What is the number of rooted planar decreasing trees on n vertices? By planar, I mean embedded in the plane (I think sometimes the terms, ordered and plane, are used interchangeably with planar).
Assume the vertices are labeled 1,2,...,n.  By decreasing, I mean the labels on the vertices along a path from the root to any other vertex will decrease.  Note that this implies the root is labeled with the integer n. 
The journal article I am reading states that there are (2n - 3)*(2n - 1)***(3)*(1) such trees.  But the article does not give a derivation.
Can someone explain how to count these trees?  
 A: Let's call these  ordered decreasing trees as the term  planar is also
used  for the  cyclic  group acting  at  the root.   Now  we have  the
following  recursive combinatorial  construction. To  assemble one  of
these we need a root node, which receives the label $n$ and an ordered
sequence of subtrees, each of some size ranging from one to $n-1$ with
a total of $n-1$ nodes (composition  into one to $n-1$ parts). We then
partition the remaining $n-1$ labels into an ordered sequence of sets,
one for each  subtree, having the matching number of  labels.  The key
observation  here  is  that  these  subtrees  say  of  some  size  $q$
correspond bijectively to ordered decreasing  trees on $q$ nodes where
the  elements of  the  set of  labels for  these  subtrees are  placed
according to the ordering induced by the source tree, which is ordered
and decreasing and has labels from $1$  to $q.$ E.g.  if we select the
labels $4,7,11$ for one of the subtrees then $4$ will replace $1$, $7$
will replace $2$ and $11$ will replace  $3$ in the source tree that is
being attached recursively.  At this point we win because  this is the
canonical  construction  that  supports   all  cartesian  products  of
exponential generating functions.
This yields for $n\ge 2$ (we have $T_1=1$) the recursive relation
$$T_n = \sum_{k=1}^{n-1} \sum_{q_1+q_2+\cdots+q_k = n-1}
{n-1\choose q_1, q_2, \ldots, q_k} \prod_{p=1}^k T_{q_p}.$$
These are  standard compositions with  no zero elements.  We introduce
the EGF as promised and obtain for
$$T(z) = \sum_{q\ge 1} T_q \frac{z^q}{q!}$$
with $n\ge 2$ the relation 
$$n! [z^n] T(z) = (n-1)! [z^{n-1}] \frac{T(z)}{1-T(z)}
\quad\text{or}\quad 
n [z^n] T(z) =  [z^{n-1}] \frac{T(z)}{1-T(z)}$$
which yields
$$[z^{n-1}] T'(z) =  [z^{n-1}] \frac{T(z)}{1-T(z)}.$$
Multiply by $z^{n-1}$ and sum over $n\ge 2$ to get
$$\sum_{n\ge 2} z^{n-1} [z^{n-1}] T'(z) =  
\sum_{n\ge 2} z^{n-1} [z^{n-1}] \frac{T(z)}{1-T(z)}.$$
Now $T'(z)$ has a constant coefficient which we miss on the left
while $T(z)/(1-T(z))$ does not and we find
$$T'(z) - 1 = \frac{T(z)}{1-T(z)}$$
so that
$$\bbox[5px,border:2px solid #00A000]{ T'(z) = \frac{1}{1-T(z)}.}$$
Solving this by separation of variables we get
$$-\frac{1}{2} (1 - T(z))^2 = z + C_1
\quad\text{or}\quad
T(z) = 1 - \sqrt{C_2-2z}.$$
Since $T(z)$ has no constant coefficient we obtain
$$\bbox[5px,border:2px solid #00A000]{ 
T(z) = 1 - \sqrt{1-2z}.}$$
Extracting coefficients from this we conclude with (again for $n\ge 2$)
$$n! [z^n] T(z) = - n! {1/2\choose n} (-1)^n 2^n
= - (1/2)^{\underline n} (-1)^n 2^n
\\ = (-1)^{n+1} 2^n \prod_{p=0}^{n-1} (1/2-p)
= (-1)^{n+1} \prod_{p=0}^{n-1} (1-2p)
= (-1)^{n-1} \prod_{p=1}^{n-1} (1-2p)
\\ = \prod_{p=1}^{n-1} (2p-1)$$
which is
$$\bbox[5px,border:2px solid #00A000]{ 
1\times 3\times\cdots\times (2n-3)}$$
and we have the claim.
 Readings.     This     set     of     notes     by     M.
Drmota    has   the
elementary combinatorial argument for this  as well as some additional
generating       functions.        We      also       find       OEIS
A001147 which  offers a  considerable number
of references. There is additional material  on page 531 of Flajolet /
Sedgewick (page number refers to PDF).
 Here is how I approached these concepts.

with(combinat);

T :=
proc(n)
    option remember;
    local k, comp, res;

    if n=1 then return 1 fi;

    res := 0;
    for k to n-1 do
        for comp in composition(n-1, k) do
            res := res +
            (n-1)!*mul(T(q)/q!, q in comp);
        od;
    od;

    res;
end;

A: T[n_] := T[n] = 
  Total[Map[Apply[Multinomial, #] Product[T[p], {p, #}] &, 
    Level[Map[Permutations[#] &, Partitions[n - 1]], {2}]]]; Table[
 T[n], {n, 1, 20}]
1, 1, 3, 15, 105, 945, 10395, 135135, 2027025, 34459425, 654729075, \
13749310575, 316234143225,...
nn = 6; A[z_] := Sum[T[n] z^n/n!, {n, 1, nn}]; 
Range[0, nn]! CoefficientList[
   Series[1/(1 - u A[z]), {z, 0, nn}], {z, u}] // Grid
{"1", "0", "0", "0", "0", "0", "0"},
{"0", "1", "0", "0", "0", "0", "0"},
{"0", "1", "2", "0", "0", "0", "0"},
{"0", "3", "6", "6", "0", "0", "0"},
{"0", "15", "30", "36", "24", "0", "0"},
{"0", "105", "210", "270", "240", "120", "0"},
{"0", "945", "1890", "2520", "2520", "1800", "720"}
