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Let $p_n$ be the offspring distribution for $B$ and define $q_n = \sum_{m\ge n} p_m (m)_n$, where $(m)_n = m!/(m-n)!$ is the descending factorial.

(new paragraph to appease MO latex scripts). Then the expected number of copies of a rooted tree $\theta$ inside the Galton Watson tree $\mathcal{T}$ is given by $\prod_{v\in\theta} q_{d_v}$ (where $d_v$ does not count the parent of $v$.

This is seen by induction on $\theta$. Given the root degree in $\mathcal{T}$ is $m$, the number of ways to select $d$ children of the root is $(m)_d$. For each of these, the expected number of ways to embed the sub-trees of $\theta$ is given by the formula (induction hypothesis). Since $\mathcal{T}$ is Galton-Watson, these are independent, and the expectation is the product of expectations.

This gives an identity $F(z) = Q(zF(z)$ for the generating function of the expected number of trees with weight $z$ for each edge. It seems that for nice $p$'s the singularity should have the same algebraic type, and so the expected number of trees in $\mathcal{T}$ grows as $C n^{-3/2} z_c^{-n}$.

In the case of the Poisson-Galton-Watson tree, it is easy to see either from the above or by staring into (probability) space that the expected number of copies of any tree $\theta$ is just $\lambda^{|\theta|}$ (still counting edges), so the expectation is just $\lambda^n C_n\sim Cn^{-3/2}(4\lambda)^n$.

Computing higher moments is probably doable in the Poisson case, but seems less fun. I will wait for additional motivation before delving into computations, but if staringinto space yields anything I'll report here.

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Let $p_n$ be the offspring distribution for $B$ and define $q_n = \sum_{m\ge n} p_m (m)_n$, where $(m)_n = m!/(m-n)!$ is the descending factorial.

(new paragraph to appease MO latex scripts). Then the expected number of copies of a rooted tree $\theta$ inside the Galton Watson tree $\mathcal{T}$ is given by $\prod_{v\in\theta} q_{d_v}$ (where $d_v$ does not count the parent of $v$.

This gives an identity $F(z) = Q(zF(z)$ for the generating function of the expected number of trees with weight $z$ for each edge. It seems that for nice $p$'s the singularity should have the same algebraic type, and so the expected number of trees in $\mathcal{T}$ grows as $C n^{-3/2} z_c^{-n}$.

In the case of the Poisson-Galton-Watson tree, it is easy to see either from the above or by staring into (probability) space that the expected number of copies of any tree $\theta$ is just $\lambda^{|\theta|}$ (still counting edges), so the expectation is just $\lambda^n C_n\sim Cn^{-3/2}(4\lambda)^n$.

Computing higher moments is probably doable in the Poisson case, but seems less fun. I will wait for additional motivation before delving into computations, but if staringinto space yields anything I'll report here.