Counting the number of rooted trees given the distance distribution at each level For a tree with a given root, let $d_i$ be the number of vertices at distance $i$ from the root. Given the distance distribution $d =[d_1, d_2, \cdots, d_k]$ ($k$ is the diameter), how many (non-isomorphic) trees can we construct with the same root?
Further, if we restrict the maximum degree of each vertex to be some constant $c$, how does the count change?
Even more, how many connected simple graphs can we construct such that at least one vertex has the distance distribution $d$.
MOTIVATION:
This is a problem in mathematical chemistry or  precisely  Inverse QSAR. I would chose $c$ to be 4 (maximum valency of atoms). Some function of distance distribution $d$ are used as "descriptor" of molecules (topological indices) and correlation are established between these descriptors and bioactivity. Now the problem is to design molecules (graph representations) that would correspond to the descriptors.
 A: Even with a distinguished root and insisting that all isomorphisms respect this
distinguished root, this will be a challenging enumeration.
For $k \leq 2$ the problem is straightforward: The count is the number of
partitions of $d_2$ into at most $d_1$ parts.  This has an upper bound of
$\binom{d_2+d_1 - 1}{d_1 - 1}$, call this quantity $q(d_2,d_1)$, and the literature doubtless has more
to say on the exact count.  Call this number $p(d_2,d_1)$, and let such a
partition be denoted by the vector $(n_1,n_2,\ldots, n_{d_1})$, where the
$n_i$ are in decreasing order and sum to $d_2$.  To handle $d_3$ will
involve a summation over all such partitions of products of terms like 
$p(k_i,n_i)$, where the $k_i$ sum up to $d_3$.  Except it won't be that
simple, as you need $k_i$ to be 0 when $n_i$ is 0, and you need to
identify certain counts when you have $n_i=n_{i+1}$, and so on.
Using $\prod q(d_{i+1},d_i)$ as an upper bound will likely be a weak
estimate, unless all the $d$'s are small, and even then I would compare
with a computer enumeration.
Of course $n(c,d_2,d_1)$ will often be much smaller than $p(d_2,d_1)$,
where this new count restricts the parts $n_i \lt c$, and there should be
some literature on $n()$ as well.  I still recommend computer enumeration
for cases of small diameter.
