform}) must divide the numerator; hence
$\tilde{R}_k(j) \in\Z$ and both factors in (\ref{rfactor}) are integers. Then, similarly to
the proof of Lemma \ref{inc}, setting $\sqrt{j+2}=2\,\mathrm{cosh}\tau$ for real $j\geq 2$ in (\ref{p2f}) gives
$$r_k(j)= \frac{\mathrm{cosh}((2k+1)\tau)}{\mathrm{cosh}\tau}
\implies \frac{\rd}{\rd\tau}\, r_k(j)=\frac{2k\,\mathrm{sinh}((2k+1)\tau)}{\mathrm{cosh}\tau} +
\frac{\mathrm{sinh}(2k\tau)}{\mathrm{cosh}^2\tau}>0
$$
for $k>0$, $\tau>0$, implying that $r_k(j)$ is a strictly increasing function of $j$ for $j\geq 2$.
Hence $\tilde{R}_k(j)>1$ for $j>2$, so when $01$ be a positive integer. The sequence $(\,s_k(n)\,)_{k\geq 0}$ contains infinitely many primes if and only if $n\neq \Tc_p(j)$ for some prime $p$ and some integer $j\geq 3$.
\end{conjecture}
In order to consider the distribution of primes in the sequence $(\,s_k(n)\,)$, it is helpful to introduce some notation.
Define a subsequence $(k_N)_{N\geq 1}$ of the positive integers by requiring that
$$
s_{k_N}(n) = N\mathrm{th}\,\,\mathrm{prime} \,\,\mathrm{term} \,\, \mathrm{in} \,\,(\,s_k(n)\,)_{k\geq 0},
$$
and, for fixed $n$, let
$$
{\cal S}_k(n) =
\{ \,\mathrm{prime}\, q \mid q<4k+1\,
\} \cup
\{ \,\mathrm{prime}\, q \mid q| \Pi_{k-1}(n)\,
\}
, $$
where $\Pi_{k-1}(n)$ is the product in (\ref{pik}).
\begin{conjecture}
If $n\geq 3$ and
$n\neq \Tc_p(j)$ for some prime $p$ and some integer $j\geq 3$,
then, as $N\to\infty$,
\beq\label{conjf}
\log\log s_{k_N}(n) \sim C\, N, \qquad
with \quad C= e^{-\gamma} \, \log \sqrt{\la} ,
\eeq
where $\la = \frac{n+\sqrt{n^2-4}}{2}$, and $\gamma$ is
the Euler--Mascheroni constant.
\end{conjecture}
The above assertion is analogous to a conjecture of Wagstaff regarding Mersenne primes \cite{wagstaff}. If $M_N$ is the $N$th prime of the form $2^k-1$, then the heuristic arguments of Wagstaff suggest that
$$
\log\log M_N \sim C' \, N, \qquad \mathrm{with} \quad C'=e^{-\gamma}\log 2.
$$
A very clear exposition of the statistical properties of Mersenne primes, with many plots,
can be found on Caldwell's website \cite{caldwell}.\footnote{More specifically, see the page
\url{https://primes.utm.edu/notes/faq/NextMersenne.html}}
When $n$ is a non-Chebyshev value, a heuristic derivation of the corresponding asymptotics of primes
in the sequences $(\,s_k(n)\,)$ can be obtained in a similar way, as we now describe.
\begin{figure} \centering
\epsfig{file=n_4_with_23_primes.eps,width=7cm}
\caption{\small{Plot of $\log\log s_{k_N}(n)$ against $N$ for the first 23 primes in the sequence for $n=4$.}}
\label{nis4}
\end{figure}
By Lemma \ref{2kp1}, if $s_k(n)$ is prime then $2k+1$ is prime, and by Lemma \ref{star},
$s_k(n)$ is then coprime to $s_j(n)$ for all $1\leq j\leq k-1$, and for large enough
$k$ it is also coprime to the discriminant $n^2-4$, hence it is coprime to $\Pi_{k-1}(n)$.
On the other hand, by Corollary \ref{rp}, if $p=2k+1$ is prime then $s_k(n)$ is coprime to
all primes $q<4k+1$: such primes are either primitive divisors of lower terms $s_j(n)$ with
$j2$ be a positive integer. The sequence $(\,r_k(n)\,)_{k\geq 0}$ contains infinitely many primes if and only if
$n\neq \Tc_p(j)$ for some prime $p$, where the integer $j\geq 3$ takes one of values specified
in Theorem \ref{tmaint}.
\end{conjecture}
The first few prime terms in the sequence $(\,r_k(3)\,)_{k\geq 0}$ are plotted in Figure \ref{nism3}; for more details see the first appendix.
\section{Conclusions}
It seems highly likely that Theorem \ref{maint} identifies all those values of $n\geq 3$ such that
the sequence $(\,s_k(n)\,)_{k\geq 0}$ contains at most one prime, and Theorem \ref{tmaint} does
the same for $(\,r_k(n)\,)_{k\geq 0}$.
The sequences corresponding to all other values of $n$
should have infinitely many prime terms, but proving this
should be at least as difficult as proving that there are infinitely many Mersenne primes.
For Lehmer numbers, the most sophisticated results currently available concern primitive divisors
\cite{bilu, stewart2013}.
The statistics of prime appearances for non-Chebyshev values of $n$ suggests a close analogy with
Mersenne primes.
For Mersenne primes, the Lucas-Lehmer test is extremely efficient \cite{bruce}. The ideas
from \cite{rod, rotk} can be adapted to yield a necessary condition for primality of $q=s_k(n)$,
which can be tested efficiently,
but to provide sufficient conditions requires the use of a Lucas test or one of its generalizations \cite{bls, pom},
for which the formulae
(\ref{qpm1a}) and (\ref{qpm1b}) are useful, since
they provide partial factorizations of $q\pm 1$.
In future we would like to consider some of the large primes that appear in these sequences,
extending the approach that was applied to the case $n=6$ in \cite{nsw}.
\section {Acknowledgements}
ANWH is supported by EPSRC fellowship EP/M004333/1.
Some results in sections 3,4 and 5 of this paper were also obtained independently
by Bradley Klee, who provided useful suggestions for an early draft, and has developed
a graphical calculator application to verify the factorizations in Theorem \ref{factors} for particular values of $p$ \cite{kleeapp}.
We are grateful to David Harvey, Robert Israel,
Don Reble, John Roberts, Igor Shparlinski and Neil Sloane for helpful comments.
We are also extremely indebted to Hans Havermann, whose
extensive numerical computations originally
inspired many of the results described here.
\section*{Appendix A: Sequences of prime appearances}
In order to study the appearance of prime terms when
$n$ is a non-Chebyshev value, for some particular small values of $n$ we calculated the possible
prime terms $q=s_{(p-1)/2}(n)$ when $p=3,5,7,11,\ldots$ is an odd prime, and then tested them for primality using the Maple {\tt isprime} command.
This uses a probabilistic test, which excludes certain composite values of $q$,
while remaining $q$ are only pseudoprimes. For all but the largest values of the index $k=(p-1)/2$, we also checked the computations
with Mathematica's {\tt PrimeQ}$[q]$ command, as well as performing a Lucas-Lehmer style test for
pseudoprimes of our own, and verified that the answer was the same,
For $n=3$, the list of the first 43 values $k$ for which $s_k(3)$ appear to be prime is OEIS
sequence \seqnum{A117522}, beginning
$$
\begin{array}{l}
2, 3, 5,6,8,9,15,18,20,23,26,30,35,39,56,156,176,251,306,308,431,548, \\
680,
2393, 2396, 2925, 3870, 4233, 5345, 6125, 6981, 7224, 9734,
17724, 18389, \\ 22253, 25584, 28001, 40835, 44924, 47411, 70028, 74045.
\end{array}
$$
The (probable) primes $s_k(3)$ corresponding to these values of $k$ are listed in sequence \seqnum{A285992}.
The log log plot of these terms is given in Figure \ref{nis3}. The slope of the best fit line for these points is
$$
C= 0.2553739565.
$$
For $n=4$, the list of the first 23 values $k$ for which $s_k(4)$ appear to be prime is
$$
\begin{array}{l}
1,2,3,6,9,14,18,146,216,293,704,1143,1530,1593,2924,7163,9176,9489, \\ 11531,39543, 50423,60720,62868,
\end{array}
$$
which are listed in OEIS sequence \seqnum{A299100}, while the corresponding values $s_k(4)$ are given in \seqnum{A299107}.
The log log plot of these terms is given in Figure \ref{nis4}. The best fit line for this set of points has slope
$$
C= 0.5196737962.
$$
For $n=5$, the list of the first 24 values $k$ for which $s_k(5)$ appear to be prime is
$$
\begin{array}{l}
2,3,5,6,8,9,15,18,23,53,114,194,564,575,585,2594,3143,4578,4970, \\9261,11508,13298, 30018,54993,
\end{array}
$$
as listed in OEIS sequence \seqnum{A299101},
with the corresponding values of $s_k(5)$ listed as sequence \seqnum{A299109}.
The log log plot of these terms is given in Figure \ref{nis5}. The best fit line for this set of points has slope
$$
C= 0.4568584420.
$$
\begin{figure} \centering
\epsfig{file=24_primes_n_is_5.eps,width=7cm}
\caption{\small{Plot of $\log\log s_{k_N}(n)$ against $N$ for the first 24 primes in the sequence for $n=5$.}}
\label{nis5}
\end{figure}
\begin{figure} \centering
\epsfig{file=n_6_with_19_primes.eps,width=7cm}
\caption{\small{Plot of $\log\log s_{k_N}(n)$ against $N$ for the first 19 primes in the sequence for $n=6$.}}
\label{nis6}
\end{figure}
\begin{figure} \centering
\epsfig{file=nm3_with_31_primes.eps,width=7cm}
\caption{\small{Plot of $\log\log r_{k_N}(n)$ against $N$ for the first 31 primes in the sequence for $n=3$.}}
\label{nism3}
\end{figure}
For $n=6$, the list of the first 25
values $k$ for which $s_k(6)$ appear to be prime is
$$
\begin{array}{l}
1,
2,
3,
9,
14,
23,
29,
81,
128,
210,
468,
473,
746,
950,
3344,
4043,
4839,
14376, \\
39521,
64563, 72984, 82899, 84338, 85206, 86121,
\end{array}
$$
as given in OEIS sequence \seqnum{A113501},
with the corresponding values of $s_k(6)$ given in sequence \seqnum{A088165}
(the prime NSW numbers \cite{nsw}).
In our initial submission of this paper, we obtained the first 19 of these values independently, before we were aware of
sequence \seqnum{A113501}, and made the log log plot of these terms is as in Figure \ref{nis6}. The best fit line for these points has slope
$$
C= 0.5434911190.
$$
Subsequently we found the web page \cite{nswnumber}, where the last six indices above are listed separately,
together with their date of discovery by Weisstein. However, on that page it is stated
unequivocally that all of the corresponding numbers $s_k(6)$ are prime, whereas presumably the largest of
these values were obtained using Mathematica's probabilistic primality test, so the most that can be claimed is that
they are pseudoprimes.
Assuming that the heuristic arguments given in section 6 above are correct, and that the small number of points plotted really gives an accurate picture
of the behaviour for large $N$, the predicted values for the ratio $\rho(n) = C/\log\sqrt{\la}$ in each case are
$$
\rho(3)\approx 0.530689, \, \rho(4)\approx 0.789203, \, \rho(5)\approx 0.583174, \, \rho(6)\approx 0.616641.
$$
Apart from the case $n=4$, all of these values are reasonably close to the number $e^{-\gamma}\approx 0.561459$ obtained from Mertens' theorem. The value for $n=4$ seems anomalous: there are fewer prime terms than predicted in this case.
However, it may be unreasonable to expect close agreement with the predicted value, given the rather small number of data points plotted in each case.
One can also consider the prime terms in the sequences
$(\,r_k(n)\,)$, $n\geq 3$, corresponding to negative values of $n$ in $s_k(n)$.
The list of the first 31 values $k$ for which $r_k(3)$ appear to be prime is
$$
\begin{array}{l}
1,
2,
3,
5,
6,
8,
11,
14,
21,
23,
41,
65,
68,
179,
215,
216,
224,
254,
284,
285,
1485, \\
2361,
2693,
4655,
4838,
7215,
12780,
15378,
17999,
18755,
25416.
\end{array}
$$
Figure \ref{nism3} is the log log plot of these terms. Note that
$(\,r_k(3)\,)$ is a bisection of the Fibonacci sequence, for which
the prime terms are listed as OEIS sequence \seqnum{A005478}.
The slope of the best fit line for these points is
$$
C=
0.3409264905.
$$
Dividing this value by $\log \sqrt{\la} = \log \left((1+\sqrt{5})/2\right)$ gives
$$
\rho(-3)\approx 0.708475,
$$
which is rather large compared with the value of $e^{-\gamma}$ expected from Mertens' theorem, suggesting that the number of primes in this sequence is initially somewhat lower than would be
expected from the heuristic argument in section 6.
\section*{Appendix B: Related sequences from the OEIS}
Here we briefly mention some other sequences in the OEIS which are related to the
considerations in this paper.
Sequence \seqnum{A294099} contains the array of values $s_k(n)$ for $n\geq 1$, $k\geq 0$,
while \seqnum{A299045} is the array of $s_k(-n)$ for the same range of $n$ and $k$.
Sequence \seqnum{A002327} consists of primes of the form $n^2-n-1$, and after sending $n\to -n$ this corresponds to prime values of the polynomial $s_2(n)=n^2+n-1$, for which the relevant values of $n$ are given by
sequence \seqnum{A045546}.
Sequence \seqnum{A000032} begins
$$
2,1,3,4,7,11,18,29,47,\ldots,
$$
and consists of the Lucas numbers denoted $\ell^+_k(1,-1)$ in section 2, which
satisfy the Fibonacci recurrence $\ell^+_{k+2}(1,-1)=\ell^+_{k+1}(1,-1)+\ell^+_k(1,-1)$.
This coincides with an interlacing of two sequences, namely
$$
\Tc_0(3), s_0(3), \Tc_1(3), s_1(3), \Tc_2(3), s_2(3),\Tc_3(3), \ldots ,
$$
so its two distinct bisections are $(\, \Tc_k(3)\,)$ and $(\, s_k(3)\,)$, given by \seqnum{A005248} and \seqnum{A002878} respectively.
Similarly, the Fibonacci sequence \seqnum{A000045} itself coincides with the interlacing
$$
\Uc_{-1}(3), r_0(3), \Uc_0(3), r_1(3), \Uc_1(3), r_2(3),\Uc_2(3), \ldots
$$
obtained from $(\, \Uc_k(3)\,)$ and $(\, r_k(3)\,)$, given by \seqnum{A001906} and \seqnum{A001519} respectively.
There are other values of $n$ for which the OEIS entry for the sequence of terms $s_k(n)$ has not been mentioned so far: $(\,s_k(4)\,)_{k\geq 0}$ is \seqnum{A001834}, $(\,s_k(5)\,)_{k\geq 0}$ is \seqnum{A030221},
$(\,s_k(7)\,)_{k\geq 0}$ is \seqnum{A033890}, $(\,s_k(8)\,)_{k\geq 0}$ is \seqnum{A057080}, and $(\,s_k(9)\,)_{k\geq 0}$ is \seqnum{A057081}.
\seqnum{A008865} is the sequence of values of $\Tc_2(j)$ for $j=1,2,3,\ldots$; the
array of values $\Tc_k(n)$ for $k\geq 1$, $n\geq 1$ is rendered as sequence
\seqnum{A298675}. The values $\Tc_p(n)$ for prime $p$ are listed in sequence \seqnum{A298878}, while the
values $\Tc_p(n)$ with $p$ an odd prime which are not also of the form $\Tc_2(m)$ for some $m$ are
given in \seqnum{A299071}.
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\end{thebibliography}
\bigskip
\hrule
\bigskip
\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B83; Secondary 11A51. \\
\noindent \emph{Keywords: }
recurrence sequence, Chebyshev polynomial, composite number, Lehmer number.
\bigskip
\hrule
\bigskip
\noindent (Concerned with sequence
\seqnum{A000032},
\seqnum{A000045},
\seqnum{A001519},
\seqnum{A001834},
\seqnum{A001906},
\seqnum{A002315},
\seqnum{A002327},
\seqnum{A002878},
\seqnum{A005248},
\seqnum{A005478},
\seqnum{A008865},
\seqnum{A030221},
\seqnum{A033890},
\seqnum{A045546},
\seqnum{A057080},
\seqnum{A057081},
\seqnum{A088165},
\seqnum{A113501},
\seqnum{A117522},
\seqnum{A269251},
\seqnum{A269252},
\seqnum{A269253},
\seqnum{A269254},
\seqnum{A285992},
\seqnum{A294099},
\seqnum{A298675},
\seqnum{A298677},
\seqnum{A298878},
\seqnum{A299045},
\seqnum{A299071},
\seqnum{A299100},
\seqnum{A299101},
\seqnum{A299107}, and
\seqnum{A299109}.)
\bigskip
\hrule
\bigskip
\vspace*{+.1in}
\noindent
Received February 8 2018;
revised versions received July 23 2018; July 31 2018; August 2 2018.
Published in {\it Journal of Integer Sequences}, August 23 2018.
\bigskip
\hrule
\bigskip
\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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