It looks like this is asking about the combinatorial class
$\mathcal{F}$ where
$$\def\textsc#1{\dosc#1\csod}
\def\dosc#1#2\csod{{\rm #1{\small #2}}}
\mathcal{F} = \mathcal{U} \times \mathcal{Z}
+ \mathcal{Z} \times \textsc{MSET}_{\ge 2}(\mathcal{F}).$$
We first compute these trees by the number of nodes represented by
$\mathcal{Z}$ classified by the number of leaves and then extract the
coefficient on the number of leaves, which are marked here with
$\mathcal{U}.$ Translating to generating functions and using the
exponential formula for the multiset operator we find
$$F(z,u) = uz +
z \left(
\exp\left(\sum_{\ell\ge 1} \frac{F(z^\ell, u^\ell)}{\ell} \right)
- 1 - F(z,u) \right).$$
We introduce $F_n(u) = [z^n] F(z,u)$ to get
$$F(z,u) = uz +
z \left(
\exp\left(\sum_{\ell\ge 1} \frac{1}{\ell}
\sum_{q\ge 0} F_q(u^\ell) z^{q\ell} \right)
- 1 - F(z,u) \right).$$
Differentiating with respect to $z$ we obtain
$$F'(z,u) = u + (F(z,u)-uz)/z \\ +
z \left(
\exp\left(\sum_{\ell\ge 1} \frac{F(z^\ell, u^\ell)}{\ell} \right)
\left(\sum_{\ell\ge 1}
\sum_{q\ge 1} q F_q(u^\ell) z^{q\ell-1} \right)
- F'(z,u) \right).$$
This is
$$F'(z,u) = F(z,u)/z \\ +
z \left( ( (F(z,u)-uz)/z + 1 + F(z,u)) \times
\left(\sum_{\ell\ge 1}
\sum_{q\ge 1} q F_q(u^\ell) z^{q\ell-1} \right)
- F'(z,u) \right)$$
or alternatively
$$F'(z,u)/z - F(z,u)/z^2 + F'(z,u) \\ =
(1-u + F(z,u) + F(z,u)/z) \times
\left(\sum_{\ell\ge 1}
\sum_{q\ge 1} q F_q(u^\ell) z^{q\ell-1} \right)
.$$
Extracting the coefficient on $z^{n-2}$ we obtain
for the LHS
$$n F_n(u) - F_n(u) + (n-1) F_{n-1}(u)
= (n-1) (F_n(u) + F_{n-1}(u)).$$
We get for the first piece on the RHS
$$(1-u) [z^{n-2}]
\left(\sum_{\ell=1}^{n-1}
\sum_{q\ge 1} q F_q(u^\ell) z^{q\ell-1} \right).$$
Here we must have $n-2 = q\ell-1$ or $n-1=q\ell.$ We find
$$(n-1) (1-u) \sum_{\ell|(n-1)}
\frac{1}{\ell} F_{(n-1)/\ell}(u^\ell).$$
We get for the second piece
$$\begin{align*} & [z^{n-2}]
\left(\sum_{\ell=1}^{n-1}
\sum_{q\ge 1} q F_q(u^\ell) z^{q\ell-1} \right)
(F(z,u) + F(z,u)/z)
\\ & = \sum_{\ell=1}^{n-1}
\sum_{q\ge 1} q F_q(u^\ell)
[z^{n-1-q\ell}] (F(z,u) + F(z,u)/z)
\\ & = \sum_{\ell=1}^{n-1}
\sum_{q=1}^{\lfloor (n-1)/\ell \rfloor} q F_q(u^\ell)
[z^{n-1-q\ell}] (F(z,u) + F(z,u)/z)
\\ & = \sum_{\ell=1}^{n-1}
\sum_{q=1}^{\lfloor (n-1)/\ell \rfloor} q F_q(u^\ell)
(F_{n-1-q\ell}(u) + F_{n-q\ell}(u)).\end{align*}$$
This gives the recurrence for $n\ge 2$ where $F_0(u) = 0$ and
$F_1(u) = u$:
$$\bbox[5px,border:2px solid #00A000]{
\begin{align*}
F_n(u) & = - F_{n-1}(u)
+ (1-u) \sum_{\ell|(n-1)}
\frac{1}{\ell} F_{(n-1)/\ell}(u^{\ell}) \\
& + \frac{1}{n-1} \sum_{\ell=1}^{n-1}
\sum_{q=1}^{\lfloor (n-1)/\ell \rfloor}
q F_q(u^\ell)
(F_{n-1-\ell q}(u)+F_{n-\ell q}(u)).
\end{align*}}$$
As an example we have
$$F_5(u) = u^4 + u^3.$$
The reader is invited to replicate these trees from the number of
leaves. We also have as another example
$$F_{10}(u) =
{u}^{9}+6\,{u}^{8}+16\,{u}^{7}+12\,{u}^{6}.$$
This says that e.g. there are sixteen of these trees on ten nodes
having seven leaves.
We get for the count of our trees the sequence
$\{F_n(1)\}_{n\ge 0}$ which is
$$0, 1, 0, 1, 1, 2, 3, 6, 10, 19, 35, 67, 127, 248, 482, 952,
\ldots $$
which points us to OEIS A001678 where a
considerable amount of material both theoretic and applied can be
found. We quote from the definition of the sequence which says it is
the "number of unordered rooted trees with $n$ nodes where nodes
cannot have out-degree $1$." This is precisely the meaning of the
combinatorial class we have used so we know we have the correct
answer.
Now to collect the trees with
$m$ leaves on some number of nodes
we first observe that we need at least
$m+1$ nodes (this gives the star
graph). On the other hand the maximum number of nodes happens in a full
binary tree on
$2m-1$ nodes. Therefore the number of trees with
$m$
leaves where
$m\ge 2$ is given by
$$G_m = \sum_{q=m+1}^{2m-1} [u^m] F_q(u).$$
We also have $G_1 = 1.$ If we desire a single formula for $m\ge 1$
and don't mind a zero term for $m\ge 2$ we may also use
$$G_m = \sum_{q=m}^{2m-1} [u^m] F_q(u).$$
This gives the sequence $\{G_m\}_{m\ge 1}$ which is
$$1, 1, 2, 5, 12, 33, 90, 261, 766, 2312, 7068, 21965, \ldots $$
Note that the diagram by OP for $m=5$ contains a duplicate (trees
number two and three). The sequence points us to OEIS
A000669 where we learn that this object is
known as a "series-reduced planted tree with $n$ leaves" and there is
much more.
Here is some Maple code for those who are interested in working
with these sequences. (The variable names originate with the MathJax
above.)
with(numtheory);
F :=
proc(n)
option remember;
local res, ell, q;
if n=0 then return 0 fi;
if n=1 then return u fi;
res :=
- F(n-1) +
(1-u)*add(1/ell*subs(u=u^ell, F((n-1)/ell)),
ell in divisors(n-1)) +
1/(n-1)*
add(add(q*subs(u=u^ell, F(q))*
(F(n-1-ell*q)+F(n-ell*q)),
q=1..floor((n-1)/ell)), ell=1..n-1);
expand(res);
end;
G := m -> local q; add(coeff(F(q), u, m), q=m..2*m-1);
Best Answer
Your solutions presuppose that the cubes are identical; this is not the case in Instant Insanity. If they are all distinguishable you are missing a factor of $N!$.
The smaller possible solution is valid under the assumption that a stack is regarded as identical with that stack produced by turning the original stack upside down. The larger solution regards these two stacks as different.