For any natural $n$, $D_n$ is degree sequence with $8$ vertices of degree $4$ and $6n$ vertices of degree $6$. I expect there is a lot of graphs with this degree sequence.
For instance, it matches a union of a complete bipartite graph $K_{4,4}$ and any $6$-regular graph with $6n$ vertices (for instance, a union of any six mutually disjoint perfect mathching on $6n$ vertices).
As far as I know, class is a non-rigorous notion of a graph theory, so I think it is OK to call by class the family of graphs whose degree sequence is $D_n$.
I guess that Hamiltonian Path problem is NP-hard for the graph with $D_n$, because
it is NP-hard even for families of graphs with much more simple structure than $6$-regular graphs. Namely, according to Wikipedia’s article:
The problem of finding a Hamiltonian cycle or path is in FNP; the analogous decision problem is to test whether a Hamiltonian cycle or path exists. The directed and undirected Hamiltonian cycle problems were two of Karp's $21$ NP-complete problems. They remain NP-complete even for undirected planar graphs of maximum degree three, for directed planar graphs with indegree and outdegree at most two, for bridgeless undirected planar $3$-regular bipartite graphs, for $3$-connected $3$-regular bipartite graphs, subgraphs of the square grid graph, and cubic subgraphs of the square grid graph.
But for the family of the graphs which you have illustrated it looks that a spiral is such a path. Namely, for the drawn graph a Hamiltonian path is
ABCDEGFHLIKJPMONSQTRVWUX
or
AGBFCHDEIOKNJPLMTXRVSWQU
Coefficient extraction:
We have
\begin{align*}
D(z)&=\frac{1}{\left(1-z\right)\left(1-z^5\right)\left(1-z^{10}\right)\left(1-z^{25}\right)\left(1-z^{50}\right)}
\end{align*}
We know from this answer
\begin{align*}
A(z)&=\frac{1}{\left(1-z\right)\left(1-z^5\right)\left(1-z^{10}\right)}\\
&=\sum_{m=0}^\infty\left(\frac{1}{4}\left\lfloor\frac{m}{5}\right\rfloor^2+\frac{5}{4}\left\lfloor\frac{m}{5}\right\rfloor
-\frac{1}{2}\left\lfloor\frac{m}{10}+\frac{1}{2}\right\rfloor+1\right)z^m\tag{1}
\end{align*}
We calculate analogously
\begin{align*}
\color{blue}{B(z)}&=\frac{1}{\left(1-z^{25}\right)\left(1-z^{50}\right)}\\
&=\sum_{q=0}^\infty z^{25q}\sum_{f=0}^\infty z^{50f}\\
&=\sum_{n=0}^\infty\left(\sum_{{25q+50f=n}\atop{q,f\geq 0}}\right)z^n\\
&=\sum_{n=0}^\infty\left(\sum_{{2q+f=n}\atop{q,f\geq 0}}\right)z^{25n}\\
&=\sum_{n=0}^\infty\left(\sum_{q=0}^{\lfloor n/2\rfloor}1\right)z^{25n}\\
&\,\,\color{blue}{=\sum_{n=0}^\infty\left(\left\lfloor \frac{n}{2}\right\rfloor+1\right)z^{25n}}\tag{2}
\end{align*}
Using the coefficient of operator $[z^t]$ to denote the coefficient of $z^t$ of a series we obtain from (1) and (2)
\begin{align*}
\color{blue}{[z^t]}&\color{blue}{D(z)}=[z^t]A(z)B(z)=[z^t]\sum_{m=0}^\infty a_mz^m\sum_{n=0}^\infty b_nz^{25n}\\
&=[z^t]\sum_{q=0}^\infty\left(\sum_{{m+25n=q}\atop{m,n\geq 0}}a_mb_n\right)z^q\\
&=[z^t]\sum_{q=0}^\infty\left(\sum_{n=0}^{\lfloor q/25\rfloor}a_{q-25n}b_n\right)z^q\\
&=\sum_{n=0}^{\lfloor t/25\rfloor}a_{t-25n}b_n\\
&=\sum_{n=0}^{\lfloor t/25\rfloor}
\left(\frac{1}{4}\left\lfloor\frac{t-25n}{5}\right\rfloor^2+\frac{5}{4}\left\lfloor\frac{t-25n}{5}\right\rfloor
-\frac{1}{2}\left\lfloor\frac{t-25n}{10}+\frac{1}{2}\right\rfloor+1\right)\\
&\qquad\qquad\quad\cdot
\left(\left\lfloor\frac{n}{2}\right\rfloor+1\right)\\
&\,\,\color{blue}{=\sum_{n=0}^{\lfloor t/25\rfloor}
\left(\frac{1}{4}\left(\left\lfloor\frac{t}{5}\right\rfloor-5n\right)^2+\frac{5}{4}\left(\left\lfloor\frac{t}{5}\right\rfloor-5n\right)
-\frac{1}{2}\left\lfloor\frac{t}{10}-\frac{5n}{2}+\frac{1}{2}\right\rfloor+1\right)}\\
&\qquad\qquad\quad\color{blue}{\cdot
\left(\left\lfloor\frac{n}{2}\right\rfloor+1\right)}\tag{3}
\end{align*}
It seems, the result (3) can be considerably simplified, since we obtain with the help of Wolfram alpha the nice representation
\begin{align*}
D(z)&=\color{blue}{1} + z + z^2 + z^3 + z^4\\
&\qquad + \color{blue}{2} z^5 + 2 z^6 + 2 z^7 + 2 z^8 + 2 z^9 \\
&\qquad+ \color{blue}{4} z^{10} + 4 z^{11} + 4 z^{12} + 4 z^{13} + 4 z^{14}\\
&\qquad+ \color{blue}{6 }z^{15} + 6 z^{16} + 6 z^{17} + 6 z^{18} + 6 z^{19}\\
&\qquad + \color{blue}{9} z^{20} + 9 z^{21} + 9 z^{22}+ 9 z^{23} + 9 z^{24}\\
&\qquad + \color{blue}{13}z^{25} + 13 z^{26} + 13 z^{27} + 13 z^{28} + 13 z^{29}\\
&\qquad + \color{blue}{18} z^{30} + 18 z^{31} + 18 z^{32} + 18 z^{33} + 18 z^{34}\\
&\qquad+ \color{blue}{24} z^{35} + 24 z^{36}+ 24 z^{37} + 24 z^{38} + 24 z^{39}\\
&\qquad + \color{blue}{31} z^{40} + 31 z^{41} + 31 z^{42} + 31 z^{43} + 31 z^{44}\\
&\qquad+ \color{blue}{39} z^{45} + 39 z^{46 }+ 39 z^{47} + 39 z^{48} + 39 z^{49}\\
&\qquad + O(z^{50})
\end{align*}
with equal coefficients in groups of five.
First-order Asymptotics:
The asymptotic estimation of OP is correct. We find in chapter IV: Complex Analysis, Rational and Meromorphic Asymptotics of Analytic Combinatorics by P. Flajolet and R. Sedgewick the
Proposition IV.2: Let $T$ be a finite set of integers without a common divisor ($\gcd(T) = 1$). The number of partitions with summands restricted to $T$ satisfies
\begin{align*}
P_t^{T}\sim\frac{1}{\tau}\,\frac{t^{r-1}}{(r-1)!},\qquad \text{ with }\tau:=\prod_{\omega\in T}\omega,\qquad r:= \mathrm{card}(T)
\end{align*}
Here we consider
\begin{align*}
[z^t]D(z)=[z^t]\frac{1}{\left(1-z\right)\left(1-z^5\right)\left(1-z^{10}\right)\left(1-z^{25}\right)\left(1-z^{50}\right)}\tag{4}
\end{align*}
For first-order asymptotic of coefficients in (4) only the pole at $z=1$, which is the nearest pole to $0$ with highest order $5$ needs to be considered.
We have according to (4) $T=\{1,5,10,25,50\},\tau=1\cdot5\cdot10\cdot25\cdot50,r=4$ from which
\begin{align*}
\color{blue}{[z^t]D(z)\sim} \frac{1}{1\cdot5\cdot10\cdot25\cdot50}\,\frac{t^4}{4!}=\color{blue}{\frac{2}{3}10^{-6}t^4}
\end{align*}
follows.
Chapter 4 of the book provides all the necessary information to derive this asymptotic estimation.
Best Answer
This is A003292 in the OEIS. There is no simple recurrence, but there is a nice generating function: $$\frac1{1-x}\frac1{1-x^2}\prod_{k=3}^\infty\frac1{(1-x^k)^2}$$ This generating function arises simply: in graphs with maximum degree $2$, each connected component is a single vertex, a single edge (two vertices), a cycle on $n\ge3$ vertices or a path on $n\ge3$ vertices. These explain the $\frac1{1-x},\frac1{1-x^2}$ and remaining factors of the generating function respectively.
We have $d_8=46$ and $d_9=70$.