Maximal knotless graphs

A graph is maximal knotless if it is edge maximal for the property of knotless embedding in $R^3$. We show that such a graph has at least $\frac74 |V|$ edges, and construct an infinite family of maximal knotless graphs with $|E|<\frac52|V|$. With the exception of $|E| = 22$, we show that for any $|E| \geq 20$ there exists a maximal knotless graph of size $|E|$. We classify the maximal knotless graphs through nine vertices and 20 edges. We determine which of these maxnik graphs are the clique sum of smaller graphs and construct an infinite family of maxnik graphs that are not clique sums.


Introduction
A graph G is maximal planar if it is edge maximal for the property of being a planar graph.That is, G is either a planar complete graph, or else adding any missing edge to G results in a non-planar graph.Maximal planar graphs are triangulations and are characterized by the number of edges: a planar graph with |V | ≥ 3 is maximal planar if and only if |E| = 3|V | − 6.
Naturally, planarity is not the only property of graphs that that can be studied with respect to edge maximality.A graph is intrinsically linked if every embedding of the graph in R 3 contains a non-split link.Some early results on maximal linkless (or maxnil) graphs-those that are edge maximal for the property of not being intrinsically linked-include a family of maximal linkless graph with 3|V | − 3 edges [J], and the fact that the graph Q(13, 3) is a splitter for intrinsic linking, a property that implies it is maximal linkless [Mh].Recently there have been several new results including families of maxnil graphs with 3|V |−3 edges (rediscovering Jørgensen's examples) [DF], with 14  5 |V | edges [A], and with 25 12 |V | edges [NPP].Lower bounds for the number of edges required for a maxnil graph have been established [A], and methods for creating new maxnil graphs via clique sum have been developed [NPP].
We extend this work with what appears to be the first study of maximal knotless graphs.A graph is intrinsically knotted (IK) if every embedding in R 3 includes a non-trivially knotted cycle, and a graph is not IK or nIK if it has a knotless embedding, that is, an embedding in which every cycle is a trivial knot.We will call a graph that is edge maximal for the nIK property maximal knotless or maxnik.
In Section 2, we establish a connection between maximal 2-apex graphs and maxnik graphs, specifically that a 2-apex graph is maxnik if and only if it is maximally 2-apex.This connection is instrumental in allowing the identification of all maxnik graphs with nine or fewer vertices, and with 20 or fewer edges.We remark that there is an analogous connection between maximal apex graphs and maxnil graphs that may be of independent interest.
We consider clique sums of maxnik graphs in Section 3, and are able to establish similar, if weaker, results to those of [NPP].Most importantly, we show that the edge sum of two maxnik graphs G 1 and G 2 on an edge e is maxnik if e is non-triangular (i.e., not part of a 3-cycle) in at least one G i .Similarly, we provide conditions that ensure that the clique sum over K 3 of two maxnik graphs is again maxnik.These results are used in Section 4 to construct new maxnik graphs from those found in Section 2.
We then turn to studying general properties of maxnik graphs in Section 4. We establish a lower bound for the number of edges in a maxnik graph of We will call a maxnik graph composite if it is the clique sum of two smaller graphs.Otherwise we say it is prime.These terms are analogous to knots, where a knot is composite if is the connected sum of two non-trivial knots, and prime otherwise.The infinite families of maxnik graphs constructed in Section 4 are all composite, as they are clique sums of smaller maxnik graphs.In Section 5, we classify the maxnik graphs found in Section 2 and construct an infinite family of prime maxnik graphs.

Classification through order nine and size 20
Theorem 2.1.
Proof.Suppose G is maxnik.If G is not connected, then, in a knotless embedding, add an edge e to connect two components.This is a knotless embedding of G + e, contradicting G being maximal knotless.
Suppose G has connectivity one with cut vertex v. Label the two components of G \ v as A and B. Let a be a neighbor of v in A and b be a neighbor of v in B. These must exist as G is connected.We will argue that G + ab is also nIK, a contradiction.
Form an embedding of G as follows: Embed A and B so that they are knotless and disjoint.Embed v on a plane separating them.Embed edges from v to A on the A side of the plane, edges from v to B on the B side, so that the embedding remains knotless.Now, isotope the rest of A (and B) until edge va (and vb) is embedded in the plane.Next add edge ab so that the triangle abv bounds a disk.Any cycle contained in A (or in B) is an unknot.Any cycle c that uses vertices from both A and B must use at least two vertices in the triangle abv.Since abv bounds a disk, this means the cycle c is a connected sum of a cycle in A and a cycle in B. Since those are unknots, c must be as well.This shows G+ab is nIK, contradicting G being maxnik.So a maxnik graph cannot have connectivity one and must be 2-connected.
Suppose G is a maxnik graph with |V | ≥ 3. Since G is connected, δ(G) > 0. If v ∈ V (G) has degree one, let u be the neighbor of v and w = v a different neighbor of u.In a knotless embedding of G, we can introduce the edge vw that closely follows the path v, u, e.This gives a knotless embedding of G + vw, contradicting the maximality of G.
Suppose G is maxnik with |V | ≥ 7. The lower bound on size is a consequence of the observation [JKM, Mt] that an IK graph has at least 21 edges.The upper bound follows as a graph with |E| ≥ 5|V | − 14 has a K 7 minor and is therefore IK [Md, CMOPRW].
In Theorem 4.3 below, we construct an infinite family of maxnik graphs, each with δ(G) = 2.
Theorem 2.2.A 2-apex graph is maxnik if and only if it is maximal 2-apex.
Proof.Let G be 2-apex.If G is not maximal 2-apex, then there is an edge e so that G + e is 2-apex, hence nIK [BBFFHL, OT].This shows that G is not maxnik.Conversely, if G is maximal 2-apex there are two cases, depending on |V |.If n = |V | < 7, then K n is 2-apex, so G = K n .But, K n is also nIK and therefore maxnik.If |V | ≥ 7, then |E| = 5|V | − 15.Since G is 2-apex, it is nIK.Adding any edge e, we have G + e with 5|V | − 14 edges.It follows that G has a K 7 minor and is IK [Mt, CMOPRW].This shows that G is maxnik.
A similar result, with essentially the same proof, holds for maxnil.
Theorem 2.3.An apex graph is maxnil if and only if it is maximal apex.
Theorem 2.4.For |V | = n ≤ 6, K n is the only maxnik graph.The only maxnik graphs for n = 7 and 8 are the three 2-apex graphs derived from triangulations on five and six vertices.
Proof.In [Mt,Proposition 1.4] it's shown that every nIK graph of order 8 or less is 2-apex.So, the maxnik graphs are the maximal 2-apex graphs.For n ≤ 6, all graphs are 2-apex, so K n is the only maximal knotless graph.For n = 7, the maximal 2-apex graph is K − 7 , formed by adding two vertices to the unique graph with a planar triangulation on five vertices, K − 5 .The two maximal planar graphs on 8 vertices are formed by adding two vertices to the two triangulations on six vertices, the octahedron and a graph whose complement is a 3-path.We will call these graphs K 8 −3 disjoint edges and K 8 − P 3 .
Let E 9 (called N 9 in [HNTY]) be the nIK nine vertex graph in the Heawood family.Figure 2 in Section 4 below shows a knotless [Mt] embedding of E 9 .
Proof.That E 9 is nIK is established in [Mt].Up to symmetry, there are two types of edges that may be added.One type yields the graph E 9 + e, shown to be IK (in fact minor minimal IK or MMIK) in [GMN].The other possible addition yields a graph that has as a subgraph F 9 in the Heawood family.Kohara and Suzuki [KS] established that F 9 is MMIK.Theorem 2.6.There are seven maxnik graphs of order nine.
Proof.The seven graphs are the five maximal 2-apex graphs with 30 edges, E 9 , and the graph G 9,29 , shown in Figure 1.Note that G 9,29 is the complement of K 1 K 2 C 6 .Theorems 2.2 and 2.5 show that six of these seven graphs are maxnik.To see that G 9,29 is as well, note that the embedding shown in Figure 1, due to Ramin Naimi [N], is knotless.Up to symmetry, there are two ways to add an edge to the graph.In either case, the new graph has a K 7 minor and is IK.
It remains to argue that no other graphs of order nine are maxnik.We know that order nine graphs with size 21 or less are either IK, the graph E 9 , or else 2-apex, see [Mt,Propositions 1.6 and 1.7].Using Theorem 2.2, this completes the argument for graphs with |E| ≤ 21.Suppose G is maxnik of order nine with |E| ≥ 22.By Theorem 2.1, we can assume |E| ≤ 30.If G is 2-apex, by Theorem 2.2, it is one of the five maximal 2-apex graphs.So, we can assume G is not 2-apex.The minor minimal not 2-apex (MMN2A) graphs through order nine are classified in [MP].With a few exceptions these graphs are also MMIK.If G has an IK minor (including a MMIK minor) it is IK and not maxnik.So, we can assume G has as a minor a graph that is MMN2A, but not MMIK.There are three such graphs.One is E 9 , the other two, G 26 and G 27 , have 26 and 27 edges.In Theorem 2.5, we showed that E 9 is maxnik.The other two are subgraphs of G 9,29 .To complete the proof, we observe that any order nine graph that contains G 26 is either a subgraph of G 9,29 or else IK and similarly for G 27 .In fact, for those that are IK, we can verify this by finding a MMIK minor, either in the K 7 or K 3,3,1,1 family, or else the graph G 9,28 described in [GMN].
Theorem 2.7.The only maxinik graph of size 20 is K − 7 .There are seven maxnik graphs with at most 20 edges.
Proof.Work above establishes this through order nine.The seven maxnik graphs with at most 20 edges are the seven on seven or fewer vertices.Suppose G of order ten or more and size 20 is maxinik.By [Mt,Theorem 2.1], G is 2-apex and therefore maximal 2-apex.But this means |E| = 5|V | − 15 ≥ 35, a contradiction.
Remark 2.8.A computer search suggests that E 9 is the only maxnik graph of size 21.The search makes use of the 92 known MMIK graphs of size 22, see [FMMNN].

Clique sums of maxnik graphs
Clique sums of maxnil graphs were studied in [NPP], and we will show similar, if weaker, versions in the case of maxnik graphs.These results are used in Section 4.
Lemma 3.1.For t ≤ 2, the clique sum over K t of nIK graphs is nIK.
Proof.Let G 1 and G 2 be nIK graphs, and let Γ(G) denote the set of all cycles in G. Let G be the clique sum of G i over a clique of size t.Let f i be an embedding of G i that contains no non-trivial knot.
Suppose t = 1.We may extend the f i to an embedding of G by embedding f 1 (G 1 ) in 3-space with z > 0, and so by isotoping vertex v from each G i to the plane z = 0 and identifying them there, we have an embedding f (G).A closed cycle in G must be contained in a single G i , and hence given c ∈ Γ(G), then c ∈ Γ(G i ) for some i.As the embeddings f i (G i ) contain no nontrivial knot, c must be the unknot, and hence G is nIK.
Suppose t = 2.We may extend the f i to an embedding of G by embedding f 1 (G 1 ) in 3-space with z > 0, and so by shrinking the edge e in each G i and then isotoping them to the plane z = 0 and identifying them there, we have an embedding f As the embeddings f i (G i ) contain no nontrivial knot, in the first case c is the unknot, and in the second, it is the connect sum of unknots and hence unknotted.Thus, G is nIK.
For H 1 , H 2 , . . ., H k subgraphs of graph G, let H 1 , H 2 , . . ., H k G denote the subgraph induced by the vertices of the subgraphs.Lemma 3.2.Let G be a maxnik graph with a vertex cut set S = {x, y}, and let G 1 , G 2 , . . ., G r denote the connected components of G \ S. Then xy ∈ E(G) and G i , S G is maxnik for all 1 ≤ i ≤ r.
Proof.As G is 2-connected by Theorem 2.1, each of x and y has at least one neighbor in each as there exists G j with i = j since S is separating, and there exists a path from x to y in G j as G j is connected.Thus in G i , G j , S G , we may contract G j to x to obtain a graph isomorphic to G i .Thus, G i is nIK.So, by Lemma 3.1, This contradicts the fact that G is maxnik, and hence xy ∈ E(G).
Suppose that one or more of the G i are not maxnik.Then add edges as needed to each G i to form graphs H i that are maxnik.Then the graph H = H 1 ∪ xy H 2 ∪ xy . . .∪ xy H r is nIK by Lemma 3.1 and contains G as a subgraph.As G is maxnik, G = H and hence G i = H i for all i, so every G i is maxnik as well.
Lemma 3.3.Let G 1 and G 2 be maxnik graphs.Pick an edge in each G i and label it e.
Proof.Suppose that e is non-triangular in G 1 and has endpoints x, y.Add an edge ab to the graph G.The graph G is nIK by Lemma 3.1.If both a, b ∈ G i for some i, then G + ab is IK, as the G i are each maxnik.Thus, we may assume that a ∈ G 1 and b ∈ G 2 .The edge e is non-triangular in G 1 , so vertex a is not adjacent to both endpoints of e.We may assume that a is not adjacent to x.As G 2 is connected, we construct a minor G of G + ab by contracting the whole of G 2 to vertex x.Note that as b ∈ G 2 , we have the edge ax in G , and in fact G = G 1 + ax.As G 1 is maxnik, G is IK and so is G + ab.Thus, G is maxnik.
Lemma 3.4.For i = 1, 2, let G i be maxnik, containing a 3-cycle C i , and admitting to a knotless embedding such that C i bounds a disk whose interior is disjoint from the graph.Then the clique sum G over K 3 formed by identifying Proof.Let f i be the knotless embedding of G i .Embed the f i (G i ) so that they are separated by a plane.We may then extend this to an embedding f (G) by isotoping the C i to the separating plane and identifying them there.
Let Γ(G) denote the set of all cycles in G.As the cycles As the embeddings f i (G i ) contain no nontrivial knot, in the first case c is the unknot, and in the second, it is the connect sum of unknots and hence unknotted.Thus, G is nIK.
Suppose deg(v) = 2.We will argue that H = H \ v is also maxnik with |E | < 7 4 |V |, contradicting our assumption that H was a minimal counterexample.Let N (v) = {w, x} and note that wx ∈ E(H).Otherwise, in an unknotted embedding of H, we could add the edge wx so that the 3cycle vwx bounds a disk.This will not introduce a knot into the embedding and contradicts the maximality of H.
As a subgraph of H, H is nIK.Suppose it is not maxnik because there is an edge ab so that H + ab remains nIK.In a knotless embedding of H + ab, we can add the vertex v and its two edges so the 3-cycle vwx bounds a disk.This will not introduce a knot into the embedding and shows that H + ab is also nIK, contradicting the maximality of H. Thus, no such graph H with a vertex of degree 2 can exist.
So, we may assume that deg(v) = 3.Here we cannot apply the techniques of [A], as Y ∇ moves do not preserve intrinsic knotting [FN].However, Lemma 4.1 allows us to show the average degree of H is actually at least 3.5, and hence H is not a counterexample.
Divide the vertices of H into 3 sets: A = {vertices of degree 3}, B = {vertices of degree > 3 that are neighbors of vertices in A}, and C = {all other vertices of H}.Form the graph H = H \ C. All vertices in C have degree 4 or greater, so it suffices to show that the vertices in each connected component of H have average degree 3.5 or higher.
A vertex a i1 of degree 3 has three neighbors, label them b i1 , b i2 , a i2 , where a i2 is a neighbor of minimal degree.If deg(a i2 ) = 3, we continue.If not, delete all edges incident on a i2 except those between a i2 and {a i1 , b i1 , b i2 }.This creates a subgraph of H with strictly fewer edges; we will abuse notation and continue to call it H . Vertex a i2 now has degree 3 in H , and we move it to set A.
If a i2 had degree greater than 3 in H, then, since it has the minimal degree among the neighbors of Consider the connected component of v in H , call it H 1 .We will calculate the total degree of the vertices in H 1 and divide by the number of vertices.Suppose there are n vertices from set A and m vertices from set B in H 1 for a total of n + m vertices.Each vertex from set A has degree 3, so the contribution to total degree from set A is 3n.Each vertex in A is adjacent to exactly 2 of the b ij , so the total degree contribution for set B is at least 2n from edges to set A. Further, H 1 is connected.As a ij is only adjacent to b i j if i = i , there must be at least m − 1 edges between the b ij , which adds 2(m − 1) to the total degree.This gives an average degree of 5n+2m−2 n+m in H 1 .However, H is 2-connected by Theorem 2.1, so there must be at least 2 edges from H 1 to its complement in H.So within H, these vertices must  0 1/2 1 3/2 2 5/2 20/7 25/8 21/ Proof.Let e be an edge of E 9 connecting a degree 4 vertex to one of degree 5. Edge e is non-triangular and there are five other edges symmetric to it.Using Lemma 3.3, take k copies of E 9 glued along edge e.The resulting graph has 7k + 2 vertices and 20k + 1 edges.Gluing on five K 3 graphs in each E 9 on the other non-triangular edges gives an additional 5k vertices and 10k edges.So, for each k ≥ 1, we have a graph G with n = 12k + 2 vertices and m = 30k + 1 edges.Then m = 30(n − 2)/12 + 1 = 5 2 n − 5.These two theorems suggest the following question.In Table 1 we give the least ratios through order nine.
Question 4.4.What is the minimal number of edges for a maxnik graph of n vertices?
For maximal planar graphs, |E| = 3|V | − 6.Similarly, maximal k-apex graphs have a fixed number of edges depending on |V |.In contrast, as with maximal linkless graphs, the number of edges in a maxnik graph can vary.In fact, with the exception of |E| = 22, for any |E| ≥ 20, there exists a maxnik graph of that size.Proof.The graph K − 7 is a maxnik of size 20 by Theorem 2.4.The graph E 9 has a knotless embedding where the 3-cycle abc bounds a disk [Mt], shown in Figure 2. As no vertex in E 9 is adjacent to all three of these vertices, we may use Lemma 3.4 to construct a maxnik graph of size 24 by taking a clique sum over K 3 of E 9 and K 4 .So, we may assume n ≥ 21 and n ∈ {22, 24}.
The graph E 9 has size 21 and 6 non-triangular edges.Let G i denote the maxnik graph obtained from i copies of E 9 by gluing along non-triangular edges.
Note that , and that G i contains at least 6 nontriangular edges for any i.Clearly a maxnik graph of size |E(G i )|+0 exists, as G i is maxnik.We may form a new maxnik graph from G i by gluing a copy of K m (for 3 ≤ m ≤ 6) along a non-triangular edge of G i by Lemma 3.3.As G i has at least 6 nontriangular edges, we can glue on up to 6 such graphs, each adding m 2 − 1 edges.Thus, to prove the result we need only to be able to form the desired values of k using six or fewer addends from the set {2, 5, 9, 14}.This is clearly possible.
In the base case i = 1, we have a maxnik graph of size |E(G 1 )| = 21, and we excluded graphs of size 22 and 24 (|E(G 1 )| + 1 and |E(G 1 )| + 3) above.Thus we may form maxnik graphs of size |E(G 1 )| + k for the k of interest as before.
Remark 4.6.A computer search shows there are no size 22 maxnik graphs.Our strategy is based on the classification through size 22 of the obstructions to 2-apex in [MP].Let's call such graphs MMN2A (minor minimal not 2-apex).All but eight of the graphs in the classification are MMIK.Two exceptions are 4-regular of order 11, the other six are in the Heawood family.
A maximal 2-apex graph has 5n − 15 edges where n is the number of vertices.By Theorem 2.2 a maxnik graph G of size 22 is not 2-apex and therefore has a MMN2A minor.Since G is nIK, it must have one of the eight exceptions as a minor.Using a computer, we verified that no size 22 expansion of any of these eight graphs is maxnik.
Theorem 4.5 implies that there are maxnik graphs of nearly every size.Note that there are maxnik graphs of any order, as there exist maximal 2-apex graphs of any order and by Theorem 2.2 these graphs are maxnik.
We have considered the minimal number and the possible number of edges in a maxnik graph.We now consider other aspects of maxnik graphs' structure, in particular, the maximal and minimal degree.Since ∆(G) = |V | − 1 for maximal 2-apex graphs, there are maxnik graphs with arbitrarily large ∆(G).
Proposition 4.7.The complete graph K 3 is the only maxnik graph with maximal degree two.
Proof.Suppose G is maxnik with ∆(G) = 2. Then |G| ≥ 3 and, by Theorem 2.1, δ(G) = 2 and G is connected.So, G is a cycle.Now, a cycle is planar, hence 2-apex, and by Theorem 2.2, G is maximal 2-apex.However, a cycle is not maximal 2-apex unless it is K 3 .
Note that Lemma 4.1 has the following two immediate corollaries.
These results motivate the following question.
Question 4.10.Do there exist regular maxnik graphs other than K n with n < 7?
A maximal 2-apex graph will have ∆(G) = |V | − 1 and δ(G) ≤ 7, so if there is such a regular maxnik graph with |V | ≥ 7, it is not 2-apex.However, through order nine, our two examples of maxnik non 2-apex graphs are both close to regular, having ∆(G) − δ(G) ≤ 2. This suggests the answer to our question is likely yes.
For δ(G), Theorem 2.1 gives a lower bound of two that is realized by the infinite family of Theorem 4.3.On the other hand, by starting with a planar triangulation of minimum degree five, we can construct graphs with δ(G) = 7 that are maximal 2-apex, and hence maxnik.At the same time, since a graph with |E| ≥ 5|V |−14 has a K 7 minor and is IK [Md, CMOPRW], a maxnik graph must have δ(G) ≤ 9.It seems likely that there are examples that realize this upper bound on δ(G).Table 2 records the range of degrees for maxnik graphs through order nine.

Prime and Composite Maxnik Graphs
We will call a graph composite if it is the clique sum of two graphs.Otherwise it is prime.These terms are analogous to knots, where a knot is composite if is the connected sum of two non-trivial knots, and prime otherwise.In this section, we classify the maxnik graphs described earlier in this paper as prime and composite.We remark that it may be of interest to study other instances of prime graphs, for example, prime maximal planar or prime maxnil.
The infinite families of maxnik graphs constructed in Section 4 are all composite, as they are clique sums of smaller maxnik graphs.
Note that K n is prime, so all maxnik graphs of order 6 or less are prime.Proof.The graph K − 7 is formed from two copies of K 6 summed over a 5clique.
The graph K 8 − P 3 is formed from K − 7 clique sum K 6 over a 5-clique, where the 5-clique contains exactly one endpoint of the missing edge.
Big-Y is formed from K 8 − P 3 clique sum K 6 over a 5-clique, where the 5-clique contains both of the terminal vertices of the 3-path.
Long-Y is formed from K 8 − 3 disjoint edges clique sum K 6 over a 5-clique.
Hat is formed from K 8 − P 3 clique sum K 6 over a 5-clique, where the 5-clique contains one terminal vertex and one (non-adjacent) interior vertex of the 3-path.
House is formed from K 8 − P 3 clique sum K 6 over a 5-clique, where the 5-clique contains one interior vertex of the 3-path.
Lemma 5.2.If G c is of the form K 2 H, then either G is prime, or G is the clique sum of two copies of K n over an n − 1 clique.
Proof.Call the two vertices of the K 2 in G c v 1 and v 2 .Suppose that G is a clique sum of G 1 and G 2 over a clique C. We cannot have both v 1 and v 2 in C, as edge v 1 v 2 is in G c .Without loss of generality, we may assume that v 1 is in G 1 \ C. So, in G c , v 1 must be adjacent to every vertex of G 2 \ C. Thus G 2 \ C is v 2 .As the only neighbor of v 1 in G c is v 2 , v 1 is adjacent to every vertex in C. Similarly for v 2 .Thus if G is composite, it is the clique sum of K n and K n over an n − 1 clique.
Proof.Each of these graphs has a complement of the form K 2 H.As these graphs are not of the form K n − a single edge, they are prime by Lemma 5.2.
Note that if G is a clique sum over a t-clique, it is not (t + 1)-connected.
Proof.The largest clique in E 9 is a 3-clique, but E 9 is 4-connected and hence must be prime.
Lemma 5.5.If G = H * K 2 , and G is 2-apex, then G is prime maxnik if and only if H is prime maximal planar.
Proof.As G is 2-apex, it is maxnik if and only if it is maximal 2-apex, and G is maximal 2-apex if and only if H is maximal planar.
If H is composite, then H is the clique sum of H 1 and H 2 over a t-clique.So G is the clique sum of H 1 * K 2 and H 2 * K 2 over a t + 2-clique, and hence G is composite.
As G is maxnik, it must be 2-connected.Hence if G is composite, it must be G 1 clique sum G 2 over a t-clique C, with t ≥ 2. Label two of the vertices in C as v 1 , v 2 .Then H is the clique sum of G 1 \ {v 1 , v 2 } and G 2 \ {v 1 , v 2 } over C \ {v 1 , v 2 }, and thus composite.
Corollary 5.6.There exist prime maxnik graphs of arbitrarily large size, and of any order ≥ 8.
Proof.The octahedron graph is max planar and 4-connected.The largest clique it contains is a 3-clique, so it is prime.New triangulations formed by repeated subdivision of a single edge are 4-connected and maximal planar, but have no 4-clique, hence are prime as well.Thus all of these graphs give prime maxnik examples when joined with K 2 .
We remark that the construction of this family of graphs is similar to the maxnil families with 3n − 3 edges due to Jørgensen [J] and 3n − 5 edges due to Naimi, Pavelescu, and Pavelescu [NPP].

74
|V |, and construct an infinite family of maxnik graphs with fewer than 5 2 |V | edges.A maximal planar graph has |E| = 3|V | − 6, and maximal k-apex graphs also have a fixed number of edges depending on |V |.In contrast, the number of edges in maxnil and maxnik graphs can vary.We show that, except for |E| = 22, given any |E| ≥ 20, there exists a maxnik graph of size |E|.
a i1 , deg(b ij ) ≥ 4 and b i1 , b i2 ∈ B. If deg(a i2 ) = 3 in H, vertices a ij are adjacent only to each other and the b ij .If either of the b ij have degree 3 in H, then H can be disconnected by deleting the other b ij .This is a contradiction as H is maxnik and must be 2-connected by Theorem 2.1.Thus, the b ij are in B.
Figure 2. A knotless embedding of E 9 .

Figure 3 .
Figure 3. Complements of the maximal 2-apex graphs of order nine.Top row, L to R: Big-Y, Long-Y, Hat; Bottom row: Pentagon-bar and House.

9 Table 1 .
The least ratios of size to order for maxnik graphs through order nine.