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\begin{center}
\uppercase{\bf Lots and Lots of Perrin-Type Primality Tests and Their Pseudo-Primes}
\vskip 20pt
{\bf Robert Dougherty-Bliss}\\
{\smallit Department of Mathematics, Rutgers University (New Brunswick)}\\
{\tt robert.w.bliss@gmail.com}\\
\vskip 10pt
{\bf Doron Zeilberger}\\
{\smallit Department of Mathematics, Rutgers University (New Brunswick)}\\
{\tt DoronZeil@gmail.com}\\
\end{center}
\vskip 20pt
\centerline{\smallit Received: , Revised: , Accepted: , Published: } % We will fill in the dates
\vskip 30pt

\centerline{\bf Abstract}
\noindent
We use {\it Experimental Mathematics} and {\it Symbolic Computation} (with
Maple), to search for lots and lots of Perrin- and Lucas- style primality
tests, and try to sort the wheat from the chaff. More impressively, we find
quite a few such primality tests for which we can explicitly construct infinite
families of pseudo-primes, rather, like in the cases of Perrin pseudo-primes
and the famous Carmichael primes, proving the mere existence of infinitely many
of them.

\pagestyle{myheadings}
\markright{\smalltt INTEGERS: 23 (2023)\hfill}
\thispagestyle{empty}
\baselineskip=12.875pt
\vskip 30pt

\section{Preface: How it all Started thanks to Vince Vatter}

It all started when we came across Vince Vatter's delightful article [14], where he gave a cute combinatorial proof, inspired by COVID, and social distancing,
of the following fact that goes back to Raoul Perrin [8] (See also [9], [10], [13], [15]).

{\bf Perrin's Observation}:  Let the integer sequence $A(n)$ defined by
\begin{align*}
    A(1) &= 0 \quad A(2) = 2 \quad A(3) = 3 \\
    A(n)&=A(n-2)+A(n-3) \qquad (\text{for $n > 3$}).
\end{align*}
Then, for every prime $p$, we have $p \mid A(p)$.

Perrin, back in 1889, was wondering whether the condition is {\it sufficient}, i.e. whether there are any {\it pseudo-primes}, i.e. {\it composite} $n$ such that
$A(n)/n$ is an integer. He could not find any, and as late as 1981, none was found $\leq 140000$ (see [1]). In 1982, Adams and Shanks [1] {\it rather quickly}
found the smallest  Perrin pseudo prime, $271441$, followed by the next-smallest, $904631$, and then they found quite a few other ones. 
Jon Grantham [4] proved that there are {\it infinitely many} Perrin pseudo-primes, and finding as many as possible of them, became a computational challenge, see Holger's paper [6].

Another, older, primality test is that based on the Lucas numbers ([11],
[12]).

\subsection*{Vince Vatter's Combinatorial Proof}

Vatter first found a {\it combinatorial interpretation} of the Perrin numbers, as the number of circular words of length $n$ in the alphabet $\{0,1\}$, that {\it avoid} the 
{\it consecutive subwords} (aka {\it factors} in formal language lingo), $\{000, 11\}$.

More formally: the number of words, $w=w_1, \dots, w_n$, in the alphabet $\{0,1\}$, such that for $1 \leq i \leq n-2$, $w_iw_{i+1}w_{i+2} \neq 000$, and also
$w_{n-1}w_n w_1 \neq 000$ and $w_{n}w_1 w_2 \neq 000$ as well as
for $1 \leq i \leq n-1$, $w_iw_{i+1} \neq 11$, and $w_nw_1 \neq 11$.

Then he argued that if $p$ is a prime, all the $p$ circular shifts are {\it different}, since otherwise there would be a non-trivial period, that can't happen since $p$ is prime. 
Since the constant words $0^p$ and $1^p$ obviously can't avoid both $00$ and $111$, Perrin's theorem follows.

This proof is reminiscent of Solomon Golomb's [3] snappy combinatorial proof of Fermat's little theorem [G] that argued that there are
$a^p-a$ non-monochromatic straight necklaces with $p$ beads of $a$ colors, and for each such necklace, the $p$ rotations are all different (see also [16 ], p. 560).

When we saw Vatter's proof we got all excited. Vatter's argument transforms {\it verbatim} to counting circular words in {\it any} (finite) alphabet, and any (finite) set of
forbidden (consecutive) patterns! More than twenty years ago one of us (DZ) wrote a paper, in collaboration with his then PhD student, Anne Edlin [2], that {\it automatically}
finds the (rational) generating function in any such scenario, hence this is a
cheap way to manufacture lots and lots of Perrin-style primality tests. We
already had a Maple package,
\href{https://sites.math.rutgers.edu/~zeilberg/tokhniot/CGJ}{CGJ}, to handle
it, so all that remained was to {\it experiment} with many alphabets and many
sets of forbidden patterns, and search for those that have only few small {\it
pseudo-primes}.

This inspired us to write our first Maple package,
\href{https://sites.math.rutgers.edu/~zeilberg/tokhniot/PerrinVV.txt}{PerrinVV.txt}.
See the front of the present article \href{
https://sites.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/perrin.html}{front
of the present article}
for many such primality tests, inspired by sets of forbidden patterns, along
with all the pseudoprimes less than a million.


\section{An Even Better Way to Manufacture Perrin-style Primality Tests}

After the initial excitement we got an {\it epiphany}, and as it turned out, it was already made, in 1990, by Stanley Gurak [5]. Take {\it any}  polynomial
$Q(x)$ with {\it integer coefficients}, and constant term $1$, and write it as
$$
Q(x)=1 \,- \, e_1\,x \, +\, e_2x^2 - \dots +(-1)^k\, e_k x^k \quad.
$$
Factor it over the complex numbers 
$$
Q(x) \,= \, (1- \alpha_1 x )  (1- \alpha_2 x )  \cdots (1- \alpha_k x )  \quad .
$$
Note that $e_1,e_2, \dots$ are the {\it elementary symmetric functions} in $\alpha_1, \dots, \alpha_k$.

Defining
$$
a(n):= \alpha_1^n \, + \, \alpha_2^n \, + \, \dots \, + \alpha_k^n \quad,
$$
it follows thanks to Newton's identities ([7]) that $\{a(n)\}$ is an {\it integer sequence}. The generating function
$$
\sum_{n=0}^{\infty} a(n)\,x^n \, = \, \frac{1}{1-\alpha_1x} \, + \, \frac{1}{1-\alpha_2x} \, + \, \dots \, + \,   \frac{1}{1-\alpha_kx} \quad ,
$$
has denominator $Q(x)$ and {\it some} numerator, let's call it $P(x)$, with {\it integer coefficients}, that Maple can easily find {\it all by itself}.

So we can {\it define} an integer sequence $\{a(n)\}$ in terms of the rational function $P(x)/Q(x)$, where $Q(x)$ is {\it any} polynomial with constant term $1$, and
$P(x)$ {\it comes out} as above:

$$
\sum_{n=0}^{\infty} a(n) x^n \, = \, \frac{P(x)}{Q(x)} \quad .
$$


We claim that each such integer sequence engenders a {\it Perrin-style}
primality test, namely
$$
a(p) \equiv e_1 (mod \,\, p).
$$
To see this, note that
$$
(\alpha_1 + \cdots + \alpha_k)^p = a(p) \,+\, p\,A(p),
$$
where
$$
A(p) = \sum_{
{{i_1 + i_2 + \cdots + i_k = p} \atop {i_1, i_2, \dots i_k < p}}
} \frac{(p-1)!}{i_1! \cdots i_k!} \alpha_1^{i_1} \cdots \alpha_k^{i_k} 
$$
is a symmetric polynomial in the $\alpha_i$. The fundamental theorem of
symmetric functions [7] implies that $A(p)$ is an integer. 
Fermat's little theorem then gives
$$
a(p) \equiv (\alpha_1 + \cdots + \alpha_k)^p = e_1^p \equiv e_1 (mod \,\, p).
$$

So this is an even easier way to manufacture lots and lots of Perrin-style primality tests, and we can let the computer search for those
that have as few small pseudo-primes as possible.

This is implemented in the
\href{https://sites.math.rutgers.edu/~zeilberg/tokhniot/Perrin.txt}{Maple
package {\tt Perrin.txt}}. Again, see the
\href{https://sites.math.rutgers.edu/~zeilberg/mamarim/mamarimhtml/perrin.html}{front of the present article}
for many such primality tests, inspired by this more general method (first suggested by Stanley Gurak [5]).

Searching for such primality tests with as few pseudo-primes less than a
million, we stumbled on the following example.

{\bf The DB-Z Primality Test}

Let 
$$
\sum_{n=0}^{\infty} a(n)\, x^n \, := \, \frac{3 x^{4}+5 x^{2}+6 x -7}{4 x^{7}+x^{4}+x^{2}+x -1} \quad,
$$
or equivalently, the integer sequence defined by the initial values $1, 3, 4,
11, 16, 30, 78$, and
\begin{equation*}
    a(n) = a(n-1)+a(n-2) +  a(n-4) +  4a(n-7)
\end{equation*}
for $n > 7$. Then $a(p) \equiv 1 \pmod{p}$ for all primes $p$.

Manuel Kauers kindly informed us that the seven smallest DB-Z pseudo-primes are
as follows.

\begin{itemize}
\item $1531398 = 2 \cdot 3 \cdot 11 \cdot 23203$ 
    
\item  $114009582 = 2 \cdot 3 \cdot 17 \cdot 1117741$ 
    
\item  $940084647 = 3 \cdot 47 \cdot 643 \cdot 10369$ 
    
\item  $4206644978 = 2 \cdot 97 \cdot 859 \cdot 25243$ 
    
\item  $7962908038 = 2 \cdot 191 \cdot 709 \cdot 29401$ 

\item $20293639091 = 11 \cdot  3547 \cdot 520123$ 

\item $41947594698= 2\cdot 3 \cdot 19 \cdot 523 \cdot 703559 $ 
\end{itemize}

(This was a computational challenge posed by us to Manuel Kauers, and we
offered to donate $100$ dollars to the OEIS in his honor. The donation was
made.)

After the first version of this paper was written, with the help of Manuel
Kauers, we discovered an even better primality test.

{\bf The DB-Kauers primality test}


Let 
$$
\sum_{n=0}^{\infty} a(n)\, x^n \, := \, \frac{9 x^{5}+16 x^{4}+10 x -6}{3 x^{6}+9 x^{5}+8 x^{4}+2 x -1}
\quad,
$$
or equivalently, the integer sequence defined by initial conditions $2, 4, 8,
48, 157, 382$ and
\begin{equation*}
    a(n)= 2a(n-1)+  8 a(n-4) +  9a(n-5) + a(n-6)
\end{equation*}
for $n > 6$. Then $a(p) \equiv 2 \pmod{p}$ for all primes $p$.

The smallest pseudoprime is $2,260,550,373 = 3 \cdot 103 \cdot 107 \cdot
68371$.

\section{Perrin-Style Primality Tests with Explicit Infinite Families of Pseudo-Primes}

We are particularly proud of the next primality test, featuring  the {\it
Companion Pell numbers} \url{https://oeis.org/A002203}. These numbers have been
studied extensively, but as far as we know using them as a {\it primality test}
is new. It is not a very good one, but the novelty is that it has an {\it
explicit} {\it doubly-infinite} set of pseudo-primes.

{\bf The Companion Pell Numbers Primality Test}
Let 
$$
\sum_{n=0}^{\infty} a(n)\, x^n \, := \, \frac{2x - 2}{x^{2}+2 x -1},
$$
or equivalently, let $a(n)$ be the integer sequence defined by
\begin{align*}
    a(1) &= 2 \quad a(2) = 6 \\
    a(n) &= 2 a(n - 1) + a(n - 2).
\end{align*}
Then $a(p) \equiv 2 \pmod{p}$ for all primes $p$.

\begin{theorem}
    The doubly-infinite family
    $$
    \{\,2^i \cdot 3^j \, | \, i \geq 3 \, ,  \, j \geq 0 \,\} \quad,
    $$
    are Companion-Pell pseudoprimes. In other words,
    $$
    \frac{a(2^i \cdot 3^j)-2}{2^i 3^j} \quad,
    $$
    is an integer if $i \geq 3$ and $j \geq 0$.
\end{theorem}

\begin{proof}
    We will proceed by a kind of double-induction where we show that if the
    conclusion holds for $n$, it also holds for $2n$ and $3n$, provided that
    $n$ is of the form stated in the theorem.

    Let
    $$
    \alpha_1:=1+\sqrt{2} \quad, \quad \alpha_2:=1-\sqrt{2}
    $$
    be the roots of the characteristic equation for $a(n)$. It is routine to check
    that $a(n) = \alpha_1^n + \alpha_2^n$. Since $\alpha_1 \alpha_2=-1$, we have
    the following recurrences:
    \begin{align}
        a(2n) &= a(n)^2 +2 (-1)^{n+1} \label{2n} \\
        a(3n) &= a(n)^3 +3 (-1)^{n+1} a(n) \label{3n} .
    \end{align}
    Now define the sequence $b(n)=a(n)-2$. Our goal is to show that $b(n)$ is
    divisible by $n$.

    If we substitute $b(n)$ into \eqref{2n} for even $n$, then we obtain
    \begin{equation*}
        b(2n) = b(n)(b(n) + 4).
    \end{equation*}
    This gives
    \begin{equation*}
        \frac{b(2n)}{2n} = \frac{b(n)}{n} \frac{b(n) + 4}{2}.
    \end{equation*}
    If an even $n$ divides $b(n)$, then both factors are integers, and
    therefore $2n$ divides $b(2n)$.

    Substituting $b(n)$ into \eqref{3n} when $n$ is even yields
    $$
    b(3n) =b(n)(b(n)^2 + 6b(n) + 12).
    $$
    Therefore
    $$
    \frac{b(3n)}{3n}=\frac{b(n)}{n}\frac{b(n)^2 + 6b(n) + 12}{3}.
    $$
    It is easy to prove that $b(n)$ is divisible by $3$ when $n$ is divisible
    by $6$. Therefore, if an $n$ divisible by $6$ divides $b(n)$, then $3n$
    divides $b(3n)$.

    Since both $b(8) / 8$ and $b(24)/24$ are integers, the theorem follows by
    induction.
\end{proof}


We now state without proofs (except for Theorem 4, where we give a sketch) a few other primality tests that have explicit infinite families of  pseudoprimes.

\begin{theorem}
    Let
    $$
    \sum_{n=0}^{\infty} a(n)\, x^n \, := \, \frac{x - 2}{2 x^{2}+x -1},
    $$
    or equivalently, let $a(n)$ be the sequence defined by
    \begin{align*}
        a(1)&=1 \quad a(2)=5 \\
        a(n)&=a(n-1) + 2a(n-2).
    \end{align*}
    Then $a(p) \equiv 1 \pmod{p}$ if $p$ is prime. Furthermore, $\{2^i \,|  i
    \geq 2\}$ are all pseudo-primes. That is, $a(2^i) \equiv 2^i \pmod{2^i}$ if
    $i \geq 2$.
\end{theorem}


\begin{theorem}
Let
$$
\sum_{n=0}^{\infty} a(n)\, x^n \, := \, \frac{2x - 2}{2 x^{2}+2 x -1},
$$
or equivalently, let $a(n)$ be defined by
\begin{align*}
    a(1)&=2 \quad a(2)=8 \\
    a(n)&= 2a(n-1) +  2a(n-2).
\end{align*}
Then $a(p) \equiv 2 \pmod{p}$ if $p$ is prime. Furthermore, the following
infinite families are all pseudo-primes:
\begin{align*}
    &\{3^i \mid i \geq 2\} \quad \{2\cdot 3^i \mid i \geq 1\} \quad \{11\cdot 81^i \mid i \geq 1\} \\
    &\{23\cdot 3^{5i} \mid i \geq 1\} \quad \{29\cdot 3^{4+12i} \mid i \geq 0\} \quad \{31\cdot 3^{16i} \mid i \geq 1\}.
\end{align*}
\end{theorem}

\begin{theorem}
Let
$$
\sum_{n=0}^{\infty} a(n)\, x^n \, := \, \frac{2 x^{2}+3}{2 x^{3}+2 x^{2}+1} \quad,
$$
or equivalently, let $a(n)$ be defined by
\begin{align*}
    a(1)&=0 \quad a(2)=-4 \quad a(3)=-6 \\
    a(n) &= -2(a(n-2) + a(n-3)).
\end{align*}
Then $a(p) \equiv 0 \pmod{p}$ if $p$ is prime. Furthermore, the following
infinite families are all pseudo-primes:
\begin{equation*}
    \{2^i \mid  i \geq 2\} \quad \{3\cdot2^{4i} \mid  i \geq 2\} \quad \{11\cdot2^{18i} \mid  i \geq 2\} \quad \{13\cdot2^{17+20i} \mid  i \geq 2\}.
\end{equation*}
\end{theorem}

\begin{proof}[Sketch of Proof]
We use the {\it C-finite ansatz} [17]. It follows from the C-finite ansatz that
$$
b(n)=a(2n)-a(n)^2
$$
satisfies {\it some} recurrence. It turns out to be
\begin{equation*}
    b(n) = 2b(n - 1) + 4b(n - 3).
\end{equation*}
We now define
$$
c(n):=\frac{b(n)}{2^{\lfloor n/2 \rfloor}},
$$
and once again the C-finite ansatz implies that $c(n)$ satisfies some
recurrence. (The even and odd terms are independently C-finite, and the
interlacing of C-finite sequences is C-finite.) It turns out to be
$$
c(n) = 2c(n-2)+ 4 c(n-4)+ 2 c(n-6)
$$
with initial conditions $-4, -4, -20, -24, -56, -76$. Note that $c(n)$ are
manifestly {\it integers}. Going back to $a(n)$ we have the recurrence
$$
a(2n)=a(n)^2+ 2^{\lfloor n/2 \rfloor} c(n) \quad,
$$
and it follows by induction that $a(2^i)/2^i$ are all integers. A similar argument goes for the other infinite families claimed.
\end{proof}

\begin{theorem}
Let
$$
\sum_{n=0}^{\infty} a(n)\, x^n \, := \, \frac{2 x^{2}+2 x -3}{x^{3}+2 x^{2}+x -1},
$$
or equivalently, let $a(n)$ be defined by
\begin{align*}
    a(1)&=1 \quad a(2)=5 \quad a(3)=10 \\
    a(n)&= a(n-1) + 2a(n-2) + a(n-3).
\end{align*}
Then $a(p) \equiv 1 \pmod{p}$ for all primes $p$. Furthermore, the following
infinite families are all pseudo-primes:
\begin{equation*}
    \{3^i \mid  i \geq 2\}  \quad \{5 \cdot 3^{6+10\,i} \mid  i \geq 0\} \quad \{5 \cdot 3^{8+10\,i} \mid  i \geq 0\}  \quad \{7 \cdot 3^{4+6\,i} \mid  i \geq 0\}.
\end{equation*}
\end{theorem}

We found nine other such primality tests with infinite explicit families of
presodoprimes. These can be viewed by typing \texttt{PDB(x);} in the Maple
package \texttt{Perrin.txt}.

For fast computations and explorations using $C$ programs, readers are welcome
to explore the authors' \href{https://github.com/rwbogl/pt}{GitHub repository}.

{\bf Acknowledgment}:  Many thanks to Manuel Kauers for his computational prowess, and to the referee for a helpful remark.

\begin{thebibliography}{1}
\footnotesize
\bibitem[1]{1}
W. Adams and D. Shanks, Strong primality tests that are not sufficient, {\it Mathematics of Computation} {\bf 39} (1982), 255-300.

\bibitem[2]{2}  A. E. Edlin and D. Zeilberger,  The Goulden-Jackson Cluster method For cyclic  Words, {\it Advances in Applied Mathematics} {\bf 25}(2000), 228-232.  \hfill\break
{\tt https://sites.math.rutgers.edu/\~{}zeilberg/mamarim/mamarimhtml/cgj.html} \quad .

\bibitem[3]{3}  S. W. Golomb, Combinatorial Proof of Fermat's ``Little'' Theorem, {\it The American Mathematical Monthly}, {\bf 63}(10),  (Dec., 1956), 718. \hfill\break
{\tt https://sites.math.rutgers.edu/\~{}zeilberg/akherim/golomb56.pdf} \quad .

\bibitem[4]{4}  J. Grantham,  There are infinitely many Perrin pseudoprimes, {\it Journal of Number Theory} {\bf 130} (2010), 1117-1128.

\bibitem[5]{5}  S. Gurak, Pseudoprimes for Higher-Order Linear Recurrence Sequences, {\it Mathematics of Computation}, {\bf 55} (1990), 783-813.


\bibitem[6]{6} S. Holger, Millions of Perrin pseudoprimes including a few giants, arXiv:2002.03756 [math.NA]

\bibitem[7]{7}  I. G. Macdonald, ``Symmetric Functions and Hall Polynomials'', Second Edition, Clarendon Press, Oxford, 1995.

\bibitem[8]{8}  R. Perrin,  Item 1484, {\it L'Interm\`ediare des math} {\bf 6} (1899), 76-77.

\bibitem[9]{9}  N. J. A. Sloane, The On-line Sequence of Integer Sequences, Sequence  A001608, \hfill\break
{\tt https://oeis.org/A001608} \quad. %Perrin numbers

\bibitem[10]{10}  N. J. A. Sloane, The On-line Sequence of Integer Sequences, Sequence  A013998, \hfill\break   
{\tt  https://oeis.org/A013998} 
%Perrin pseudo-primes

\bibitem[11]{11}  N. J. A. Sloane,  The On-line Sequence of Integer Sequences, Sequence  A005854, \hfill\break   
{\tt  https://oeis.org/A005845} \quad. %Bruckman-Lucas pseudo-primes

\bibitem[12]{12}  N. J. A. Sloane, The On-line Sequence of Integer Sequences, Sequence  A000032, \hfill\break   
{\tt   https://oeis.org/A000032} \quad. %Lucas numbers

\bibitem[13]{13}  I. Stewart, Tales of a Neglected Number, {\it Mathematical Recreations, Scientific American} {\bf 274}(6) (1996), pp. 102-103.

\bibitem[14]{14}  V. Vatter, Social Distancing, Primes, and Perrin Numbers, , {\it Math Horizons}, {\bf 29}(1) (2022). \hfill\break
{\tt https://sites.math.rutgers.edu/\~{}zeilberg/akherim/vatter23.pdf} \quad .

\bibitem[15]{15}  Wikipedia,  Perrin Number, {\tt https://en.wikipedia.org/wiki/Perrin$\_$number} \quad .

\bibitem[16]{16}  D. Zeilberger,  Enumerative and Algebraic Combinatorics, in: {\it Princeton Companion to Mathematics} (edited by W.T. Gowers),
Princeton University Press, 2008, 550-561. \hfill\break
{\tt https://sites.math.rutgers.edu/\~{}zeilberg/mamarim/mamarimPDF/enu.pdf} \quad .

\bibitem[17]{17}  D. Zeilberger,  The C-finite ansatz, {\it Ramanujan Journal} {\bf 31} (2013), 23-32. \hfill\break
{\tt https://sites.math.rutgers.edu/\~{}zeilberg/mamarim/mamarimhtml/cfinite.html} \quad .

\end{thebibliography}

\end{document}
