Perfect 1-factorisations of $K_{16}$

TL;DR

This paper presents a comprehensive computer enumeration of perfect 1-factorisations of the complete graph K_{16}, revealing new automorphism groups, invariants, and implications for Latin squares and larger graph orders.

Contribution

It provides the first complete enumeration of P1Fs of K_{16}, introduces a new invariant for distinguishing them, and explores their properties related to Latin squares and larger orders.

Findings

Found 3155 P1Fs of K_{16}

89 P1Fs have non-trivial automorphism groups

No atomic Latin squares of order 15 from these P1Fs

Abstract

We report the results of a computer enumeration that found that there are 3155 perfect 1-factorisations (P1Fs) of the complete graph $K_{16}$. Of these, 89 have a non-trivial automorphism group (correcting an earlier claim of 88 by Meszka and Rosa). We also (i) describe a new invariant which distinguishes between the P1Fs of $K_{16}$, (ii) observe that the new P1Fs produce no atomic Latin squares of order 15 and (iii) record P1Fs for a number of large orders that exceed prime powers by one.

Equations

$$126,170,244,344,530,730,1332,1370,1850,2198,2810,3126,4490,6860,6890,11450,11882,12168,$$

$$15626,16808,22202,24390,24650,26570,29792,29930,32042,38810,44522,50654,51530,52442,63002,$$

$$72362,76730,78126,79508,103824,148878,161052,205380,226982,300764,357912,371294,493040,$$

$$571788,1030302,1092728,1225044,1295030,2048384,2248092,2476100,2685620,3307950,3442952,$$

$$4330748,4657464,5735340,6436344,6967872,7880600,9393932,11089568,11697084,13651920,$$

$$15813252,18191448,19902512,22665188.$$

$$F_1=\{ab,cd,ef,gh,ij,kl,mn,op\}and{F}_2=\{ap,bc,de,fg,hi,jk,lm,no\}.$$

$$\{$$

$$N_{15}=\left[\begin{array}[]{ccccccccccccccc}1&12&2&7&10&\pagecolor[gray]{.6}8&5&15&\pagecolor[gray]{.6}13&\pagecolor[gray]{.6}3&6&4&14&11&9\\ 12&2&13&3&8&11&9&6&15&14&4&7&5&1&10\\ 2&13&3&14&4&9&12&10&7&15&1&5&8&6&11\\ 7&3&14&4&1&5&10&13&11&8&15&2&6&9&12\\ 10&8&4&1&5&2&6&11&14&12&9&15&3&7&13\\ 8&11&9&5&2&6&3&7&12&1&13&10&15&4&14\\ 5&9&12&10&6&\pagecolor[gray]{.6}3&7&4&\pagecolor[gray]{.6}8&\pagecolor[gray]{.6}13&2&14&11&15&1\\ 15&6&10&13&11&7&4&8&5&9&14&3&1&12&2\\ 13&15&7&11&14&12&8&5&9&6&10&1&4&2&3\\ 3&14&15&8&12&1&13&9&6&10&7&11&2&5&4\\ 6&4&1&15&9&13&2&14&10&7&11&8&12&3&5\\ 4&7&5&2&15&10&14&3&1&11&8&12&9&13&6\\ 14&5&8&6&3&15&11&1&4&2&12&9&13&10&7\\ 11&1&6&9&7&4&15&12&2&5&3&13&10&14&8\\ 9&10&11&12&13&14&1&2&3&4&5&6&7&8&15\\ \end{array}\right].$$

Full text

We also (i) describe a new invariant which distinguishes between the P1Fs of $K_{16}$, (ii) observe that the new P1Fs produce no atomic Latin squares of order 15 and (iii) record P1Fs for a number of large orders that exceed prime powers by one.

1 Introduction

A $k$-factor in a graph is a $k$-regular spanning subgraph. In particular, a $1$-factor (also called a perfect matching) is a set of edges in a graph which cover every vertex exactly once. A $1$-factorisation is a collection of 1-factors which partitions the edges of the graph. Suppose that we have a 1-factorisation ${\cal F}$ of some graph. The union of any two distinct 1-factors in ${\cal F}$ is a 2-factor, that is, a collection of cycles. If, regardless of which two 1-factors we choose in ${\cal F}$, their union is a single cycle, then we say that ${\cal F}$ is perfect. Throughout, we will use the abbreviation P1F for perfect $1$-factorisation. For background reading on P1Fs, including definitions of terms not included here, we refer to [8, 9].

The P1Fs of complete graphs up to $K_{14}$ have been known for some time [3]. The main purpose of this note is to report on a computer enumeration of the next case, namely P1Fs of $K_{16}$. We also check the Latin squares associated with the new P1Fs (§3) and find they are not atomic, discuss invariants which distinguish non-isomorphic P1Fs (§4), and record a number of new P1Fs for large orders that are one more than a prime power (§5).

Kotzig’s P1F conjecture famously asserts that P1Fs of $K_{n}$ exist for all even $n$. Only three infinite families are known [1]. They cover all orders of the form $n=p+1$ or $n=2p$ where $p$ is an odd prime. Sporadic constructions including those reported in §5, together with [7, 8, 10, 13], demonstrate existence for $n\leqslant 56$ and the following orders:

[TABLE]

Kotzig’s P1F conjecture remains open for other orders. The smallest unsolved cases are

$\{64,66,70,76,78,88,92,96,100\}$.

2 The catalogue

We labelled the 16 vertices of our complete graph a,b,$\dots$,p, and assumed without loss of generality that every P1F contained the two factors

[TABLE]

Next, we generated all 1-factors of $K_{16}$ containing the edge ac. Up to isomorphism, this gave us 1647 options for the initial three 1-factors $F_{1},F_{2},F_{3}$. In each isomorphism class we chose $F_{3}$ to be the lexicographically least option. Then for each of these isomorphism representatives we did the following. First, we generated and stored the set ${\cal T}$ of all 1-factors that were compatible with $F_{1},F_{2},F_{3}$, in the sense that $F_{i}\cup F$ is a Hamilton cycle for all $F\in{\cal T}$ and $i\in\{1,2,3\}$. The number of 1-factors we needed to store at this stage depended on the choice of $F_{1},F_{2},F_{3}$ and ranged from 56 816 to 59 312. We also computed and stored which pairs of these 1-factors in ${\cal T}$ were compatible (made a Hamilton cycle). A list of “active” 1-factors was maintained, which initially consisted of ${\cal T}$.

We then used backtracking to add one 1-factor at a time from the active list. Each new factor was forced to include a particular edge that was judged to be the best option of the edges remaining to be used. The criterion was to choose the edge that was contained in the fewest active 1-factors, thereby limiting the branching that our search undertook. Having chosen a new 1-factor, we checked that no relabelling of the current partial 1-factorisation would give a lexicographically smaller set of 3 initial factors than the one currently being processed. Assuming that to be the case, we updated the list of active 1-factors by removing any which were incompatible with the new 1-factor, and continued the search. Any P1Fs that we found were canonically labelled (see §4), then output. The result was

Theorem 1**.**

There are $3155$ P1Fs of $K_{16}$ up to isomorphism. Of these, $89$ have a non-trivial automorphism group.

The 3155 P1Fs can be downloaded from [11]. Note that the same catalogue has been independently and concurrently generated by Mariusz Meszka [5]. The P1Fs with non-trivial automorphism groups all had cyclic automorphism groups, with generators as follows:

• •

A unique starter induced P1F, with automorphism group generated by a permutation with cycle type $15^{1}1^{1}$.

• •

A unique even-starter induced P1F, with automorphism group generated by a permutation with cycle type $14^{1}1^{2}$.

• •

4 P1Fs with automorphism group generated by a permutation with cycle type $7^{2}1^{2}$.

• •

5 P1Fs with automorphism group generated by a permutation with cycle type $5^{3}1^{1}$.

• •

19 P1Fs with automorphism group generated by a permutation with cycle type $3^{5}1^{1}$.

• •

59 P1Fs with automorphism group generated by a permutation with cycle type $2^{7}1^{2}$.

These numbers agree with an earlier enumeration by Meszka and Rosa [6], except that we found one extra P1F with automorphism group of order 7. That P1F can be obtained by developing the following two 1-factors under the permutation (abcdefg)(hijklmn):

[TABLE]

together with the 1-factor $\{$ah, bi, cj, dk, el, fm, gn, op$\}$.

3 Connections with Latin squares

In this section we explore whether the newly discovered P1Fs of $K_{16}$ produce interesting Latin squares. First we need to give some definitions.

A Latin square of order $n$ is an $n\times n$ matrix in which each row and column is a permutation of some (fixed) symbol set of size $n$, say $\{1,2,\dots,n\}$. It is often useful to think of a Latin square of order $n$ as a set of $n^{2}$ triples $(i,j,k)$ where $k$ is the symbol that appears in cell $(i,j)$. Each Latin square then has six conjugate squares obtained by (uniformly) permuting the coordinates of each triple. Writing Latin squares of order $n$ as sets of triples also provides for a natural action of $\mathcal{S}_{n}\wr\mathcal{S}_{3}$ on the Latin squares. The species (sometimes called main class) of a Latin square is its orbit under this action.

A Latin subrectangle is a rectangular submatrix of a Latin square $L$ in which the same symbols occur in each row. If $R$ is a $2\times\ell$ Latin subrectangle of $L$, and $R$ is minimal in that it contains no $2\times\ell^{\prime}$ Latin subrectangle for $2\leqslant\ell^{\prime}<\ell$, then we say that $R$ is a row cycle of length $\ell$. Column cycles and symbol cycles can be defined similarly, and the operations of conjugacy on $L$ interchange these objects.

A Latin square of order $n$ is row-Hamiltonian if every row cycle has length $n$, symbol-Hamiltonian if every symbol cycle has length $n$, and column-Hamiltonian if every column cycle has length $n$. These three types of square are related by conjugacy. A Latin square is atomic if it is row-Hamiltonian, symbol-Hamiltonian and column-Hamiltonian. In other words, a square is atomic if all its conjugates are row-Hamiltonian.

A $1$-factorisation ${\cal F}$ of $K_{n}$ is equivalent to a symmetric unipotent Latin square ${\cal U}({\cal F})$, where we define ${\cal U}({\cal F})[i,j]=k$ if the edge $ij$ appears in the $k$-th factor of ${\cal F}$, and ${\cal U}({\cal F})[i,i]=n$ for all $i$. From ${\cal U}({\cal F})$ we can form $n$ idempotent symmetric Latin squares of order $n-1$, by “folding” a row and the corresponding column onto the main diagonal. To be precise, we form ${\cal I}({\cal F},j)$ from ${\cal U}({\cal F})$ by replacing the entry in cell $(i,i)$ by ${\cal U}({\cal F})[i,j]$ for each $i$, and then deleting row $j$ and column $j$. The 1-factorisation ${\cal F}$ is perfect if and only if for each $j$ the corresponding Latin square ${\cal I}({\cal F},j)$ is symbol-Hamiltonian [12]. There is some interest in whether ${\cal I}({\cal F},j)$ has the stronger property of being atomic. Atomic Latin squares are known to exist for some composite orders, but no example is known of an atomic Latin square whose order is not a prime power [10]. This makes order 15, the smallest case where existence of an atomic Latin square is open, particularly interesting.

Each P1F ${\cal F}$ corresponds to $k$ species of Latin squares ${\cal I}({\cal F},j)$, where $k$ is the number of orbits of the automorphism group of ${\cal F}$. Hence P1Fs with automorphism groups generated by permutations of cycle type $15^{1}1^{1},14^{1}1^{2},7^{2}1^{2},5^{3}1^{1},3^{5}1^{1},2^{7}1^{2},1^{16}$, respectively produce $2,3,4,4,6,9,16$ species of Latin squares. Hence, overall our 3155 P1Fs of $K_{16}$ produce 49 742 species containing symmetric symbol-Hamiltonian Latin squares. Checking a representative of each species, we established:

Theorem 2**.**

There exists no symmetric atomic Latin square of order $15$.

The nearest that we got to an atomic Latin square was $N_{15}$, where

[TABLE]

This square is derived from the unique even-starter induced P1F of $K_{16}$. It is one of the two species of Latin squares derived from that P1F that inherit an automorphism group of order 14. Indeed, it is clear that $N_{15}$ has an automorphism applying the cycle $(1,2,\dots,14)$ simultaneously to rows, columns and symbols. It is also clear that $N_{15}$ is not atomic, because it has a row-cycle of length 3 (highlighted). However, of the $\binom{15}{2}=105$ pairs of rows in $N_{15}$, there are 91 that produce a Hamilton row-cycle. The only pairs which fail are the 14 images under the automorphism group of the pair of rows containing the highlighted row-cycle.

4 Invariants

The literature contains several invariants that can be used for distinguishing non-isomorphic P1Fs. The best known of these are the train and tri-colour vector [9]. In our discussion we will mention several complete invariants for P1Fs of $K_{16}$. That is, invariants which coincide for two P1Fs if and only if the P1Fs are isomorphic.

The train itself is a complete invariant for the P1Fs of $K_{16}$. However, the indegree sequence of the train is not; it partitions the 3155 P1Fs into 3104 equivalence classes. The P1Fs that do not have unique indegree sequences fall into 47 pairs and 2 triples. We now explicitly give the two triples. The format is similar to that used in $(\ref{e:newaut2})$, except that we compress factors by omitting spaces and punctuation, and use only a space to separate factors. The first triple shares the indegree sequence $[573,784,336,86,19,2]$:

abcdefghijklmnop acbedgfihljmkonp adblcheofngmikjp aebnckdjfoglhpim afbicpdmeghojlkn

agbdcnemfjhkiolp ahbpcodkelfmgijn aibmcedpflgkhnjo ajbgcidnehfklomp akbfcmdoepgnhjil

albhcgdfejinkpmo ambjcfdhekgoipln anbocjdleifhgpkm aobkcldienfpgjhm apbcdefghijklmno

abcdefghijklmnop acbedgfihkjnlpmo adbkcienfhgpjmlo aebncgdfhlikjomp afbhcjdmeoglipkn

agbfckdoejhpimln ahblcfdkemgoinjp aibmcpdnehfogkjl ajbdclepfmgihnko akbjcndielfpgmho

albgcodjeifkhmnp ambpchdlekfngjio anbocmdhegfjilkp aobicedpflgnhjkm apbcdefghijklmno

abcdefghijklmnop acbedgfihljnkpmo adbfchejgmioknlp aebjcpdlfmgohnik afblckdoeghjinmp

agbncmdkelfjhoip ahbpcndfeigjkmlo aibmcjdpenfhglko ajbkcgdiemfohpln akbocedhfngpiljm

albicfdmeognhkjp ambdclehfkgijonp anbhcodjepflgkim aobgcidnekfphmjl apbcdefghijklmno

Meanwhile, the second triple shares the indegree sequence $[584,765,338,94,18,1]$:

abcdefghijklmnop acbedgfihmjokpln adbicpeoflgnhjkm aebgcldfhpiojmkn afbhckdnejgoimlp

agbjcodkeifmhlnp ahbpcgdienfkjlmo aiblcndoepfjgmhk ajbnchdmegfpilko akbfcjdpemgihnlo

albocmdjehfngkip ambkcedhfoglinjp anbdcfelgjhoikmp aobmcidlekfhgpjn apbcdefghijklmno

abcdefghijklmnop acbedgfihljnkpmo adbmcjehfpgkioln aebhckdofngijlmp afbjcndleogmhpik

agbfcedmhkiljonp ahbncgdfeijpkmlo aibdcmejfhgoknlp ajbocidnepfkglhm akbpcodjegflhnim

albgcpdkemfohjin amblchdienfjgpko anbicldpekfmgjho aobkcfdhelgnipjm apbcdefghijklmno

abcdefghijklmnop acbedgfihkjnlpmo adblciemfhgojpkn aebjcfdngmhoilkp afbdcoeigkhpjmln

agbpcndkehflimjo ahbmcjdpenfkgilo aibkcldheofjgnmp ajbgcedmfphlinko akbhcmdjelfngpio

albockdiepfmgjhn ambfchdoejgliknp anbicpdlegfohjkm aobncgdfekhmipjl apbcdefghijklmno

Among the pairs of P1Fs whose train indegree sequences coincide, there is (only) one that involves two P1Fs whose automorphism groups have different order. Specifically the following rigid P1F has indegree sequence $[598,748,332,102,18,2]$, which coincides with one of the P1Fs with automorphism group of order 2 reported in [6]:

abcdefghijklmnop acbedgfjhlinkomp adbmcpeiflgohkjn aebdclfogihmjpkn afbjcndkelgmhpio

agblcjdhekfnipmo ahbpcodnejfkglim aibhcfdoemgnjlkp ajbgcedmfhiklonp akbicgdpeofmhjln

albocmdjepfigkhn ambkchdfengpiljo anbfckdieghojmlp aobncidlehfpgjkm apbcdefghijklmno

The train is a complete invariant for P1Fs of $K_{16}$, but comparing trains is an instance of digraph isomorphism, which is not known to be possible in polynomial time. That makes it desirable to identify features of the train that are easily computed (like the indegree sequence), but still have enough information to be a complete invariant (unlike the indegree sequence). One such property is the following. For each vertex $v$ of the train define $p(v)$ to be the length of the shortest directed path from $v$ to any vertex $w$ that is in a directed cycle. Since every vertex has outdegree 1, it is a simple matter to construct a path starting at $v$ and following the unique outgoing arc until we reach a vertex $w$ that we have previously seen on the path. The distance from $v$ to the first occurrence of $w$ is, by definition, $p(v)$. This also shows that $p(v)$ is well defined. Note that if $v$ itself is in a cycle then $p(v)=0$. Also, for a general $1$-factorisation the cycle involving $w$ may be a loop. However, for a P1F of $K_{n}$ (where $n>2$) this cannot happen, because loops only occur on vertices that are otherwise isolated. To see this, consider two vertices $u=(\{a,b\},F)$ and $v=(\{c,d\},G)$ of the train where $F,G$ are 1-factors and $a,b,c,d$ are vertices of the graph being factorised. If there was an arc $uv$ as well as a loop on $v$ that would mean (up to swapping $c$ and $d$) that $ac,bd\in F$ and $ab,cd\in G$, which produces a 4-cycle in $F\cup G$.

We found that for $i=0,1,2,3,4,5$ the counts of how many vertices have $p(v)=i$ together form a complete invariant for the P1Fs of $K_{16}$. Counting only up to $i=4$ does not suffice, since there is a single pair of (rigid) P1Fs whose first difference occurs at $i=5$. They are

abcdefghijklmnop acbedgfihmjokpln adbmcfeigjhlkonp aebdcnflgkhjiomp afbicjdkenglhpmo

agblcodhejfnipkm ahbkcpdjeofmgnil aibncldoepfjgmhk ajbfckdnemgiholp akbjcedlfpgohnim

albpchdiegfojmkn ambhcgdfekinjplo anbocidmehfkgpjl aobgcmdpelfhikjn apbcdefghijklmno

which has counts $[139,19,15,14,17,17]$, and

abcdefghijklmnop acbedgfihljokmnp adbicefmgjhoknlp aebkcndhfogpiljm afbpcjdlekgihnmo

agbjchdpemfkinlo ahbockdmeiflgnjp aibgcmdoepfjhkln ajbfcldnehgoikmp akbmcidfeoglhpjn

albhcpdiejfngmko amblcgdjenfhiokp anbdcoelfpgkhjim aobncfdkeghmipjl apbcdefghijklmno

which has counts $[139,19,15,14,17,22]$.

Another well studied invariant of P1Fs is the tricolour vector [9]. It was already reported in [6] that the tricolour vector does not fully distinguish the P1Fs of $K_{16}$ with non-trivial automorphism group. We found that it partitions the set of all P1Fs of $K_{16}$ into 2320 equivalence classes.

Next we propose a new invariant that performs better than either the train indegree sequence or tri-colour vector on the P1Fs of $K_{16}$. It is based on the vertex cycles which were used for switchings among 1-factorisations in [4]. These cycles correspond exactly to the row-cycles in ${\cal U}({\cal F})$ other than the row-cycles of length 2 that include entries on the main diagonal. Finding the length of all row-cycles in ${\cal U}({\cal F})$ is easily done in cubic time. The vector of tallies of how many row-cycles there are of each length is a complete invariant for P1Fs of $K_{16}$.

Another invariant (that is presumably more discriminating in general), is to count for each row of ${\cal U}({\cal F})$ how many row-cycles of each length involve that row. The resulting list of $n$ vectors should then be sorted lexicographically to accommodate relabelling of the vertices of ${\cal F}$. We found that for the P1Fs of $K_{16}$ this invariant was complete even if we only counted cycles of length 3 and 4. Counting only the cycles of length 3 was not a complete invariant, but did separate the 3155 P1Fs into 3102 equivalence classes.

The invariant involving $p(v)$ as described above can be calculated in polynomial time, but not obviously in cubic time. By comparison, the indegree sequence of the train, tri-colour vector and our new invariant based on vertex cycle lengths are all easily computed in time cubic in the order of the graph being factorised. The indegree sequence and tri-colour vector are not complete invariants for P1Fs of $K_{16}$ and it seems likely that the vertex cycle lengths and path lengths will not be a complete invariant when $n$ gets larger. However, it is possible in polynomial time to canonically label a P1F, which does provide a complete invariant for every order. We may define our canonical form to contain the factors $F_{1}$ and $F_{2}$ from $(\ref{e:F1F2})$, and to be the lexicographically least possibility under that assumption. Starting with any P1F, there are only quadratically many choices for the factors that will become $F_{1}$ and $F_{2}$. For each of those choices, there are only linearly many ways that the two factors can map to $F_{1}$ and $F_{2}$ (given that their union is a single cycle). Thus we can check all relabellings in polynomial time and choose the canonical one. Once we have a canonical relabelling, isomorphism checking is a triviality. The fact that isomorphism testing for P1Fs can by done in polynomial time has long been known [2, 6]. The P1Fs given explicitly above, except the one given in $(\ref{e:newaut2})$, have been given in the canonical form just described. All P1Fs in the catalogue [11] are also in this form.

5 New orders of P1Fs

The paper [10] gave constructions for many new orders of P1Fs using the quotient coset starter method. Subsequent to publishing that paper the author discovered many more examples and put them on his webpage, where they have been visible for more than 12 years. In the interests of committing them to a more permanent part of the literature, we take this opportunity to record them here. The P1Fs of order 1092728 and 1225044 were apparently previously found by Volker Leck, but we are not aware of them having been published.

The following produce P1Fs for complete graphs of orders $q+1=p^{3}+1$, where $p$ is prime. They can be considered as extra entries for Table 6 in [10], and that paper should be consulted for the meaning of the notation.

$p=101$, $q=1030301$, $\zeta(x)=x^{3}+x+3$, $\tilde{c}=[813092,759910,233271,3]$

$p=103$, $q=1092727$, $\zeta(x)=x^{3}+x+4$, $\tilde{c}=[828376,896]$

$p=107$, $q=1225043$, $\zeta(x)=x^{3}+x+9$, $\tilde{c}=[1107573,151]$

$p=109$, $q=1295029$, $\zeta(x)=x^{3}+x+6$, $\tilde{c}=[271574,645911,1082655,4]$

$p=127$, $q=2048383$, $\zeta(x)=x^{3}+x+15$, $\tilde{c}=[840749,23]$

$p=131$, $q=2248091$, $\zeta(x)=x^{3}+x+3$, $\tilde{c}=[2096100,298]$

$p=139$, $q=2685619$, $\zeta(x)=x^{3}+x+7$, $\tilde{c}=[436598,2118]$

$p=149$, $q=3307949$, $\zeta(x)=x^{3}+x+14$, $\tilde{c}=[1861398,3141536,1357853,1]$

$p=151$, $q=3442951$, $\zeta(x)=x^{3}+x+5$, $\tilde{c}=[1492322,66]$

$p=163$, $q=4330747$, $\zeta(x)=x^{3}+x+4$, $\tilde{c}=[2015256,4602]$

$p=167$, $q=4657463$, $\zeta(x)=x^{3}+x+3$, $\tilde{c}=[3183263,109]$

$p=179$, $q=5735339$, $\zeta(x)=x^{3}+x+4$, $\tilde{c}=[2740965,1219]$

$p=191$, $q=6967871$, $\zeta(x)=x^{3}+x+3$, $\tilde{c}=[4789910,1160]$

$p=199$, $q=7880599$, $\zeta(x)=x^{3}+x+13$, $\tilde{c}=[3457494,2368]$

$p=211$, $q=9393931$, $\zeta(x)=x^{3}+x+24$, $\tilde{c}=[5457264,1168]$

$p=223$, $q=11089567$, $\zeta(x)=x^{3}+x+9$, $\tilde{c}=[4722613,4305]$

$p=227$, $q=11697083$, $\zeta(x)=x^{3}+x+9$, $\tilde{c}=[9051956,1442]$

$p=239$, $q=13651919$, $\zeta(x)=x^{3}+x+11$, $\tilde{c}=[1597504,5918]$

$p=251$, $q=15813251$, $\zeta(x)=x^{3}+x+7$, $\tilde{c}=[9285089,11965]$

$p=263$, $q=18191447$, $\zeta(x)=x^{3}+x+8$, $\tilde{c}=[8313030,2840]$

$p=271$, $q=19902511$, $\zeta(x)=x^{3}+x+4$, $\tilde{c}=[6563520,170]$

$p=283$, $q=22665187$, $\zeta(x)=x^{3}+x+24$, $\tilde{c}=[2245440,3574]$

The following produce P1Fs for complete graphs of orders $q+1=p^{5}+1$, where $p$ is prime. They can be considered as extra entries for Table 7 in [10].

$p=19$, $q=2476099$, $\zeta(x)=x^{5}+x+9$, $\tilde{c}=[949007,791]$

$p=23$, $q=6436343$, $\zeta(x)=x^{5}+x+3$, $\tilde{c}=[1045440,7580]$

Acknowledgements

This research was supported by the Monash eResearch Centre and eSolutions-Research Support Services through the use of the MonARCH HPC Cluster. The authors are also grateful to Mariusz Meszka and Alex Rosa for helpful discussion, including pointers to the literature.