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\begin{center}
\vskip 1cm{\LARGE\bf 
Some Remarks Concerning the \\
\vskip .1in
Lucas-Lehmer Primality Test
}
\vskip 1cm
\large
E.~L.~Roettger\\
Department of General Education \\
Mount Royal University\\
4825 Mount Royal Gate SW\\
Calgary, AB T3E 6K6\\
Canada \\
\href{mailto:eroettger@mtroyal.ca}{\tt eroettger@mtroyal.ca} \\
\ \\
H.~C.~Williams \\
Department of Mathematics and Statistics \\ 
University of Calgary\\
2500 University Drive NW\\
Calgary, AB  T2N 1N4 \\
Canada  \\
\href{mailto:hwilliam@ucalgary.ca}{\tt hwilliam@ucalgary.ca} 
\end{center}

\vskip .2 in


\begin{abstract}
In 1876 Lucas devised a test for the primality of Mersenne numbers $M_n$, where $n\equiv \modd{3} {4}$.
He did so by calculating a recursive  sequence $(s_k)$ with $s_{k+1}=s_k^2-2,$ using an initial value of $s_1=3$.
Later, Lehmer refined Lucas' techniques and constructed a test for all Mersenne numbers
with odd $n$ ($n\geq 3$); again he used the sequence
$(s_k)$, but with an initial value of $4$.  We investigate what initial values can be used in similar primality tests 
and for which values of $n$ they can be used to test $M_n$.  We also
present some computational results.
\end{abstract}


\section{Introduction}\label{Intro}



A perfect number $P$ is defined to be an integer which is equal to the sum of its aliquot (proper) divisors, 
i.e., the divisors less than $P$.  For example, $P=6=1+2+3$.  It is well known that if $P$ is even, 
then it is both necessary and sufficient that $P=2^{n-1}M_n$, where $M_n=2^n-1$ and $M_n$ is a prime.
Such primes are known today as Mersenne primes and integers of the form $2^n-1$ are called Mersenne numbers.
Clearly, in order for $M_n$ to be a prime when $n&gt;2$, it is necessary that $n$ be odd; indeed, $n$ must be a prime.

In 1876 Lucas announced that he had proved that $M_{127}$ is a prime.  This was the first time that a number of this 
magnitude had been substantiated as
 a prime; furthermore, Lucas was able to achieve this without recourse to the tedious and 
time consuming process of trial division.  A detailed discussion of Lucas' work, including references, can be found in Chapters
3 and 5 of \cite{Wil98} and Section 3.1 of \cite{BW23}.

Suppose for a given value of $s_1$ (the {\it seed}), we compute the sequence $(s_n)$ recursively by
\[
s_{i+1}=s_i^2-2 \quad \mbox{($i\geq 1$).}
\]
Indeed, sequences satisfying this recursion can be found in the {\it On-line Encyclopedia of
Integer Sequences} (OEIS) \cite{Slo24}; in particular we mention that when $s_1=3$, as in the theorem below, we have
\seqnum{A001566}.
Lucas proved the following result:
\begin{theorem}\label{earlyLucas}
If $n\equiv \modd{3} {4}$, $n\geq 3$ and $s_1=3$, then $M_n$ is a prime if and only if  
\[
s_{n-1}\equiv 0 \pmod{M_n}.
\]
\end{theorem}
\noindent For example, consider $M_7=127$.  We have $s_1=3$, $s_2=7$, $s_3=47$, $s_4=2207\equiv \modd{48} {M_7}$,
$s_5\equiv 48^2-2\equiv \modd{16} {M_7}$, $s_6\equiv 16^2-2\equiv \modd{0} {M_7}$.  
Thus, $127$ is a prime by Theorem \ref{earlyLucas}.

It was an early version of Theorem \ref{earlyLucas} that Lucas used to establish the primality of $M_{127}$.
He proved this result by making use of the properties of the Fibonacci and Lucas numbers.
However, there are values of $n\equiv \modd{1} {4}$, such as $n=5$, for which $M_n$ is a prime.
Possibly, in order to deal with the case of $n\equiv \modd{1} {4}$, Lucas began his investigations into the features of 
$(U_n)$, $(V_n)$; these are respectively generalizations of the Fibonacci and Lucas numbers and are today referred to as 
the Lucas sequences.

If we select numbers $P$, $Q$ such that $D=P^2-4Q\not=0$, we define
\begin{equation}\label{UnVn}
U_n= U_n(P,Q)=\frac{\alpha^n-\beta^n}{\alpha-\beta} \quad \mbox{and} \quad V_n=V_n(P,Q)=\alpha^n+\beta^n,
\end{equation}
where $\alpha$, $\beta$ are the zeros of $x^2-Px+Q$.  Note that $\alpha+\beta=P$ and $\alpha\beta=Q$.
Also, $U_0=0$, $U_1=1$, $V_0=2$, $V_1=P$.  In addition both $(U_n)$ and $(V_n)$ satisfy the second order linear
recurrence
\[
X_{n+2}=PX_{n+1}-QX_n \quad \mbox{($n\geq 0$).}
\]
Thus, if $P$, $Q$ are integers, then $(U_n)$, $(V_n)$ are both sequences of integers when $n\geq 0$.

By putting $P=4$, $Q=1$, Lucas was able to show that if $n&gt;1$, $n\equiv \modd{1} {4}$ and $s_1=4$, then 
$M_n$ is a prime if $s_{n-1}\equiv \modd{0} {M_n}$.  Lucas' presentation of this result is somewhat confusing, 
but it seems that he was in possession of all the facts needed to prove it.  Later, D.~H.~Lehmer \cite{Leh30} 
was able to prove the following result concerning \seqnum{A003010}:
\begin{theorem}\label{LucasLehmerTest}
If  $n\equiv \modd{1} {2}$, $n\geq 3$ and $s_1=4$, then $M_n$ is a prime if and only if
\[
s_{n-1}\equiv 0 \pmod{M_n}.
\]
\end{theorem}
\noindent This is the famous Lucas-Lehmer test for the primality of Mersenne numbers.  
It has been used to establish the primality of all known Mersenne numbers with $n&gt;127$.
Lucas tried to prove this by putting $m=\frac{n-1}{2}$, $P=2^{m+1}$, $Q=-1$, but 
while he got very close, he didn't quite complete the proof.
Observe that $P^2\equiv \modd{2} {M_n}$.  It is pointed out by Robinson in \cite{Rob54} that a similar test can be performed using
the starting seed of $s_1=10$ (sequence \seqnum{A135927} in the OEIS); indeed, 
as discovered by Lehmer (see \S \ref{IS}),
any of the entries in \seqnum{A018844} is a valid starting value 
for a Lucas-Lehmer primality test like the one above.

Thus, we see by Theorem \ref{earlyLucas} that we can use $s_1=3$ to determine the primality of $M_n$ whenever
$n\equiv \modd{3} {4}$, but by Theorem \ref{LucasLehmerTest}, we can use $s_1=4$ to establish the primality of $M_n$
when $n$ is simply odd.  The only real difference between the tests is the selection of the value of $s_1$.
This suggests the following problem:  Given some particular $s_1$, for what values of $n\geq 3$ does 
Assertion \ref{assertionA} hold?
\begin{assertion}\label{assertionA}
The Mersenne number $M_n$ is a prime if and only if 
$s_{n-1}\equiv \modd{0} {M_n}$.
\end{assertion}
\noindent We have seen that Assertion \ref{assertionA} holds for $n\equiv \modd{3} {4}$ whenever $s_1=3$ and that it holds for 
$n\equiv \modd{1} {2}$ whenever $s_1=4$.
When Assertion \ref{assertionA} is true for a certain $s_1$, we say that $s_1$ is an {\it acceptable seed} for $M_n$.
Since $n-1\geq 2$ and $s_2=s_1^2-2$, we can always replace $s_1$ by $|s_1|$ with no loss of generality.  
Thus, we may assume that an acceptable seed is always nonnegative. 

Because $M_n$ can be a prime only when $n$ is a prime, Jansen \cite{Jan12} calls a seed $s_1$ a 
{\it universal starting value} when it is acceptable for all (with a finite number of exceptions) $M_n$
whenever $n$ is a prime.  Here we will be somewhat more general and define $s_1$ to be a {\it universal seed}
if it is acceptable for all $M_n$ with $n$ odd and $n\geq 3$.  Thus, $4$ and $10$ are universal seeds.
In what follows we will find infinite families of universal seeds and we will also show how to determine
those values of $n$ for which a given integral value of $s_1$ is an acceptable seed for $M_n$.

\section{The Lehmer sequences}\label{TLS}

In \cite{Leh30} Lehmer extended the Lucas sequences by permitting $P=\sqrt{R}$, where $R$ and $Q$ are coprime 
integers.  By doing this he was able to have $D=R-4Q\equiv \modd{\mbox{$2$ or $3$}} {4}$.  We have
\[
U_n=U_n(\sqrt{R},Q), \quad V_n=V_n(\sqrt{R},Q),
\]
but if $\sqrt{R}$ is not an integer, $(U_n)$ and $(V_n)$ are no longer sequences of integers; however, when we set
\begin{align*}
&amp; \bar{U}_n=\bar{U}_n(\sqrt{R},Q)= \begin{cases}
U_n/\sqrt{R}, &amp; \text{if $2\mid n$;}\\
U_n, &amp; \text{if $2\nmid n$,}\\
\end{cases}\\
&amp;\bar{V}_n=\bar{V}_n(\sqrt{R},Q)= \begin{cases}
V_n, &amp; \text{if $2\mid n$;}\\
V_n/\sqrt{R}, &amp; \text{if $2\nmid n$,}\\
\end{cases}\\
\end{align*}
then for $n\geq 0$ $(\bar{U}_n)$ and $(\bar{V}_n)$ are sequences of integers.  We can deduce this by noting that
\begin{align*}
&amp;\bar{U}_0=0, \quad \bar{U}_1=1, \quad \bar{U}_2=1, \quad \bar{U}_3=R-Q,\\ 
&amp;\bar{V}_0=2, \quad \bar{V}_1=1, \quad \bar{V}_2=R-2Q, \quad \bar{V}_3=R-3Q.
\end{align*}
Also, it is easy to establish from (\ref{UnVn}) that both $(\bar{U}_n)$ and $(\bar{V}_n)$ satisfy the fourth order linear recurring sequence
\begin{equation}\label{4thLehmer}
X_{n+4}=(R-2Q)X_{n+2}-Q^2X_n \quad \mbox{($n\geq 0$).}
\end{equation}
Thus, we find by induction that $(\bar{U}_n)_{n\geq 0}$ and $(\bar{V}_n)_{n\geq 0}$ 
are sequences of integers.
In what follows we will only consider $\bar{U}_n$ and $\bar{V}_n$ when $n\geq 0$.

Lehmer deduced a number of properties of $(\bar{U}_n)$ and $(\bar{V}_n)$, but we will only mention those that will be 
useful in this investigation.  We first note the simple identities:
\begin{align}
&amp;\bar{U}_{2n}=\bar{V}_n\bar{U}_n,\label{LehmerU2n}\\
&amp;\bar{V}_{2n}=\begin{cases}
\bar{V}_n^2-2Q^n, &amp; \text{if $2\mid n$;}\\
R\bar{V}_n^2-2Q^n, &amp; \text{if $2\nmid n$.}\label{LehmerV2n}\\
\end{cases}
\end{align}
Also, when $n\geq 0$, we have $\gcd(\bar{U}_n,\bar{V}_n)\mid 2$.
If we put $R^{\prime}=R-4Q=D$ and $Q^{\prime}=-Q$, then 
$\bar{V}_0(R^{\prime},Q^{\prime})=2=\bar{V}_0(R,Q)$, 
$\bar{V}_2(R^{\prime},Q^{\prime})=R^{\prime}-2Q^{\prime}=\bar{V}_2(R,Q)$ and 
$(\bar{V}_n(R^{\prime},Q^{\prime}))$ satisfies (\ref{4thLehmer}) when $2\mid n$.
Thus, whenever $n$ is even we have 
$\bar{V}_n(R^{\prime},Q^{\prime})=\bar{V}_n(R,Q)$ or
\begin{equation}
\bar{V}_n(R,Q)=\bar{V}_n(D,-Q),\label{LehmerRQDminusQ}
\end{equation}
a result also noted by Lehmer.

For a given odd prime $p$, we define the {\it rank of appearance of $p$ in $(\bar{U}_n)$} to be the 
least positive integer $r=r(p)$ such that $p\mid \bar{U}_r$.  Lehmer showed (Theorem 1.8) that if 
$r=r(p)$, then $p\mid \bar{U}_n$ if and only if $r(p) \mid n$.  Of course, it is possible that $r(p)$ might not exist.
However, he also showed (Theorem 1.9) that if $p$ is odd and $p\nmid RQ$, then $r(p)$ does exist and 
$r(p)\mid p-\epsilon \sigma$, where $\epsilon$ is the Legendre symbol $(D|p)$ and $\sigma$ is the 
Legendre symbol $(R|p)$.  
In addition (Theorem 4.9), if $p$ is an odd prime and $p\nmid RQ$, then 
$p\mid \bar{V}_{(p-\epsilon\sigma)/2}$ if and only if $\sigma=-\tau$, where $\tau$ is the 
Legendre symbol $(Q|p)$.
Let $p$ be any odd prime such that $p\nmid Q$ and $p\mid R$.  
In this case it can be shown by using the reasoning in Section 1.4 of \cite{Leh30} that $r(p)$ exists 
and $r(p)=2p$.
Furthermore if $p\nmid Q$ and $p\mid D$, then since $p\mid R-4Q$, we cannot have $p\mid R$.
Thus $r(p)$ exists and $r(p)\mid p-\epsilon \sigma$, where $\epsilon = (D|p)=0$.
Thus, in this case, $r(p)=p$.

From these results we can prove the following primality result concerning $M_n$.
\begin{lemma}\label{criteriontest}
Let $n$ be odd and $n\geq 3$.  If $\gcd(M_n,Q)=1$ and $M_n\mid \bar{V}_{\frac{M_n+1}{2}}(R,Q)$, 
then $M_n$ must be a prime.  
\end{lemma}
\begin{proof}
By (\ref{LehmerU2n}) we must have $M_n\mid \bar{U}_{M_n+1}(R,Q)$.
If $p$ is any prime divisor of $M_n$, then $p\nmid Q$ and $r=r(p)$ must exist in $(\bar{U}_n(R,Q))$.
If $p\mid R$, then $r=2p$ and $2p\mid M_n+1$, which means that $2p\mid 2^n$,
an impossibility.  Thus, $p\nmid QR$ and $r\mid p-\epsilon \sigma$.
Since $r\mid M_n+1$ and $p \mid \bar{V}_{\frac{M_n+1}{2}}(R,Q)$, we cannot have $r\mid \frac{M_n+1}{2}$
because in this case $p\mid \gcd(\bar{V}_{\frac{M_n+1}{2}},\bar{U}_{\frac{M_n+1}{2}})\mid 2$ and $p=2$,
which means that $r=2^n$.
Since $r\mid p-\epsilon \sigma$, we must have $p=k2^n\pm 1\geq 2^n-1$.
But $2^n&gt;3$ means that $2^{2n}&gt;3\cdot 2^n \rightarrow (2^n-1)^2&gt;M_n$.  Thus, $p&gt;\sqrt{M_n}$ and therefore
$M_n$ must be a prime.
\end{proof}


We also have a simple condition which guarantees that $M_n\mid \bar{V}_{\frac{M_n+1}{2}}(R,Q)$ when $M_n$ is a prime.
\begin{lemma}\label{criteriontest2}
Let $n$ be odd, $n\geq 3$ and $M_n$ be a prime.  If the Legendre symbols
\[
(DR|M_n)=(QR|M_n)=-1,
\]
then $M_n\mid \bar{V}_{\frac{M_n+1}{2}}(R,Q)$.
\end{lemma}
\begin{proof}
If $M_n$ is a prime, then $(DR|M_n)=-1$ implies that $\epsilon \sigma=-1$.  
Also, since $\tau=-\sigma$, we have $M_n\mid \bar{V}_{\frac{M_n+1}{2}}(R,Q)$.
\end{proof}


We next produce a criterion for $M_n$ to be a divisor of $\bar{V}_{\frac{M_n+1}{2}}(R,Q)$
when
 $\gcd(Q,M_n)=1$,
 by providing a reformulation of the sequence $\bar{V}_{2^i}(R,Q)$ in terms of the sequence $s_i$.
 We first observe that $\frac{M_n+1}{2}=2^{n-1}$.  Suppose we put 
\[
s_i\equiv \bar{V}_{2^i}(R,Q) Q^{-2^{i-1}}\pmod{M_n},
\]
where $QQ^{-1}\equiv \modd{1} {M_n}$ and $i\geq 1$.  By (\ref{LehmerV2n}) we have
\[
\bar{V}_{2^{i+1}}(R,Q)\equiv \bar{V}_{2^i}^2(R,Q)-2Q^{2^i}\pmod{M_n}.
\]
Thus,
\[
Q^{2^i}s_{i+1}\equiv (Q^{2^{i-1}}s_i)^2-2Q^{2^i}\pmod{M_n}
\]
and
\[
s_{i+1}\equiv s_i^2-2 \pmod{M_n}.
\]
It follows that if $s_1\equiv \bar{V}_2(R,Q)/Q \equiv \modd{RQ^{-1}-2} {M_n}$, then
\[
s_{n-1}\equiv Q^{-2^{n-2}}\bar{V}_{2^{n-1}}(R,Q)\pmod{M_n}.
\]

If we combine this observation with Lemmas \ref{criteriontest} and \ref{criteriontest2}, we get the following generalization of 
Theorem \ref{LucasLehmerTest}.
\begin{theorem}\label{generalLLT}
Let $n$ be odd and $n\geq 3$.  If the Jacobi symbols  $(RD|M_n)=(RQ|M_n)=-1$ and 
$s_1\equiv \modd{RQ^{-1}-2} {M_n}$, then $M_n$ is a prime if and only if $s_{n-1}\equiv \modd{0} {M_n}$,
where
\[
s_{i+1}\equiv s_i^2-2\pmod{M_n}.
\]
\end{theorem}
\noindent If we consider the case of $R=2$, $Q=-1$, we have $D=6$.
We observe that 
\[
M_n\equiv -1 \pmod{8}.
\]
Hence  $(R|M_n)=1$, $(Q|M_n)=-1$, $(D|M_n)=-(M_n|3)$. 
We have  $-(M_n|3)=-1$ since $n$ is odd and $M_n\equiv \modd{1} {3}$.
Thus,
$(DR|M_n)=(QR|M_n)=-1$, $s_1\equiv \modd{ -4} {M_n}$,  and we can put $s_1=4$ to get
 Theorem \ref{LucasLehmerTest}.
Notice that because $P^2\equiv \modd{2} {M_n}$ in Lucas' attempted proof of Theorem \ref{LucasLehmerTest},
this result of Lehmer is essentially the same as that of Lucas, except that Lehmer completed the proof.
While Lehmer did not mention Theorem \ref{generalLLT} explicitly, he might have been aware of it.
In the next three sections we will use Theorem \ref{generalLLT} to produce other values of $s_1$ for which theorems 
like Theorem \ref{LucasLehmerTest} also hold.



\section{Rational seeds and permissible triplets}\label{RS}

Suppose $a$, $b$ are integers such that $\gcd(b,M_n)=1$.  If $s_1\equiv \modd{ab^{-1}} {M_n}$ and $s_1$ is an 
acceptable seed for $M_n$, we say that $s_1=\frac{a}{b}$ is an acceptable {\it rational} seed for $M_n$.
In this section we will show that there exists an infinitude 
of distinct universal rational seeds for any given $M_n$ with $n$ odd and $n\geq 3$.
In \cite{Jan12} Jansen gave $s_1=\frac{2}{3}$ and $s_1=\frac{626}{363}$ as examples of rational universal seeds.
Indeed, as shown later in 
Section \ref{URS}, Jansen's seeds can be considered as $s_1\equiv \modd{RQ^{-1}-2} {M_n}$ for some $R$, $Q$. 
Although it was not part of the main goal of \cite{Jan12}, he also provided a technique for finding an infinitude of
rational and even irrational universal starting values.  These latter seeds can be written as polynomials
over the rationals $\mathbb{Q}$ in $\sqrt{2}$.  
To express $\sqrt{2}$ as an integer modulo $M_n$, he suggested
using $2^{\frac{n+1}{2}}$ which satisfies $(2^{\frac{n+1}{2}})^2\equiv \modd{2} {M_n}$.
In this paper we will focus our attention on rational seeds only.  We will next show how to find an infinitude of rational 
universal seeds, but we will make use of more elementary methods than those employed in \cite{Jan12}.


\begin{lemma}\label{rationallemma}
Let $n$ be odd, $n\geq 3$ and $M_n$ be a prime.
If $r$, $d$, $q$, $X$, $Y$, $Z$ are integers such that $\gcd(M_n,Y)=1$, 
\begin{equation}\label{dioratlemma}
4qY^2=rX^2-dZ^2
\end{equation}
and the Legendre symbols $(rd|M_n)=(qr|M_n)=-1$, then
\[
M_n\mid \bar{V}_{\frac{M_n+1}{2}}(R,Q)
\]
where $R=rX^2$, $Q=qY^2$ ($D=dZ^2$).
\end{lemma}
\begin{proof}
If $M_n\mid X$, then $4qY^2\equiv \modd{-dZ^2} {M_n}$.  Since $(-4dq|M_n)=(-4|M_n)(rd|M_n)(qr|M_n)=-1$, we must have
$M_n\mid Y$, which is impossible.  Similarly, $M_n\nmid Z$.  Thus,
\[
(DR|M_n)=(QR|M_n)=-1
\]
and $M_n\mid \bar{V}_{\frac{M_n+1}{2}}(R,Q)$ by Lemma \ref{criteriontest2}.
\end{proof}
We can now combine the results of Theorem \ref{generalLLT} and Lemma \ref{rationallemma} to obtain 
 a universal rational seed $s_1$: we need first to find values of $r$, $q$, $d$
such that $(rd|M_n)=(rq|M_n)=-1$ for all odd $n\geq 3$.  Since $(-1|M_n)=-1$ when $n\geq 2$, this means that we must have
\begin{equation}\label{rqdleg}
(r|M_n)=(-d|M_n)=(-q|M_n)
\end{equation}
for all odd $n\geq 3$.  Also, since (\ref{dioratlemma}) must have integer solutions $X$, $Y$, $Z$, there is no loss of generality in 
assuming $\gcd(r,q,d)=1$.

We have already observed that if $n$ is odd and $n\geq 3$, then 
\[
(3|M_n)=-1, \quad (-1|M_n)=-1, \quad (2|M_n)=1, \quad \mbox{and} \quad (6|M_n)=-1.
\]
We can therefore restrict the values of $r$, $d$, $q$ to be in $\mathcal{S}=\{-6,-3,-2,-1, 1, 2, 3, 6\}$ because, in these cases
the value of the Jacobi symbol does not depend on $n$.
However, we could also use, if they exist, values of $a$ not in $\mathcal{S}$ such that  $(a|M_n)$ has a fixed value
not depending on $n$ ($\geq 3$).
A computer search up to $10^8$ for such a number 
eliminated all possible values of $a$ except for $a=76627=19\cdot37\cdot109\equiv \modd{-1} {4}$.
Notice that for this remaining value of $a$ we have $2^{36}\equiv \modd{1} {a}$.  We then determined that $(a|M_n)=-1$ for all 
odd $n\geq 3$ by verifying that 
\[
(M_n|19)(M_n|37)(M_n|109)=1
\]
for all odd $n$ such that $3\leq n\leq 35$.
This same number had been found earlier, but
in a somewhat different context, by Vardhana (see \cite[p.~13]{Jan12}).


We define a {\it permissible} triplet to be a triplet $(r,q,d)$ such that $r$, $q$, $d\in \mathcal{S}$, $\gcd(r,q,d)=1$, (\ref{rqdleg}) holds, 
and (\ref{dioratlemma}) has integer solutions.  Notice that if $(a,b,c)$ is a permissible triplet, then so is $(-a,-b,-c)$
and $(c,-b,a)$.  We may therefore always assume that $r&gt;0$.  
Furthermore, if $q&lt;0$, then we must have $d&gt;0$ and if $d&lt;0$, then we must have $q&gt;0$.
Since $\#\mathcal{S}=8$, there $512=8^3$ possible permissible triplets, 
but many of these can be eliminated.
For example, if $r=3$, $q=d=1$, then (\ref{rqdleg}) holds and (\ref{dioratlemma}) becomes
\[
Z^2=3X^2-4Y^2.
\]
We may assume here that $\gcd(X,Y,Z)=1$.
Since $3\mid Z^2+Y^2$, we can only have $3\mid Z$ and $3\mid Y$, but this means that $3\mid X$, which is not
possible.  Similarly, if $r=1$, $q=3$, $d=-1$, we get
\[
-Z^2=X^2-12Y^2,
\]
which also has no integer solutions. 
Indeed if (\ref{dioratlemma}) reduces to a Diophantine equation of the form $az^2=x^2+y^2$, where $a$ is a fixed integer
such that $3\mid a$, but $9\nmid a$, then (\ref{dioratlemma}) cannot have integer solutions.

If we put $r=2$, $q=3$, $d=-1$, we have 
(\ref{rqdleg});  we put
\[
R=2X^2, \quad Q=3Y^2, \quad D=-Z^2,
\]
and (\ref{dioratlemma}) becomes
\[
12Y^2=2X^2+Z^2.
\]
Here we must have $2\mid Z$ and $2\mid X$ from which we deduce that
\[
(Z/2)^2=3Y^2-2(X/2)^2.
\]

We now consider the Diophantine equation
\begin{equation}
z^2=3y^2-2x^2. \label{dioeq}
\end{equation}
We can find by using standard techniques (see, for example, Chapter 7, Theorem 4 of \cite{Mor69}) that
all integral solutions of (\ref{dioeq}) with $\gcd(x,y,z)=1$ are given by
\begin{align*}
&amp;x=(3t^2+2u^2-6tu)/\delta=(3(t-u)^2-u^2)/\delta\\
&amp;y=(3t^2+2u^2-4tu)/\delta=(2(t-u)^2+t^2)/\delta\\
&amp;z=(3t^2-2u^2)/\delta.
\end{align*}
Here $t$, $u$ are coprime integer parameters and 
\[
\delta=\gcd(3t^2-2u^2, 3t^2+2u^2-4tu, 3t^2+2u^2-6tu).
\]
It is easy to show that $\delta$ must be a divisor of $6$.
We see, then, that 
for $r$, $q$, $d\in \mathcal{S}$,
(\ref{dioratlemma}) will have integer solutions only if we can put it in the form (\ref{dioeq}).


By subjecting each of the 512 possible triplets $(r,q,d)$ to the above checks, we found that only 24 permissible triplets remain.  These are:
\begin{gather*}
(1,-2,3), (1,-2,6), (1,3,-2), (1,3,6), (1,6,-2), (1,6,3), (2,1,3), (2,-1,6),\\ (2,3,6), (2,6,3), (3,1,-6), (3,2,-6),
\end{gather*}
with the remaining 12 produced by using the transformations $(a,b,c)\rightarrow(c,-b,a)$ and $(a,b,c)\rightarrow (-a,-b,-c)$.
For example, $(1,-2,3)\rightarrow(3,2,1)$ and
 $(3,2,-6)\rightarrow (-6,-2,3)\rightarrow (6,2,-3)$.

\section{Universal rational seeds}\label{URS}
Returning to the permissible triple $(2,3,-1)$, let
 $t$, $u$ be integer parameters such that $\gcd(t,u)=1$ and 
\begin{align*}
&amp;X=2(3t^2+2u^2-6tu)/\delta,\\
&amp;Y=(3t^2+2u^2-4tu)/\delta,
\end{align*}
we see by Theorem \ref{generalLLT} that if $\gcd(M_n, 3t^2+2u^2-4tu)=1$, then $M_n$ is a prime if and only if
\[
s_{n-1}\equiv 0 \pmod{M_n}
\]
when
\[
s_1\equiv RQ^{-1}-2 \equiv \frac{8(3t^2+2u^2-6tu)^2}{3(3t^2+2u^2-4tu)^2}-2\pmod{M_n}.
\]
If we put $t=u=1$, we get $Q=3$,
\[
s_1\equiv \frac{8}{3}-2=\frac{2}{3} \pmod{M_n};
\]
if we put $t=3$, $u=4$, we get $Q=3\cdot 11^2$.
Since in this latter case we have $\gcd(M_n,Q)=1$ for odd $n$,
we can use
\[
s_1\equiv \frac{8(-13)^2}{3(11)^2}-2=\frac{626}{363} \pmod{M_n}.
\]

Indeed, it is well known (see, for example, Chapter 1 of Cox \cite{Cox89}) 
that if $p$ is any prime such that $p\equiv \modd{3} {8}$, then there must exist integers $t$, $u$ 
such that $p=t^2+2(t-u)^2$.
Since $p\equiv \modd{3} {8}$, we have $(2|p)=-1$
which means that $p\nmid 2^n-1$ or $\gcd(p,M_n)=1$ whenever $n$ is odd.
Since there are infinitely many primes $p$ satisfying $p\equiv \modd{3} {8}$, it follows that 
there exists an infinitude of pairs $(t,u)$, with $\gcd(t,u)=1$ and $p=3t^2+2u^2-4tu$
such that the values assumed by $3t^2+2u^2-4tu$ are all distinct and 
\[
\gcd(M_n, 3(3t^2+2u^2-4tu))=1.
\]
For such pairs we have
\[
\gcd(3t^2+2u^2-4tu, 3t^2+2u^2-6tu)=\gcd(3t^2+2u^2-4tu, 2tu)=\gcd(p,2tu)=1.
\]
It follows that there must exist an infinitude of distinct universal rational seeds.


We now consider the permissible triple $(3,1,2)$.
Note that $(3,1,2)=(c,-b,a)$ when $(a,b,c)=(2,-1,3)$, so 
\begin{equation}
2Z^2=3X^2-4Y^2\label{2Zhat2}
\end{equation}
has an infinitude of integer solutions.
 Indeed,
 all integer solutions of (\ref{2Zhat2}) with $\gcd(X,Y,Z)=1$ can be written as
\begin{align*}
&amp;X=2(3t^2+2u^2-4tu)/\delta,\\
&amp;Z=2(3t^2+2u^2-6tu)/\delta,\\
&amp;Y=(3t^2-2u^2)/\delta,
\end{align*}
where $t$, $u$ are coprime integer parameters and 
\[
\delta=\gcd(3t^2+2u^2-4tu, 3t^2+2u^2-6tu, 3t^2-2u^2).
\]
Suppose we put
\[
s_1=R/Q-2=3X^2/Y^2-2
      =\frac{12(3t^2+2u^2-4tu)^2}{(3t^2-2u^2)^2}-2
\]
for any integer parameters $t$ and $u$.  We see that $s_1$ is an acceptable rational seed for 
all $M_n$ with $n$ odd and $n\geq 3$.

Let
\[
\mathcal{C}=\{c: c=3t^2-2u^2; t,u\in \mathbb{Z}; \gcd(t,u)=1; \gcd(c, M_n)=1 \mbox{ for all odd $n$}\}.
\]
Note that $1$ ($t=u=1$), $-2$ ($t=4$, $u=5$), $-5$ ($t=1$, $u=2$), $19$ ($t=3$, $u=2$) are in
$\mathcal{C}$, so $\mathcal{C}$ is not empty. Indeed, we can assert that $\# \mathcal{C}$ is infinite. 
Let $p$ be any prime such that $p\equiv 5$ or $\modd{19} {24}$.
We have $(6|p)=1$, and consequently there exists some integer $s$ such that $s^2\equiv \modd{6} {p}$.
Let $\mathbb{K}$ denote the real quadratic field $\mathbb{Q}(\sqrt{6})$ with discriminant 24.
The maximal order $\mathcal{O}$ of $\mathbb{K}$ is given by $\mathcal{O}=\mathbb{Z}+\sqrt{6}\mathbb{Z}$ 
and since $s^2\equiv \modd{6} {p}$, we see that the ideal $\mathfrak{p}=p\mathbb{Z}+(s+\sqrt{6})\mathbb{Z}$
is a prime ideal of $\mathcal{O}$ with norm $N(\mathfrak{p})=p$.
Also, $\mathfrak{t}=2\mathbb{Z}+\sqrt{6}\mathbb{Z}$ is a prime ideal of $\mathcal{O}$ with $\mathfrak{t}^2=$
the principal ideal ($2$) and $N(\mathfrak{t})=2$.
Since the class number of $\mathbb{K}$ is $1$, we know that the ideal $\mathfrak{m}=\mathfrak{t}\mathfrak{p}$
must be principal.  Thus, there must exist some $u^\prime$, $t \in \mathbb{Z}$ and
$\mu=u^\prime+t\sqrt{6} \in \mathcal{O}$ such that $\mathfrak{m}=(\mu)$ and 
\[
2p=N(\mathfrak{m})=|N(\mu)|=|{u^\prime}^2-6t^2|.
\]
Putting $u=u^\prime/2$, we have
\[
p=|2u^2-3t^2|
\]
and $\gcd(t,u)=1$.
Since $(2|p)=-1$, we must, as noted earlier, have $\gcd(p,M_n)=1$ for all odd $n$. 
Thus, either $p$ or $-p$ must be in $\mathcal{C}$ and as there exists an infinitude of $p$, we see that $\#\mathcal{C}$ is infinite.


For each $c\in\mathcal{C}$, we have
\[
s_1=\frac{12(3t^2+2u^2-4tu)^2}{(3t^2-2u^2)^2}-2=\frac{12(c+4u(u-t))^2}{c^2}-2.
\]
If we let $s$ be any prime divisor of $\gcd(c, c+4u(u-t))$, it is easy to show that $s\mid t$ and $s\mid u$,
which is impossible by selection of $c$.  Thus, we must have $\gcd(c, c+4u(u-t))=1$, and therefore there must exist an 
infinitude of distinct rational values of $s_1$, each corresponding to some $c\in \mathcal{C}$.

We now make use of a result that goes back to Euler.  Suppose that $a$, $b$, $c$ are integers where $ab&gt;1$ and 
$ab$ is not a perfect integral square.  It is well known that the Pellian equation (see \cite[Corollary 1.10]{JW09} or \cite[p.~377]{Dic34})
\begin{equation}\label{Pellian}
x^2-aby^2=1
\end{equation}
has an infinitude of integer solutions $(x_k,y_k)$ given by
\[
x_k+\sqrt{ab}y_k=(x_1 +\sqrt{ab}y_1)^k,
\]
where $(x_1,y_1)$ is the {\it fundamental solution} 
(that solution $(x,y)$ for which $x$, $y&gt;0$ and $x+y\sqrt{ab}$ is least) of (\ref{Pellian}).
For example, if $ab=6$, then it is easy to verify that $x_1=5$, $y_1=2$.
The next result is an observation of Euler (see \cite[p.~408]{Dic34}).
\begin{theorem}\label{PellTheorem}
Let $a$, $b$ be specified as above.  If $(T_1,U_1)$ is any given solution of the Diophantine equation
\begin{equation}\label{DioEq}
aT^2-bU^2=c,
\end{equation}
then there exists an infinitude of distinct solutions of (\ref{DioEq}) given by $(T_k,U_k)$, where
\[
\sqrt{a}T_k+\sqrt{b}U_k=(\sqrt{a}T_1+\sqrt{b}U_1)(x_1+\sqrt{ab}y_1)^{k-1}.
\]
That is, $T_k=T_1 x_{k-1}+bU_1y_{k-1}$, $U_k=aT_1y_{k-1}+U_1x_{k-1}$.
\end{theorem}
\noindent Notice that the sequences $(T_k)$ and $(U_k)$ both satisfy the linear recurrence
\[
W_{k+2} =2 x_1 W_{k+1}-W_k.
\]

Suppose now  that $t_1$, $u_1$ are some positive integers such that
\[
3t_1^2-2u_1^2=c
\]
for some constant $c\in \mathcal{C}$.  In this case we find that if we put 
\[
\sqrt{3}t_k+\sqrt{2}u_k=(\sqrt{3}t_1+\sqrt{2}u_1)(5+2\sqrt{6})^{k-1} \quad \mbox{($k\geq 1$)},
\]
we get
\[
t_2=5t_1+4u_1, \quad u_2=6t_1+5 u_1
\]
and
\[
t_{j+2}=10t_{j+1}-t_j, \quad u_{j+2}=10u_{j+1}-u_j \quad \mbox{($j\geq 1$).}
\]
Also, for any $k\geq 1$
\[
3t_k^2-2u_k^2=c.
\]
Furthermore, it is clear that the sequences $(u_k)$ and $(u_k-t_k)$ 
are both strictly increasing with increasing $k$.

Putting 
\[
Y_k=(3t_k^2-2u_k^2)/\delta=c/\delta,
\]
we get
\begin{align*}
X_k&amp;= 2(3t_k^2+2u_k^2-4t_ku_k)/\delta\\
&amp;=2(c+4u_k(u_k-t_k))/\delta.
\end{align*}
Thus, if for any fixed $k\geq 1$, we put 
\begin{equation}\label{seedRQ}
s_1= RQ^{-1}-2 =3X_k^2 Y_k^{-2}-2= 12(c+4u_k(u_k-t_k))^2/c^2-2,
\end{equation}
then $s_1$ is a universal rational seed for $M_n$.
Notice that $c+4u_k(u_k-t_k)$ is a strictly increasing function of $k$; thus, for each fixed $c\in \mathcal{C}$
there exists an infinitude of distinct universal rational seeds $s_1$ 
for  $M_n$.
Since there exists an infinitude of distinct $c\in\mathcal{C}$, we see that 
(\ref{seedRQ}) is a doubly parametric set 
of distinct rational universal seeds $s_1$ for $M_n$ for all odd $n\geq 3$.

Suppose we put 
\begin{align*}
&amp;\delta f_1(t,u)=3t^2+2u^2-6tu\\
&amp;\delta f_2(t,u)=3t^2+2u^2-4tu\\
&amp;\delta f_3(t,u)=3t^2-2u^2,
\end{align*}
where, as earlier, $\delta=\gcd(3t^2+2u^2-6tu, 3t^2+2u^2-4tu, 3t^2-2u^2)$.
We have 
\begin{equation}\label{gcdfi}
\gcd(f_1,f_2,f_3)=1
\end{equation}
and
\begin{equation}\label{relationshipfi}
f_3^2=3f_2^2-2f_1^2,
\end{equation}
where we write $f_i$ for $f_i(t,u)$ ($i=1,2,3$).
It turns out that we can find four families of universal rational seeds.  These can be expressed as:
\[
s_1=8f_1^2/3f_2^2-2
\]
for each of the permissible triples $(2,3,-1)$,  $(3,2,-6)$,  $(6,1,-3)$,  $(1,6,2)$;
\[
s_1=12f_2^2/f_3^2-2
\]
for $(1,3,6)$,  $(1,-2,6)$,  $(2,-1,3)$,  $(2,6,3)$,  $(3,-6,2)$,  $(3,1,2)$,  $(6,-3,1)$,  $(6,2,1)$;
\[
s_1=4f_3^2/3f_2^2-2
\]
for $(1,3,2)$, $(2,6,-1)$, $(3,1,-6)$, $(6,2,-3)$;
\[
s_1=6f_2^2/f_1^2-2
\]
for $(1,-2,3)$, $(1,6,3)$, $(2,-1,6)$, $(2,3,6)$, $(3,-6,1)$, $(3,2,1)$, $(6,-3,2)$, $(6,1,2)$.
Each of these families represents an infinitude of distinct universal rational seeds.

If $s_1=8f_1^2/3f_2^2-2$ or $s_1=4f_3^2/3f_2^2-2$ and $s_1$ is an integer, then $3\mid f_1$ or $3\mid f_3$.
In either case, because of (\ref{relationshipfi}) we must have $3\mid \gcd(f_1,f_2,f_3)$, 
which is impossible by (\ref{gcdfi}).  
Thus, if $s_1=12 f_2^2/f_3^2-2$ is an integer, then $f_3\mid 2f_2$.  If $2\mid f_3$, then by 
 (\ref{relationshipfi}) we get $2\mid f_2$ and then $2\mid f_1$, which is impossible.
 Hence $f_3\mid f_2$ and by  (\ref{relationshipfi}) $f_3\mid f_1$, which means that $|f_3|=1$ and we get
 \[
 s_1=12 f_2^2-2=8f_1^2+2.
 \]
 Similarly, if $s_1=6f_2^2/f_1^2-2\in\mathbb{Z}$, then
 \[
 s_1=6f_2^2-2=2f_3^2+2.
 \]


In the next section we constrain the values of admissible seeds to be integral.





\section{Integral seeds}\label{IS}

We have already seen that we can assume an acceptable seed for $M_n$ when $n\geq 3$ is nonnegative.  
For integral values of $s_1$, we have $|s_i|\leq 2$ for all $i\geq 1$ whenever $s_1=0$, $1$, $2$.
Thus we may assume further that a universal integral seed $s_1$ for $M_n$ ($n\geq 3$) must be such that
$s_1\geq 3$

In section 5 of \cite{Leh30} Lehmer considered the case of $n$ odd, $n\geq 3$ and $Q=1$, 
$R=-X^2$, $-2X^2$, $3X^2$, $6X^2$ and $D=Z^2$, $2Z^2$, $-3Z^2$, $-6Z^2$.
On referring to (\ref{LehmerRQDminusQ}) (by interchanging the values of $R$ and $D$ we can get $Q=-1$)
he found that the only Diophantine equations (\ref{dioratlemma}) that had solutions which could be produced from these values
of $R$ and $D$, subject to the constraint that $D-R=-4$, are 
\[
2Z^2-3X^2=-4 \quad \mbox{($R=3X^2$, $D=2Z^2$),}
\]
and
\[
2Z^2-6X^2=-4\quad \mbox{($R=6X^2$, $D=2Z^2$).}
\]
These correspond to the permissible triples $(3,1,2)$ and $(6,1,2)$.
There can be no solution of (\ref{dioratlemma}) for $(6,1,-3)$ or $(3,1,-6)$ because $3$ must be a divisor of $Q$ in these cases and this
 is not possible because $Q=1$.
Each of these Diophantine equations has an infinitude of integer solutions.  In the case of the first equation, we get
\[
X=2w_k, \quad Z=z_k \quad\mbox{($k\geq 1$),}
\]
where
\[
3w_k^2-2z_k^2=1.
\]
Here we can put $w_1=1$, $w_2=9$ and $w_{j+2}=10w_{j+1}-w_j$ ($j\geq 1$)
and
\[
s_1=12w_k^2-2.
\]
In the case of the second equation, we get 
\[
X=w_k\quad\mbox{($k\geq 1$),}
\]
where $w_1=1$, $w_2=3$, $w_{j+2}=4w_{j+1}-w_j$ ($j\geq 1$) and
\[
s_1=6w_k^2-2.
\]
Thus, there exists an infinitude of universal integral seeds for all $M_n$ with $n$ odd and $n\geq 3$.  
For such $s_1$, we have
\[
s_1\in \{4, 10, 52, 724, 970, 10084, 95050, \ldots\}.
\]


If we consider $s_1=3$, we know that $s_1$ is acceptable for all $M_n$ with $n\equiv \modd{3} {4}$.
We now consider the problem of finding other values of $s_1$ which are acceptable for such $M_n$.  
We first observe that 
\[
M_n=2^{4m+3}-1\equiv 2\pmod{5}.
\]
Thus, $(M_n|5)=(2|5)=-1$.  We can then put $R=5X^2$, $Q=1$ and $D=Z^2$ and derive the Diophantine equation
\[
Z^2-5X^2=-4.
\]

The well-known Fibonacci numbers $(F_k)_{k\geq 0}$ and the Lucas numbers $(L_k)_{k\geq 0}$ each satisfy the linear 
recurrence 
\[
X_{k+2}=X_{k+1}+X_k,
\]
with $F_0=0$, $F_1=1$, $L_0=2$, $L_1=1$.  Furthermore, we also know that 
\[
L_k^2-5F_k^2=4(-1)^k.
\]
It follows that if we put $s_1=5F_k^2-2=L_k^2+2$ for odd $k$, then 
\[
s_1\in \{3, 18, 123, 843, \ldots\}
\]
is acceptable for all $M_n$ with $n\equiv \modd{3} {4}$.
However, these are not the only such seeds.  If we put $R=10X^2$, $Q=1$, $D=Z^2$, then we 
can show that if $w_0=0$, $w_1=6$ and 
\[
w_{k+2}=38w_{k+1}-w_k,
\]
then $s_1=40 w_j^2-2$ ($j\geq 1$) is also such a seed.  Furthermore, such a value for $s_1$ can never be 
of the form $5F_k^2-2$ for any $k\geq 1$.

In the next section we will turn our attention to the 
problem of finding those values of $n$ for which a given integral $s_1$ is an acceptable seed
for $M_n$.



\section{Acceptable integral seeds}\label{AIS}

We begin by stating a simple criterion for a given $s_1$ to be an acceptable integral seed for $M_n$.
This criterion is given in a more general setting as Theorem 2.1 of \cite{Jan12}.
\begin{theorem}\label{acceptableseed}
Let $s_1$ be a given integer, then $s_1$ is an acceptable seed for $M_n$ if and only if
\[
(s_1+2|M_n)=-1 \quad \mbox{and} \quad (s_1-2|M_n)=1.
\]
\end{theorem}
\begin{proof}
Let  $R$, $Q$ be integers such that $\gcd(Q,M_n)=1$ and
$s_1+2\equiv \modd{RQ^{-1}} {M_n}$.
We have
\begin{align*}
&amp;Q(s_1+2)\equiv R\pmod{M_n},\\
&amp;Q(s_1-2)\equiv R-4Q=D \pmod{M_n}.
\end{align*}
Thus,
 \[
(RQ|M_n)=(Q^2(s_1+2)|M_n)=(s_1+2|M_n)
\]
and
\[
(RD|M_n)=(Q^2(s_1+2)(s_1-2)|M_n)=(s_1+2|M_n)(s_1-2|M_n).
\]
We have $(RQ|M_n)=-1$ and $(RD|M_n)=-1$ if and only if
\[
(s_1+2|M_n)=-1\quad \mbox{and} \quad (s_1-2|M_n)=1,
\]
and therefore by Theorem \ref{generalLLT}, Assertion \ref{assertionA} holds for $M_n$.
\end{proof}

If $N$ is any positive integer, we can write $N=m^2k$, where $m^2$ is the largest possible square divisor of $N$.  \
We call the positive squarefree integer $k$ here the {\it squarefree part} of $N$ and denote this by
\[
k=\sfp(N).
\]
For example, $1=\sfp(9)$, $5=\sfp(20)$, $3=\sfp(27)$, $10=\sfp(360)$.  We now define for a given positive integer $N$
\begin{equation*}
N^*= \begin{cases}
\sfp(N), &amp; \text{when $2\nmid \sfp(N)$;}\\
\sfp(N)/2, &amp; \text{when $2\mid \sfp(N)$.}\\
\end{cases}
\end{equation*}
Note that  $N^*$ is positive, odd and if $\gcd(N,M_n)=1$, then 
\[
(N|M_n)=(N^*|M_n).
\]


We next put $s_1^+=(s_1+2)^*$ and $s_1^-=(s_1-2)^*$ and observe that 
$\gcd(s_1^-,s_1^+)=1$.  We can write
\begin{equation}\label{s1negplus}
2^\alpha m_1^2s_1^-=s_1-2, \quad 2^\beta m_2^2 s_1^+=s_1+2,
\end{equation}
where $m_1$, $m_2$ are odd integers and $\alpha$, $\beta\geq 0$.
Also, if $s_1$ is an acceptable seed for $M_n$, then by Theorem \ref{acceptableseed}
\[
-1=(s_1+2|M_n)=(2^\beta m_2^2 s_1^+|M_n)=(s_1^+m_2^2|M_n).
\]
Thus, if $s_1^+=1$, we cannot have $s_1$ as an acceptable seed for $M_n$.
For example, $s_1=7$, $14$, $16$, $23,\ldots$ cannot be acceptable seeds for $M_n$ with odd $n\geq 3$.

We next prove a theorem which is suggested by the results in Section \ref{IS}.

\begin{theorem}\label{infinitemore}
Let $s_1$ be any acceptable seed for $M_n$ and define $s_1^-$, $s_1^+$ as above.
There exists an infinitude of distinct acceptable seeds $S_1$ for $M_n$ such that
$S_1^-=s_1^-$, $S_1^+=s_1^+$.
\end{theorem}
\begin{proof}
We let $\alpha$, $\beta$, $s_1^-$, $s_1^+$ be defined as in (\ref{s1negplus}).
If we put $a=2^\alpha s_1^-$ and $b=2^\beta s_1^+$, then $(m_1, m_2)$ is a solution of
\begin{equation}\label{m1m2abc}
ax^2-by^2=c,
\end{equation}
where $c=-4$.  Since we must have $s_1^+&gt;1$, we must have $ab&gt;1$ and $ab$ cannot be a
perfect integral square.  By Theorem \ref{PellTheorem}, we know there must exist an infinitude 
of distinct solutions to (\ref{m1m2abc}).  If we let $(X,Y)$ be one of these, we can put
\[
S_1=2^\alpha s_1^- X^2+2=2^\beta s_1^+Y^2-2.
\]
Hence,
\[
S_1^-=(S_1-2)^*=s_1^- \quad \mbox{and} \quad S_1^+=(S_1+2)^*=s_1^+.
\]
\end{proof}





Let 
\[
s_1^+=(s_1+2)^*=\prod_{i=1}^j q_i \quad \mbox{and}\quad s_1^-=(s_1-2)^*=\prod_{i=1}^k r_i,
\]
where $r_i$ ($i=1,2,\ldots,k$), $q_i$ ($i=1,2,\ldots,j$) are distinct odd primes.  For any odd prime $p$ we define
$\eta_n(p)$ to be the value of the Legendre symbol $(2^n-1|p)=(M_n|p)$.  By quadratic reciprocity we have
\begin{align*}
(s_1^+|M_n)&amp;=(-1)^{\frac{s_1^+-1}{2}}(M_n|s_1^+)\\
                    &amp;=(-1)^{\frac{s_1^+-1}{2}}\prod_{i=1}^j\eta_n(q_i)
\end{align*}
and
\[
(s_1^-|M_n)=(-1)^{\frac{s_1^--1}{2}}
                    \prod_{i=1}^k\eta_n(r_i).
\]

It follows from Theorem \ref{acceptableseed} that $s_1$ is an acceptable seed for $M_n$ if and only if 
\begin{equation}
s_1^-=1\quad \mbox{or} \quad \prod_{i=1}^k\eta_n(r_i)=(-1)^{\frac{s_1^--1}{2}}\label{etar}\quad \mbox{when $s_1^-&gt;1$}
\quad \mbox{and} \quad
\prod_{i=1}^j\eta_n(q_i)=(-1)^{\frac{s_1^++1}{2}}.
\end{equation}

For any odd integer $m$, we define $\omega(m)$ to be the order of $2$ modulo $m$;
that is, $\omega(m)$ is the least positive integer $k$ such that $2^k\equiv \modd{1} {m}$.
Such a value of $k$ always exists and $k\mid \phi(m)$, where $\phi$ is the the Euler totient function.
It is easy to establish the following proposition.
\begin{proposition}
If $m_1$ and $m_2$ are odd integers such that $\gcd(m_1,m_2)=1$, then
\[
\omega(m_1m_2)=\lcm(\omega(m_1)\omega(m_2)).
\]
\end{proposition}

We next observe that 
\[
\eta_{n}(p)=\eta_m(p)
\]
if $n\equiv \modd{m} {\omega(p)}$.
For any positive integer $m$, we define
\[
\Omega(m)= \begin{cases}
2\omega(m), &amp; \text{when $2\nmid \omega(m)$;}\\
\omega(m), &amp; \text{otherwise}.\\
\end{cases}
\]
Put $\Omega=\Omega(s_1^-s_1^+)$ and
\[
\mathcal{A}(s_1)=\{\modd{n} {\Omega}: \mbox{$n$ odd, $1\leq n&lt; \Omega$, (\ref{etar})  holds}\}.
\]
It is not difficult to see that $s_1$ is an acceptable seed for $M_n$ if and only if
$n\equiv\modd{m} {\Omega}$ for some $m\in \mathcal{A}(s_1)$.  For example,
consider $s_1=37$; we have $s_1^-=35$ and $s_1^+=39$.
Also, $\omega(5)=4$, $\omega(7)=3$, $\omega(3)=2$ and $\omega(13)=12$.
Hence, $\omega(s_1^-)=\omega(s_2^+)=12$ and $\Omega=12$.  
We consider the odd values of $\modd{n} {12}$ and find that 
(\ref{etar}) 
holds only for $n\equiv \modd{7} {12}$.
Thus, $\mathcal{A}(37)=\{7\}$. It follows that 37 is an acceptable seed for $M_n$ if and only if 
$n\equiv \modd{7} {12}$.  During the process of testing whether 
(\ref{etar})
holds, we found that $\eta_3(7)=(2^3-1|7)=0$ and
$\eta_9(7)=(2^9-1|7)=0$.  
Indeed, we have the following proposition.
\begin{proposition}
Let $p$ be either $r_i$ ($i=1,2,\ldots,k$) or $q_i$  ($i=1,2,\ldots,j$).  If $\eta_n(p)=0$, then $\gcd(n,\Omega)&gt;1$. 
\end{proposition}
\begin{proof}
We have $\omega(p)\mid \Omega$ and $p\mid 2^n-1$.  Since $\omega(p)&gt;1$ and $\omega(p)\mid n$,
we have $\gcd(n,\Omega)&gt;1$.
\end{proof}
\noindent
As we shall see below, the converse of this result is not true.

We also observe that if $d=\gcd(n,\Omega)$, then $2^d-1\mid 2^n-1$ and $M_n$ cannot be a prime.
Thus, before testing (\ref{etar}),
we could eliminate all values of $n$ from 
$\mathcal{A}(s_1)$ when $\gcd(n,\Omega)\not=1$.
We can now define
\[
\mathcal{A}^{\prime}(s_1)=\{\modd{n} {\Omega}: \mbox{$\gcd(n,\Omega)=1$, $1\leq n&lt; \Omega$, (\ref{etar}) holds}\}.
\]
We see that $\mathcal{A}^{\prime}(s_1)\subseteq \mathcal{A}(s_1)$, but 
$\mathcal{A}^{\prime}(s_1)\not= \mathcal{A}(s_1)$ for infinitely many values of $s_1$.  For example,
if $s_1=9$, then $s_1^-=7$, $s_1^+=11$ and $\Omega=30$.  Thus,
\[
\mathcal{A}(s_1)=\{5,11,29\}, \quad\mbox{but} \quad \mathcal{A}^{\prime}(s_1)=\{11,29\}.
\]

Suppose $2\mid m$.  In what follows we will denote by $\mathcal{T}(m)$, resp.~$\mathcal{R}(m)$, the set of all odd
residues modulo $m$, resp.~the set of reduced residues modulo $m$.  That is,
\begin{align*}
&amp;\mathcal{T}(m)=\{\modd{n} {m}: \mbox{$n$ odd, $1\leq n&lt;m$}\},\\
&amp;\mathcal{R}(m)=\{\modd{n} {m}: \mbox{$\gcd(n,m)=1$, $1\leq n&lt;m$}\}.
\end{align*}
Since $m$ is even, we have $\#\mathcal{T}(m)=m/2$ and $\#\mathcal{R}(m)=\phi(m)$. Hence
\begin{equation}\label{sizesAT}
\#A(s_1)\leq \#\mathcal{T}(\Omega)=\Omega/2
\end{equation}
and
\begin{equation}\label{sizesAprimeT}
\#A^{\prime}(s_1)\leq \#\mathcal{R}(\Omega)=\phi(\Omega).
\end{equation}
If we define $l(s_1)=\#A(s_1)$, resp.~$l^{\prime}=\#A^{\prime}(s_1)$ and put
\[
q=q(s_1)=\frac{2l(s_1)}{\Omega}, \quad q^{\prime}=q^{\prime}(s_1)=\frac{l^{\prime}(s_1)}{\phi(\Omega)}
\]
we see by (\ref{sizesAT}) and (\ref{sizesAprimeT}) that $q$, $q^{\prime}\leq 1$.  
Furthermore, $q$ represents the proportion of all possible values of $n$ ($n\geq 3$, $n$ odd) which
can be acceptable seeds for $M_n$.  Thus, if $q(s_1)=1$, then $s_1$ is a universal seed for $M_n$.
Since $\mathcal{R}(\Omega)\subseteq\mathcal{T}(\Omega)$, we see that if $q(s_1)=1$, then $q^{\prime}(s_1)=1$.
It follows that if $q^{\prime}(s_1)\not=1$, then $q(s_1)\not=1$, an observation we will make use of in Section \ref{TCQ}.



\section{Some computational results}\label{SCR}

We begin with some examples.
If $s_1=4$, then $s_1^-=1$ and $s_1^+=3$.  We get $\Omega=2$,
 $\mathcal{A}(4)=\{1\}$, $l=1$ and $q=q(4)=1$.
If $s_1=3$, then $s_1^-=1$ and $s_1^+=5$.  We get $\Omega=4$,
 $\mathcal{A}(3)=\{3\}$, $l=1$ and $q=q(3)=1/2$.
That is, $3$ is an acceptable seed for $M_n$ for $1/2$ of all the possible 
values of $n$ ($n\equiv \modd{3} {4}$).
If $s_1=22$, then $s_1^-=5$, $s_1^+=3$, $\Omega=4$, $\mathcal{A}(22)=\{1\}$;
that is, $22$ is an acceptable seed for $1/2$ of all the possible $M_n$ 
(in this case those for which $n\equiv \modd{1} {4}$).
If  $s_1=20$, then $s_1^-=1$ and $s_1^+=11$.  We get $\Omega=10$,
 $\mathcal{A}(20)=\{1, 5, 9\}$, $l=3$ and $q=q(20)=3/5$.
Thus, $s_1=20$ is an acceptable seed for $M_n$ for $3/5$ of all the possible values of $n$.  
This is not as good as the case of 
 $s_1=4$, but is better than $s_1=3$.

The above process was programmed on a computer, and the following tables were produced.


\begin{longtable}[H]
{|c|c|c|c|c|c|c|}\hline
$s_1$ &amp; $s_1^-$ &amp;$s_1^+$ &amp; $\Omega$ &amp; $\mathcal{A}(s_1)$ &amp; $l$ &amp; $q$ \\\hline
$3$ &amp; $1$ &amp;$5$ &amp; $4$ &amp; $\{3\}$ &amp; $1$ &amp; $1/2$ \\\hline
$4$ &amp; $1$ &amp;$3$ &amp; $2$ &amp; $\{ 1\}$ &amp; $1$ &amp; $1$ \\\hline
$9$ &amp; $7$ &amp;$11$ &amp; $30$ &amp; $\{ 5, 11, 29\}$ &amp; $3$ &amp; $1/5$ \\\hline
$10$ &amp; $1$ &amp;$3$ &amp; $2$ &amp; $\{1 \}$ &amp; $1$ &amp; $1$ \\\hline
$11$ &amp; $1$ &amp;$13$ &amp; $12$ &amp; $\{ 3, 5, 11\}$ &amp; $3$ &amp; $1/2$ \\\hline
$12$ &amp; $5$ &amp;$7$ &amp; $12$ &amp; $\{1 \}$ &amp; $1$ &amp; $1/6$ \\\hline
$13$ &amp; $11$ &amp;$15$ &amp; $20$ &amp; $\{13, 17 \}$ &amp; $2$ &amp; $1/5$ \\\hline
$15$ &amp; $13$ &amp;$17$ &amp; $24$ &amp; $\{13, 19, 21 \}$ &amp; $3$ &amp; $1/4$ \\\hline
$17$ &amp; $15$ &amp;$19$ &amp; $36$ &amp; $\{3, 15, 19, 27, 35 \}$ &amp; $5$ &amp; $5/18$ \\\hline
$18$ &amp; $1$ &amp;$5$ &amp; $4$ &amp; $\{3 \}$ &amp; $1$ &amp; $1/2$ \\\hline
$19$ &amp; $17$ &amp;$21$ &amp; $24$ &amp; $\{17, 23 \}$ &amp; $2$ &amp; $1/6$ \\\hline
$20$ &amp; $1$ &amp;$11$ &amp; $10$ &amp; $\{1, 5, 9 \}$ &amp; $3$ &amp; $3/5$ \\\hline
$21$ &amp; $19$ &amp;$23$ &amp; $198$ &amp; $\{5, 7, 13, 23, 29, 41, 49, 67, 79, 85, 95, $ &amp; $20$ &amp; $20/99$ \\\hline
$$ &amp; $$ &amp;$$ &amp; $$ &amp; $ 101, 115, 133, 137, 139, 151, 155, 167, 173\}$ &amp; $$ &amp; $$ \\\hline
$22$ &amp; $5$ &amp;$3$ &amp; $4$ &amp; $\{1 \}$ &amp; $1$ &amp; $1/2$ \\\hline
$24$ &amp; $11$ &amp;$13$ &amp; $60$ &amp; $\{3, 17, 23, 27, 47, 53 \}$ &amp; $6$ &amp; $1/5$ \\\hline
$25$ &amp; $23$ &amp;$3$ &amp; $22$ &amp; $\{3, 9, 15, 17, 21 \}$ &amp; $5$ &amp; $5/11$ \\\hline
$27$ &amp; $1$ &amp;$29$ &amp; $28$ &amp; $\{5, 7, 9, 11, 15, 25, 27 \}$ &amp; $7$ &amp; $1/2$ \\\hline
$28$ &amp; $13$ &amp;$15$ &amp; $12$ &amp; $\{ 1,9\}$ &amp; $2$ &amp; $1/3$ \\\hline
$31$ &amp; $29$ &amp;$33$ &amp; $140$ &amp; $\{3, 13, 17, 23, 47, 57, 73, 77, 87, 97, 103, 107, 113, 133 \}$ &amp; $14$ &amp; $1/5$ \\\hline
$32$ &amp; $15$ &amp;$17$ &amp; $8$ &amp; $\{3 \}$ &amp; $1$ &amp; $1/4$ \\\hline
$33$ &amp; $31$ &amp;$35$ &amp; $60$ &amp; $\{37, 47, 49, 59 \}$ &amp; $4$ &amp; $2/15$ \\\hline
$35$ &amp; $33$ &amp;$37$ &amp; $180$ &amp; $\{5, 15, 25, 29, 35, 41, 49, 51, 61, 65, 69, 71, 85, 89, 99, $ &amp; $27$ &amp; $3/10$ \\\hline
$$ &amp; $$ &amp;$$ &amp; $$ &amp; $ 101, 105, 121, 125, 135, 141, 149, 159, 161, 169, 171, 179 \}$ &amp; $$ &amp; $$ \\\hline
$36$ &amp; $17$ &amp;$19$ &amp; $72$ &amp; $\{1, 9, 15, 17, 33, 39, 55, 57, 63, 71 \}$ &amp; $10$ &amp; $5/18$ \\\hline
$37$ &amp; $35$ &amp;$39$ &amp; $12$ &amp; $\{ 7\}$ &amp; $1$ &amp; $1/6$ \\\hline
$38$ &amp; $1$ &amp;$5$ &amp; $4$ &amp; $\{3 \}$ &amp; $1$ &amp; $1/2$ \\\hline
$39$ &amp; $37$ &amp;$41$ &amp; $180$ &amp; $\{3, 9, 11, 23, 31, 37, 43, 57, 83, 91, 103, 109, 111, $ &amp; $18$ &amp; $1/5$ \\\hline
$$ &amp; $$ &amp;$$ &amp; $$ &amp; $117, 129, 131, 151, 163 \}$ &amp; $$ &amp; $$ \\\hline
$40$ &amp; $19$ &amp;$21$ &amp; $18$ &amp; $\{ 5, 11\}$ &amp; $2$ &amp; $2/9$ \\\hline
$41$ &amp; $39$ &amp;$43$ &amp; $84$ &amp; $\{5, 15, 23, 27, 29, 35, 41, 47, 51, 63, 65, 71, 75, 77, 83 \}$ &amp; $15$ &amp; $5/14$ \\\hline
$42$ &amp; $5$ &amp;$11$ &amp; $20$ &amp; $\{ 1, 5, 9\}$ &amp; $3$ &amp; $3/10$ \\\hline
$43$ &amp; $41$ &amp;$5$ &amp; $20$ &amp; $\{ 7, 15, 19\}$ &amp; $3$ &amp; $3/10$ \\\hline
$44$ &amp; $21$ &amp;$23$ &amp; $66$ &amp; $\{1, 7, 13, 19, 49 \}$ &amp; $5$ &amp; $5/33$ \\\hline
$45$ &amp; $43$ &amp;$47$ &amp; $322$ &amp; $\{ 3, 25, 39, 81, 87, 95, 129, 157, 171, 179, 185, 199, $ &amp; $22$ &amp; $22/161$ \\\hline
 &amp;  &amp; &amp;  &amp; $213, 221, 249,  255, 263, 269, 277, 291, 305, 311 \}$ &amp;  &amp;  \\\hline
$46$ &amp; $11$ &amp;$3$ &amp; $10$ &amp; $\{3, 7 \}$ &amp; $2$ &amp; $2/5$ \\\hline
$49$ &amp; $47$ &amp;$51$ &amp; $184$ &amp; $\{7, 9, 17, 31, 55, 57, 63, 73, 89, 97, 103, 105, $ &amp; $22$ &amp; $11/46$ \\\hline
$$ &amp; $$ &amp;$$ &amp; $$ &amp; $113, 119, 135, 137, 143, 145, 151, 159, 169, 183 \}$ &amp; $$ &amp; $$ \\\hline
$51$ &amp; $1$ &amp;$53$ &amp; $52$ &amp; $\{5, 7, 9, 11, 17, 23, 27, 31, 33, 37, 39, 49, 51 \}$ &amp; $13$ &amp; $1/2$ \\\hline
$52$ &amp; $1$ &amp;$3$ &amp; $2$ &amp; $\{1 \}$ &amp; $1$ &amp; $1$ \\\hline
$53$ &amp; $51$ &amp;$55$ &amp; $40$ &amp; $\{ 3, 5, 21, 27, 29\}$ &amp; $5$ &amp; $1/4$ \\\hline
$54$ &amp; $13$ &amp;$7$ &amp; $12$ &amp; $\{1, 7 \}$ &amp; $2$ &amp; $1/3$ \\\hline
$55$ &amp; $53$ &amp;$57$ &amp; $468$ &amp; $\{ 13, 25, 29, 41, 43, 47, 65, 67, 77, 95, 97, 119, 133, 139,  $ &amp; $52$ &amp; $2/9$ \\\hline
 &amp;&amp; &amp;  &amp;                                        $149, 151, 157, 169, 175, 185, 191, 203, 209, 211, 221, $ &amp;  &amp;  \\\hline
 &amp;  &amp; &amp;  &amp;                                      $223, 227, 229, 263, 275, 281, 295, 301, 313, 331, 337, $ &amp; &amp;  \\\hline
 &amp; &amp; &amp;  &amp;                                        $347, 353, 355, 365, 367, 383, 385, 389, 407, 409, 419,  $ &amp; &amp;  \\\hline
 &amp; &amp; &amp;  &amp; $437, 445, 457, 461, 463\}$ &amp; &amp;\\\hline
 \caption{Values for $s_1$ up to $55$.}
\label{Setsupto55}
\end{longtable}



\begin{longtable}[H]
{|c|c|c|c|c|c||c|c|c|c|c|c|}\hline
$s_1$ &amp; $s_1^-$ &amp;$s_1^+$ &amp; $\Omega$  &amp; $l$ &amp;       $q$ &amp;$s_1$ &amp; $s_1^-$ &amp;$s_1^+$ &amp; $\Omega$  &amp; $l$ &amp; $q$ \\\hline
$3$ &amp; $1$ &amp;$5$ &amp; $4$  &amp; $1$ &amp; $1/2$ &amp;                  $598$ &amp; $149$&amp; $3$ &amp; $148$ &amp; $37$ &amp; $1/2$         \\\hline
$4$ &amp; $1$ &amp;$3$ &amp; $2$  &amp; $1$ &amp; $1$ &amp;               $650$    &amp; $1$         &amp; $163$             &amp; $162$            &amp; $41$        &amp; $41/81$        \\\hline
$10$ &amp; $1$ &amp;$3$ &amp; $2$  &amp; $1$ &amp; $1$&amp;               $673$ &amp; $671$&amp; $3$ &amp; $60$ &amp; $15$ &amp; $1/2$         \\\hline
$11$ &amp; $1$ &amp;$13$ &amp; $12$  &amp; $3$ &amp; $1/2$&amp;             $678$ &amp; $1$&amp; $85$ &amp; $8$ &amp; $2$ &amp; $1/2$           \\\hline
$18$ &amp; $1$ &amp;$5$ &amp; $4$ &amp; $1$ &amp; $1/2$&amp;                $724$ &amp; $1$&amp; $3$ &amp; $2$ &amp; $1$ &amp; $1$           \\\hline
$20$ &amp; $1$ &amp;$11$ &amp; $10$  &amp; $3$ &amp; $3/5$&amp;          $731$ &amp; $1$&amp; $733$ &amp; $244$ &amp; $61$ &amp; $1/2$       \\\hline
$22$ &amp; $5$ &amp;$3$ &amp; $4$  &amp; $1$ &amp; $1/2$&amp;               $786$ &amp; $1$&amp; $197$ &amp; $196$ &amp; $49$ &amp; $1/2$       \\\hline
$27$ &amp; $1$ &amp;$29$ &amp; $28$ &amp; $7$ &amp; $1/2$&amp;        $843$ &amp; $1$&amp; $5$ &amp; $4$ &amp; $1$ &amp; $1/2$     \\\hline
$38$ &amp; $1$ &amp;$5$ &amp; $4$  &amp; $1$ &amp; $1/2$&amp;          $862$ &amp; $215$&amp; $3$ &amp; $28$ &amp; $7$ &amp; $1/2$ \\\hline
$51$ &amp; $1$ &amp;$53$ &amp; $52$  &amp; $13$ &amp; $1/2$&amp;      $884$    &amp;    $1$  &amp;    $443$   &amp; $442$    &amp; $111$    &amp;$111/221$    \\\hline
$52$ &amp; $1$ &amp;$3$ &amp; $2$  &amp; $1$ &amp; $1$&amp;               $963$ &amp; $1$&amp; $965$ &amp; $96$ &amp; $24$ &amp; $1/2$ \\\hline
$66$ &amp; $1$&amp; $17$ &amp; $8$ &amp; $2$ &amp; $1/2$ &amp;        $970$ &amp; $1$&amp; $3$ &amp; $2$ &amp; $1$ &amp; $1$   \\\hline
  $74$ &amp; $1$&amp; $19$ &amp; $18$ &amp; $5$ &amp; $5/9$ &amp;        $1012$ &amp; $505$&amp; $3$ &amp; $100$ &amp; $30$ &amp; $3/5$    \\\hline
  $83$ &amp; $1$&amp; $85$ &amp; $8$ &amp; $2$ &amp; $1/2$ &amp;             $1026$ &amp; $1$&amp; $257$ &amp; $16$ &amp; $4$ &amp; $1/2$         \\\hline
  $100$ &amp; $1$&amp; $51$ &amp; $8$ &amp; $2$ &amp; $1/2$ &amp;        $1060$ &amp; $1$&amp; $59$ &amp; $58$ &amp; $15$ &amp; $15/29$   \\\hline
 $102$ &amp; $1$&amp; $13$ &amp; $12$ &amp; $3$ &amp; $1/2$ &amp;        $1081$ &amp; $1079$&amp; $3$ &amp; $492$ &amp; $123$ &amp; $1/2$  \\\hline
 $106$ &amp; $13$&amp; $3$ &amp; $12$ &amp; $3$ &amp; $1/2$ &amp;           $1091$ &amp; $1$&amp; $1093$ &amp; $364$ &amp; $91$ &amp; $1/2$ \\\hline
 $123$ &amp; $1$&amp; $5$ &amp; $4$ &amp; $1$ &amp; $1/2$ &amp;                $1174$ &amp; $293$&amp; $3$ &amp; $292$ &amp; $73$ &amp; $1/2$    \\\hline
  $145$ &amp; $143$&amp; $3$ &amp; $60$ &amp; $15$ &amp; $1/2$ &amp;             $1227$ &amp; $1$&amp; $1229$ &amp; $1228$ &amp; $307$ &amp; $1/2$     \\\hline
  $146$ &amp; $1$&amp; $37$ &amp; $36$ &amp; $9$ &amp; $1/2$ &amp;       $1252$ &amp; $1$&amp; $627$ &amp; $90$ &amp; $23$ &amp; $23/45$          \\\hline
  $164$ &amp; $1$&amp; $83$ &amp; $82$ &amp; $21$ &amp; $21/41$ &amp;      $1298$ &amp; $1$&amp; $13$ &amp; $12$ &amp; $3$ &amp; $1/2$         \\\hline
$171$ &amp; $1$&amp; $173$ &amp; $172$ &amp; $43$ &amp; $1/2$ &amp;      $1348$ &amp; $673$&amp; $3$ &amp; $48$ &amp; $14$ &amp; $7/12$  \\\hline
$202$ &amp; $1$&amp; $51$ &amp; $8$ &amp; $2$ &amp; $1/2$ &amp;        $1354$ &amp; $1$&amp; $339$ &amp; $28$ &amp; $10$ &amp; $5/7$    \\\hline
$214$ &amp; $53$&amp; $3$ &amp; $52$ &amp; $13$ &amp; $1/2$ &amp;         $1371$ &amp; $1$&amp; $1373$ &amp; $1372$ &amp; $343$ &amp; $1/2$  \\\hline
$227$ &amp; $1$&amp; $229$ &amp; $76$ &amp; $19$ &amp; $1/2$ &amp;     $1446$ &amp; $1$&amp; $181$ &amp; $180$ &amp; $45$ &amp; $1/2$  \\\hline
 $244$ &amp; $1$&amp; $123$ &amp; $20$ &amp; $6$ &amp; $3/5$ &amp;          $1450$ &amp; $181$&amp; $3$ &amp; $180$ &amp; $45$ &amp; $1/2$            \\\hline
$291$ &amp; $1$&amp; $293$ &amp; $292$ &amp; $73$ &amp;  $1/2$ &amp;     $1460$ &amp; $1$&amp; $731$ &amp; $56$ &amp; $14$ &amp; $1/2$    \\\hline
 $292$ &amp; $145$&amp; $3$ &amp; $28$ &amp; $8$ &amp;  $4/7$ &amp;           $1523$ &amp; $1$&amp; $61$ &amp; $60$ &amp; $15$ &amp; $1/2$  \\\hline
$298$ &amp; $37$&amp; $3$ &amp; $36$ &amp; $9$ &amp; $1/2$ &amp;               $1602$ &amp; $1$&amp; $401$ &amp; $200$ &amp; $50$ &amp; $1/2$      \\\hline
$340$ &amp; $1$&amp; $19$ &amp; $18$ &amp; $5$ &amp;  $5/9$ &amp;         $1684$ &amp; $1$&amp; $843$ &amp; $70$ &amp; $19$ &amp; $19/35$  \\\hline
$382$ &amp; $95$&amp; $3$ &amp; $36$ &amp; $9$ &amp; $1/2$ &amp;          $1732$ &amp; $865$&amp; $3$ &amp; $172$ &amp; $44$ &amp; $22/43$ \\\hline
$394$ &amp; $1$&amp; $11$ &amp; $10$ &amp; $3$ &amp; $3/5$ &amp;      $1766$ &amp; $1$&amp; $221$ &amp; $24$ &amp; $6$ &amp; $1/2$      \\\hline
  $402$ &amp; $1$&amp; $101$ &amp; $100$ &amp; $25$ &amp; $1/2$ &amp;           $1802$ &amp; $1$&amp; $451$ &amp; $20$ &amp; $6$ &amp; $3/5$             \\\hline
$452$ &amp; $1$&amp; $227$ &amp; $226$ &amp; $57$ &amp; $57/113$ &amp;           $1851$ &amp; $1$&amp; $1853$ &amp; $72$ &amp; $18$ &amp; $1/2$         \\\hline
$484$ &amp; $241$&amp; $3$ &amp; $24$ &amp; $6$ &amp; $1/2$ &amp;           $1924$ &amp; $1$&amp; $107$ &amp; $106$ &amp; $27$ &amp; $27/53$\\\hline
$486$ &amp; $1$&amp; $61$ &amp; $60$ &amp; $15$ &amp; $1/2$ &amp;              $1938$ &amp; $1$&amp; $485$ &amp; $48$ &amp; $12$ &amp; $1/2$        \\\hline
   $531$ &amp; $1$&amp; $533$ &amp; $60$ &amp; $15$ &amp; $1/2$&amp;              $1942$ &amp; $485$&amp; $3$ &amp; $48$ &amp; $12$ &amp; $1/2$      \\\hline
    $580$ &amp; $1$&amp; $291$ &amp; $48$ &amp; $14$ &amp; $7/12$    &amp;               &amp;                &amp;        &amp;         &amp;        &amp; \\\hline

 
\caption{Values for $s_1$ up to $2000$, where $q\geq 1/2$.}
\label{qbetween1and2}
\end{longtable}



\begin{longtable}[H]
{|c|c|c|c|c|c||c|c|c|c|c|c|}\hline
$s_1$ &amp; $s_1^-$ &amp;$s_1^+$ &amp; $\Omega$  &amp; $l$ &amp; $q$ &amp;$s_1$ &amp; $s_1^-$ &amp;$s_1^+$ &amp; $\Omega$  &amp; $l$ &amp; $q$ \\\hline
$4$ &amp; $1$&amp; $3$ &amp; $2$ &amp; $1$ &amp; $1$ &amp;                   $1252$ &amp; $1$&amp; $627$ &amp; $90$ &amp; $23$ &amp; $23/45$  \\\hline
$10$ &amp; $1$&amp; $3$ &amp; $2$ &amp; $1$ &amp; $1$ &amp;                    $1348$ &amp; $673$&amp; $3$ &amp; $48$ &amp; $14$ &amp; $7/12$  \\\hline
$20$ &amp; $1$&amp; $11$ &amp; $10$ &amp; $3$ &amp; $3/5$ &amp;             $1354$ &amp; $1$&amp; $339$ &amp; $28$ &amp; $10$ &amp; $5/7$ \\\hline
$52$ &amp; $1$&amp; $3$ &amp; $2$ &amp; $1$ &amp; $1$ &amp;               $1684$ &amp; $1$&amp; $843$ &amp; $70$ &amp; $19$ &amp; $19/35$ \\\hline
$74$ &amp; $1$&amp; $19$ &amp; $18$ &amp; $5$ &amp; $5/9$ &amp;             $1732$ &amp; $865$&amp; $3$ &amp; $172$ &amp; $44$ &amp; $22/43$   \\\hline
$164$ &amp; $1$&amp; $83$ &amp; $82$ &amp; $21$ &amp; $21/41$ &amp;     $1802$ &amp; $1$&amp; $451$ &amp; $20$ &amp; $6$ &amp; $3/5$  \\\hline        
$244$ &amp; $1$&amp; $123$ &amp; $20$ &amp; $6$ &amp; $3/5$ &amp;             $1924$ &amp; $1$&amp; $107$ &amp; $106$ &amp; $27$ &amp; $27/53$ \\\hline    
$292$ &amp; $145$&amp; $3$ &amp; $28$ &amp; $8$ &amp; $4/7$ &amp;               $2180$ &amp; $1$&amp; $1091$ &amp; $1090$ &amp; $273$ &amp; $273/545$   \\\hline
$340$ &amp; $1$&amp; $19$ &amp; $18$ &amp; $5$ &amp;  $5/9$ &amp;                        $2644$ &amp; $1321$&amp; $3$ &amp; $60$ &amp; $22$ &amp; $11/15$  \\\hline
$394$ &amp; $1$&amp; $11$ &amp; $10$ &amp; $3$ &amp; $3/5$ &amp;                       $2698$ &amp; $336$&amp; $3$ &amp; $42$ &amp; $12$ &amp; $4/7$ \\\hline
$452$ &amp; $1$&amp; $227$ &amp; $226$ &amp; $57$ &amp; $57/113$ &amp;                 $3044$ &amp; $1$&amp; $1523$ &amp; $1522$ &amp; $381$ &amp; $381/761$  \\\hline
  $580$ &amp; $1$&amp; $291$ &amp; $48$ &amp; $14$ &amp; $7/12$            &amp;   $3202$     &amp; $1$    &amp;  $89$      &amp; $22$       &amp; $8$  &amp; $8/11$ \\\hline
  $650$ &amp; $1$  &amp; $163$   &amp; $162$    &amp; $41$ &amp; $41/81$  &amp;  $3466$   &amp;    $433$     &amp;     $3$  &amp;    $72$      &amp;   $22$      &amp; $11/18$ \\\hline
$724$ &amp; $1$&amp; $3$ &amp; $2$ &amp; $1$ &amp; $1$ &amp;                            $3530$   &amp;    $1$         &amp; $883$    &amp;   $882$      &amp; $221$ &amp;$221/441$ \\\hline        
  $884$  &amp;    $1$  &amp;    $443$  &amp; $442$ &amp; $111$  &amp;$111/221$&amp;   $3748$ &amp; $1873$&amp; $3$ &amp; $936$ &amp; $242$ &amp; $121/234$  \\\hline
$970$ &amp; $1$&amp; $3$ &amp; $2$ &amp; $1$ &amp; $1$ &amp;            $4052$   &amp;   $1$ &amp;    $2027$   &amp;    $2026$  &amp;   $507$  &amp;$507/1013$ \\\hline       
$1012$ &amp; $505$&amp; $3$ &amp; $100$ &amp; $30$ &amp; $3/5$ &amp;        $4420$ &amp; $1$&amp; $2211$ &amp; $330$ &amp; $83$ &amp; $83/165$ \\\hline
 $1060$ &amp; $1$&amp; $59$ &amp; $58$ &amp; $15$ &amp; $15/29$&amp;                $4610$ &amp; $1$&amp; $1153$ &amp; $288$ &amp; $76$ &amp; $19/36$ \\\hline
\caption{Values for $s_1$ up to $5000$, where $q&gt;1/2$.}
\label{qlessthan2}
\end{longtable}


On examining Table \ref{qlessthan2}, we see that seeds $s_1$ producing values of $q$ such that $1&gt;q&gt;1/2$ do occur, 
but are not frequent.  It is of some interest to consider the case of $s_1=3202$ where $\Omega=22$ and $q=8/11$.  Here we have
\[
\mathcal{A}(s_1)=\{3,5,7, 9, 13, 15, 17, 19\},
\]
which is the set of all possible integers $k$ in $\mathcal{T}(\Omega)$, except for $k=1$, $11$ and $k=21$ ($=22-1$).
Note that for any odd prime $p$ such that $\omega(p)\mid \Omega$
\begin{equation}\label{Legendreomega}
(2^{\Omega-n}-1|p)=(-2^n|p)(2^n-1|p)=(-2|p)(2^n-1|p).
\end{equation}
Since for $p\equiv\modd{1} {8}$, we have $(-2|p)=1$, we see that
\[
(2^{\Omega-n}-1|p)=(2^n-1|p)
\]
in this case.  Thus, putting $n=1$, we find that
\[
(2^{21}-1|89)=(2^1-1|89)=1\not=(-1)^{\frac{89+1}{2}}.
\]
Also, $89\mid 2^{11}-1$.
Hence, {\it a priori} we could not have $1$, $11$, $21\in \mathcal{A}(s_1)$ when $s_1=3202$, but all other possibilities are in $\mathcal{A}(s_1)$,
a most remarkable situation.  It would be of some interest to find some $s_1$ for which $q\not=1$ and $q&gt;8/11$.

\section{The case of $q=1$}\label{TCQ}

A short inspection of Table \ref{qbetween1and2} reveals that the only values of $s_1$ with $s_1\leq 2000$ for which $q=1$ 
are those found by Lehmer.  It is an interesting problem to determine whether there can exist values of $s_1$ not 
predicted by Lehmer for which $q=q(s_1)=1$.  That is, do there exist values for $s_1$ such that 
$s_1^-\not=1$ or $s_1^+\not=3$ for which $q(s_1)=1$?

To proceed any further, we need two simple lemmas.
\begin{lemma}\label{simplelemmaone}
If $s_1^-\equiv \modd{-1} {4}$, then (\ref{etar}) cannot hold for all $n\in \mathcal{R}(\Omega)$.
\end{lemma}
\begin{proof}
We note that $1\in \mathcal{R}(\Omega)$ and 
\[
\eta_1(r_i)=(2^1-1|r_i)=(1|r_i)=1.
\]
Thus,
\[
\prod_{i=1}^{k}\eta_1(r_i)=1\not=(-1)^{\frac{s_1^--1}{2}},
\]
and therefore (\ref{etar}) cannot hold for $n=1$.
\end{proof}

\begin{lemma}\label{simplelemmatwo}
If $s_1^+\equiv \modd{1} {4}$, then (\ref{etar}) cannot hold for all $n\in \mathcal{R}(\Omega)$.
\end{lemma}
\begin{proof}
Similar to that of the previous result.
\end{proof}

It is now a simple matter to prove the following theorem.

\begin{theorem}\label{q=1result}
If $q^{\prime}(s_1)=1$, then $s_1^-\equiv \modd{1} {4}$ and $s_1^+\equiv \modd{-1} {4}$.
\end{theorem}

\begin{corollary}\label{odds1}
If $s_1$ is odd, then $q^{\prime}(s_1)\not=1$.
\end{corollary}
\begin{proof}
From (\ref{s1negplus}), we have
\[
2^\alpha m_1^2 s_1^-=s_1-2, \quad 2^\beta m_2^2 s_1^+=s_1+2,
\]
where $m_1$ and $m_2$ are odd integers and integers $\alpha$, $\beta\geq 0$. 
Since $s_1$ here is odd, we have $\alpha=\beta=0$, 
both $m_1$ and $m_2$ are odd 
and $(m_1m_2)^2\equiv \modd{1} {4}$.  Also,
\[
m_1^2m_2^2s_1^-s_2^+=s_1^2-4
\]
means that 
\[
s_1^-s_2^+\equiv 1\pmod{4} \quad \mbox{or}\quad s_1^-\equiv s_1^+\pmod{4}.
\]
\end{proof}




In order for $q^{\prime}(s_1)=1$, we need 
\begin{equation}\label{snegplusmod4}
s_1^-\equiv 1 \pmod{4} \quad \mbox{and} \quad s_1^+\equiv -1 \pmod{4}
\end{equation}
If (\ref{snegplusmod4}) holds, then (\ref{etar}),  becomes
\begin{equation}\label{newetar}
\prod_{i=1}^k \eta_n(r_i)=1
\quad \mbox{and}\quad
\prod_{i=1}^j \eta_n(q_i)=1,
\end{equation}
respectively.  Since $\Omega-1\in \mathcal{R}(\Omega)$ and by (\ref{Legendreomega})
\[
\eta_{\Omega-n}(p)=(-2|p)\eta_n(p),
\]
we see that if $k=1$, then $s_1^-=r_1\equiv \modd{1} {4}$ and if $r_1\equiv \modd{5} {8}$, then
\[
\eta_{\Omega-1}(r_1)=-1.
\]
Thus, (\ref{newetar}) cannot hold for $n=\Omega-1$ when $s_1^-$ is a prime and $s_1^-\equiv \modd{5} {8}$.
Similarly (\ref{newetar}) cannot hold for $n=\Omega-1$ when $s_1^+$ is a prime and $s_1^+\equiv \modd{7} {8}$.
In either case $q_1(s_1)\not=1$.

More generally, suppose that $p$ is any odd prime and observe that $2^n-1$ is never a perfect integral square when $n&gt;1$.
Since for any given $a$ not a perfect square, the probability that $(a|p)=1$ is about $1/2$, we expect that for any $n\geq 3$
the likelihood that $\eta_n(p)=\epsilon$, where $\epsilon \in \{1,-1\}$ to be about $1/2$.
Suppose that $p_1$, $p_2, \ldots, p_m$ are $m$ distinct odd primes. 
Put $\omega=\omega(p_1p_2\cdots p_m)$ and let each of $\epsilon_1$, $\epsilon_2,\ldots, \epsilon_m$ have fixed values 
selected from $\{1,-1\}$. 
For a particular $n\in \mathcal{R}(\omega)$, the probability that 
\[
\eta_n(p_i)=\epsilon_i \quad \mbox{($i=1,2,\ldots,m$)}
\]
is about $1/2^m$.  Also, since there are 2 possible values for $\eta_n(p_i)$ ($\not=0$) 
for $i=1$, $2,\ldots, m-1$ and only one possible value for $\eta_n(p_m)$ such that 
\begin{equation}\label{eta1prob}
\prod_{i=1}^m \eta_{p_i}(n)=1,
\end{equation}
there can be only $2^{m-1}$ possible sets of values of $(\epsilon_1, \epsilon_2,\ldots,\epsilon_m)$
such that (\ref{eta1prob}) holds.
Thus, we anticipate that the probability that (\ref{eta1prob}) is true for any fixed $n\in \mathcal{R}(\omega)$ 
should be about
\[
2^{m-1}\times \frac{1}{2^m}=\frac{1}{2}.
\]
Under the not unreasonable assumption that for distinct $n_1$, $n_2\in \mathcal{R}(\omega)$, the events
\[
\prod_{i=1}^m \eta_{n_1}(p_i)=1\quad \mbox{and}\quad \prod_{i=1}^m \eta_{n_2}(p_i)=1
\]
are independent, we would predict the probability that 
\[
\prod_{i=1}^k \eta_{n}(r_i)=1 \quad \mbox{for all}\quad n\in \mathcal{R}(s_1^-)
\]
is $\approx (1/2)^{\phi(\Omega_1)}$ and the probability that 
\[
\prod_{i=1}^j \eta_{n}(q_i)=1 \quad \mbox{for all}\quad n\in \mathcal{R}(s_1^+)
\]
is $\approx (1/2)^{\phi(\Omega_2)}$, where 
 $\Omega_1=\Omega(s_1^-)$, $\Omega_2=\Omega(s_1^+)$.
 These probabilities will generally tend to be rather small.

The above reasoning suggests that for most values of $s_1$ there should be values of $n\in\mathcal{R}(\Omega)$ such that either statement in 
(\ref{newetar}) is false.  We programmed a computer to search for all even $s_1$ 
up to some bound $\mathcal{B}$ for which $q^{\prime}(s_1)\not=1$.  
The computer examined each value of $s_1$ satisfying (\ref{snegplusmod4}) and 
eliminated this candidate as soon as it found some value of $n\in\mathcal{R}(\Omega_1)$ such that 
 (\ref{newetar}) is false
 or some value of $n\in \mathcal{R}(\Omega_2)$ such that 
 (\ref{newetar}) is false.  We ran the program up to $\mathcal{B}=10^8$ and found only the following where the second statement in 
values of $s_1$ for which $q^{\prime}(s_1)$ (or $q(s_1)$) could be $1$:
\[
s_1=4, 10, 52, 724, 970, 10084, 95050, 140452, 1956244, 9313930, 27246964.
\]
Indeed, as we have seen in Section \ref{IS}, Lehmer had shown $q(s_1)=1$ for all of these values of $s_1$. Thus,
 it might appear that if $q(s_1)=1$, then $s_1^-=1$ and $s_1^+=3$.
 By the results at the end of Section \ref{URS},
this would certainly be the case if the only integers $a$ such that the Jacobi symbol $(a|M_n)$ has a fixed value
for all odd $n\geq 3$ must be in $\mathcal{S}$, but we have seen in Section \ref{RS} that this is not true;
thus, it might well be that Lehmer did not find all the values for $s_1$ for which $q(s_1)=1$.
Of course, if we allow $s_1$ to be rational, 
then as shown in Section \ref{URS},
there are vastly more possibilities.




\section{Conclusion}\label{Con}
We have seen that there exists an infinitude of rational universal seeds $s_1$; 
in fact, we can produce simple parametric formulas for such seeds.
We have also produced a simple process which will tabulate those values of $n$ for which a given integral 
$s_1$ is acceptable for $M_n$.  Furthermore, we found those integers $s_1$ ($s_1\leq 5000$) such that $q(s_1)$ satisfies
$1/2&lt;q(s_1)&lt;1$.

Lehmer found an infinite set of integral $s_1$ for which $q(s_1)=1$.  These all have the same property that $s_1^-=1$
and $s_1^+=3$.  A computer search for all integral $s_1$ up to to $10^8$ for which $q(s_1)=1$ did not produce any other $s_1$;
however, it is possible that there might exist some $s_1$ which Lehmer did not find 
such that $q(s_1)=1$.  This needs further investigation. 

We close by mentioning the following interesting result, which can be easily deduced from Theorem 3 
of Berrizbeitia and Berry \cite{BB03}.
\begin{theorem}
If $n$ is odd, $n\geq 3$ and $s_1$ is the rational seed $6/5$, then $M_n$ is prime if and only if
\begin{align*}
&amp;s_{n-1}\equiv 0\pmod{M_n}\quad \mbox{when} \quad n\equiv 3\pmod{4}\\
&amp;s_{n-2}\equiv 0\pmod{M_n}\quad \mbox{when} \quad  n\equiv 1\pmod{4}.
\end{align*}
\end{theorem}
The proof of the first part of this result can be easily deduced by putting $R=4$ and $Q=5$ above, but 
the second part requires the properties of the biquadratic residue symbol.
It would be of some interest to investigate the existence of similar theorems with different $s_1$  (and $n$).


\section{Acknowledgments}
We are very grateful to an anonymous referee, whose careful reading and several suggestions 
have resulted in a much more presentable version of this paper from that originally submitted.  
We also wish to thank Jeffrey Shallit for bringing Jansen's interesting thesis \cite{Jan12} to our attention.

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\bibitem{BB03}
Pedro Berrizbeitia and T.~G.~Berry, Biquadratic reciprocity and a Lucasian primality test, {\it Math. Comput.} {\bf 73} (2003), 1559--1564.

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David A.~Cox, {\it Primes of the Form $x^2+ny^2$}, Wiley-Interscience, 1989.


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Leonard Eugene Dickson, {\it History of the Theory of Numbers},
Carnegie Institute of Washington, Volume {\bf II}, G.~E.~Stechert \&amp; Co., 1934.


\bibitem{JW09}
Michael J.~Jacobson,~Jr.~and Hugh C.~Williams, {\it Solving the Pell Equation},
CMS Books in Mathematics, Springer, New York, NY, 2009.


\bibitem{Jan12}
B.~J.~H.~Jansen, {\it Mersenne primes and class field theory}, PhD thesis, 
 Mathematical Institute, Faculty of Science, Leiden University, 2012.
\newblock Available at \url{https://hdl.handle.net/1887/20310}.

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\end{thebibliography}


\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B37; Secondary 11Y11, 11B50.

\noindent \emph{Keywords: } 
linear recurrence, Lucas function, primality testing.

\bigskip
\hrule
\bigskip


\noindent (Concerned with sequences
\seqnum{A001566},
\seqnum{A003010},
\seqnum{A018844}, and
\seqnum{A135927}.)

\bigskip
\hrule
\bigskip


\vspace*{+.1in}
\noindent
Received July 25 2024;
revised version received  September 27 2024; March 22 2025.
Published in {\it Journal of Integer Sequences}, March 23 2025.

\bigskip
\hrule
\bigskip

\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{https://cs.uwaterloo.ca/journals/JIS/}.
\vskip .1in


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