How many Distinct Hamiltonian Maximal Planar Graphs are there (n vertices) and could this representation help? If we make a regular polygon with n vertices (n edges) and triangulate on the inside with n-3 edges, then triangulate on the outside with (n-3) edges (or draw dotted lines inside again), a Maximal Planar Graph is formed.  Edges shouldn't be repeated and there's no loops or directions.
How many distinct graphs of this type are there?
It's connected to an earlier question where it was asked 'How many Distinct Maximal Planar Graphs are there? How many distinct Maximal Planar Graphs exist with $n$ vertices?
Will Orrick gave the OEIS numbers A000109.  It was then wondered if there was a known formula for those numbers or bounds on them.  It's conjectured that graphs of the type in this question might constitute most Maximal Planar Graphs, so a formula for the 'polygon' types might be an approximate formula or a lower bound for all types of Maximal Planar Graphs.
Dividing the polygon on the inside can be done in C(n-2) ways, where C(n) are the Catalan numbers A000108, so a connection was looked for - and A000109 seem to be close to C(n-2)*2^(n-13) at least for n=9 to 23.
So, the answer to this question would be of interest and any thoughts on connecting the A000108 and A000109 numbers.  Lots of ways have been tried so far, e.g. since the next Catalan number can be formed by adding the product of ones before, e.g. C(4) = C(3)*C(1) + C(2)*C(2) + C(1)*C(3), perhaps something similar happens for A000109, or by incorporating numbers from both sequences.  Lots of coincidences (probably) have been found including the 233 number in A000109 say X(10) is C(9-2) - the sum of the Catalan numbers before it.
Think I'm going to go crazy looking for patterns any longer!  Any suggestions please!
 A: This is too long to be a comment, but only provides a modest reduction in the $C_{n-2}^2$ upper bound.
For $n\ge3$ let $a_n$ be the number of unordered pairs of triangulations of the $n$-gon with no common diagonals. Let $g(x)=a_3x^3+a_4x^4+a_5x^5+a_6x^6+\ldots$ be the generating function. The first few coefficients are $a_3=\frac{1}{2}$, $a_4=1$, $a_5=5$, $a_6=34$ (see A257887). The non-integer value of $a_3$ makes sense in light of the property that there is one unordered pair of triangulations for every set of two ordered pairs of triangulations in which the triangulations trade places, except in the case of the triangle, where the only choice is for both triangulations to be empty. The OEIS omits the $a_3$ term. To understand why $a_6=34$ let $T$, $Z$, and $F$ denote the triangle, Z-shaped, and fan triangulations described in one of your comments. Then there is one $TT$ pair, six $TF$ pairs, $12$ $ZF$ pairs, nine $ZZ$ pairs, and six $FF$ pairs. The $TT$ pair and three of the $ZZ$ pairs give the octahedron; all others give the graph $K_6$ minus a path of length $4$ described by Dan Uznanski in a comment here. This example shows that one must account, not only for rotations and reflections of a pair of triangulations, but also for different choice Hamiltonian circuit, which can result in different triangulation pairs for the same graph.
Danièle Huguet and Dov Tamari in La structure polyédrale des complexes de parenthésages, J. Combin. Inform. System Sci., 3(2):69–81, 1978 showed that the generating function satisfies
$$
x=\sum_{j=0}^\infty C_j^2\left(x-\frac{2}{x}g(x)\right)^{j+1}.
$$
In other words, $g(x)$ can be obtained by inverting the formal power series
$$
\phi(x)=\sum_{j=0}^\infty C_j^2x^{j+1},
$$
which means finding the formal power series $u(x)$ such that $\phi(u(x))=x$. Then $g(x)=\frac{x}{2}(x-u(x))$. I have not been able to find a copy of Huguet and Tamari's paper, but a derivation is also given in

Dimbinaina Ralaivaosaona, Jean Bernoulli Ravelomanana, and Stephan Wagner, Counting Planar Tanglegrams, in James Allen Fill and Mark Daniel Ward eds., 29th International Conference on Probabilistic,  Combinatorial and Asymptotic Methods for the Analysis of Algorithms (AofA 2018), Leibniz International Proceedings in Informatics (LIPIcs) 110:32:1–32:18, 2018.

Ralaivaosaona, Ravelomanana, and Wagner point out that $\phi(x)$ is the series expansion of a complete elliptic integral and use this to do an asymptotic analysis. They find that $a_n$ grows like a constant times $n^{-3}\left(\frac{4\pi}{4-\pi}\right)^n$. The upper bound $C_{n-2}^2$ grows like a constant times $n^{-3}\cdot16^n$. The revised upper bound has slower growth since $\frac{4\pi}{4-\pi}\approx14.64$.
A: Here is the work so far.  In what follows the number of distinct Hamiltonian Maximal Planar Graphs with n vertices is $X(n)$.  It should follow A000109 https://oeis.org/A000109/list up to $n=11$ when the first non Hamiltonian occurs.
Also used are the number of distinct ways to divide a polygon https://oeis.org/A000207 .   $A(n)$ will be a number from this list corresponding to the number of distinct ways to triangulate a polygon with $n$ sides (offset by $2$ from the list in the link) .  There is a formula for these, not written here.
Since a Maximal Planar Hamiltonian graph can be represented by a regular polygon triangulated by 'inside' edges and also triangulated again with 'outside' edges, without repeating edges, the following is suggested.
$$
\begin{array}{c|cc}
n & A(n) & X(n)\\
\hline
3 & 1 & 1\\
4 & 1 & 1\\
5 & 1 & 1\\
6 & 3 & 2\\
7 & 4 & 5\\
8 & 12 & 14\\
9 & 27 & 50\\
10 & 82 & 233\\
11 & 228 & 1249\\
\end{array}
$$
Dividing the pentagon on the inside can be done in $1$ way with a fan $F$ from one vertex.
there is only one distinct way to do it on the outside, another fan, $1\times1=1$
For the hexagon let's start with the maximum number of edges possible from either set of edges, leaving a vertex, it's $3$, making a fan $F$.  Let the hexagon be $ABCDEF$.  If the $3$ inside (say) edges leave node $B$.  The outside edges can't repeat any of these, one of the outside edges must also join $A$ to $C$, to ensure it's triangulated and not repeat edges.  This leaves a pentagon to triangulate by the outside edges, but using one less edge (one was used for $AC$) - the number of distinct ways for that is $A(5)$
There are two remaining ways to divide the hexagon with inside edges, a triangle $T$ and a $Z$ shape.
Next the case where the maximum number from one set of edges leaving a vertex is $2$.  If we draw diagrams using the $2$ remaining ways to divide the hexagon but not using a fan $F$, we find an extra distinct case from this, so $X(6)=A(5)+1$.

In a similar way using tracing paper, with the $4$ ways to divide a heptagon, it's found that $X(7) = A(6) + 2$.  The $A(6)$ is the number using a Fan of $4$ edges at one corner of the heptagon, similar to above, and the $2$ comes by combining the other $3$ ways, each of the $2$ new cases came when one of the set of edges has $3$ edges from one set leaving a vertex.  The case where the maximum number is $2$ edges leaving a vertex gave no extra distinct cases.
So a pattern is now being looked for of this type
$$
X(n) = A(n-1)+aX(n-1)+bX(n-2)+cX(n-3)\ldots
$$
where $a$, $b$, $c$ etc...are integer coefficients.
The pattern isn't clear and might involve the factors of the number $n$, perhaps the Catalan numbers?  So a definite reason isn't given here for the coefficients, perhaps it's coincidence, but it's possible to build up the $X(n)$ in this way e.g.
$$
\begin{array}{r|rcl|l}
n & X(n) &\!=\!& A(n-1)+aX(n-1)+\ldots &\\
\hline
6 & 2 &\!\!=\!\!& 1 + 1 &\\
7 & 5 &\!\!=\!\!& 3 + 2 &\\
8 & 14 &\!\!=\!\!& 4 + 2\times5 &\\
9 & 50 &\!\!=\!\!& 12 + 2\times14 + 2\times5 &\\
10 & 233 &\!\!=\!\!& 27 + 3\times50 + 4\times14 &\\
11 & 1248 &\!\!=\!\!& 82 + 4\times233 + 3\times50 + 6\times14 & \text{1248 instead of 1249 as there is one}\\
& & & & \text{non Hamiltonian graph for $n=11$.}
\end{array}
$$
