3,4)*{\overbrace{\qquad}^i}};
			\endxy.  \]
			Using Theorem \ref{th:recursion planar}, the associated polynomials are $P(T)(t) = \frac{(-1)^i}{i+1} t^j(B_{i+1}(t)-B_{i+1})$ and $P(\widetilde{\Sigma}(T))(t) = \frac{(-1)^j}{j+1} t^i(B_{j+1}(t)-B_{j+1})$. Equating the corresponding Eulerian coefficients, we obtain the identity
			\[ \frac{1}{i+1} \sum_{k=1}^{i+1}(-1)^k \binom{i+1}{k} B_{i+1-k}B_{j+k} =  \frac{1}{j+1} \sum_{k=1}^{j+1}(-1)^k \binom{j+1}{k} B_{j+1-k}B_{i+k}.        \]
		\end{remark}
		As a final corollary to Theorem \ref{th:recursion planar}, we notice that 
		\begin{corollary}\label{cor:coefficients of tall trees} For the tall tree $T_{n+1} =  
			\left.  \xy
			{\ar@{-}(0,-8)*{\bullet};(0,-3)*{\circ}};
			{\ar@{.}(0,-3)*{\circ};(0,2)*{\circ}};
			{\ar@{-}(0,2)*{\circ};(0,7)*{\circ}};
			\endxy \right\}\mbox{\scriptsize $n+1$} $ with $(n+1)$ vertices, the associated polynomial and Eulerian coefficient are (where $(t)^k:= t(t+1)\cdots (t+k-1)$ is the rising factorial, $(t)^0:=1$)
			\[ P(T_{n+1})(t) = \frac{(t)^{\lceil \frac{n}{2}\rceil}(1-t)^{\lfloor \frac{n}{2}\rfloor} }{n!},\qquad\qquad E_{T_{n+1}} = \frac{\lfloor \frac{n}{2}\rfloor! \lceil\frac{n}{2}\rceil!}{(n+1)!}.  \]   
		\end{corollary} 
		\begin{example} For instance, \[ P(T_1)(t) = 1,\qquad P(T_2)(t)=t,\qquad P(T_3)(t)  = \frac{1}{2} t(1-t),\qquad P(T_4)(t)=\frac{1}{6} t(1-t)(1+t),  \]
			\[  P(T_5)(t)=\frac{1}{24} t(1-t)(1+t)(2-t),\qquad  P(T_6)(t)=\frac{1}{120} t(1-t)(1+t)(2-t)(2+t),\quad \ldots  \]
			\[E_{T_1}=1,\quad E_{T_2}=\frac{1}{2},\quad E_{T_3}=\frac{1}{6},\quad E_{T_4}=\frac{1}{12},\quad E_{T_5}=\frac{1}{30},\quad E_{T_6}=\frac{1}{60},\quad\ldots \]
		\end{example}
		\begin{proof} The statement about the polynomials follows inductively from \ref{th: product rule planar}, by showing that $\overrightarrow{\Delta}P(T_{n+1})(t)=P(T_n)(-t)$, where the $P(T_{n+1})(t)$ are as in the claim of the corollary and $\overrightarrow{\Delta}:\mathbb{K}[t]\to\mathbb{K}[t]$, $\overrightarrow{\Delta}p(t):=p(t+1)-p(t)$, is the forward difference operator. This is a direct computation, left to the reader (it is convenient to consider the cases $n=2m$ and $n=2m+1$ separately).
			
			To prove the claim about the coefficients, we look at the coefficient of $t$ in the polynomial \[P(T_{2m+2})(t)=\frac{1}{(2m+1)!}t(1-t)(1+t)\cdots(m-t)(m+t)=\frac{1}{(2m+1)!}t(1-t^2)\cdots(m^2-t^2).\]
By a straightforward computation, this is $\frac{(m!)^2}{(2m+1)!}$. On the other hand, Lemma \ref{lem:recuplanar} (or rather, its translation in the context of planar rooted trees) implies that this coefficient is precisely the Eulerian coefficient of the tall tree $T_{2m+1}$\footnote{More generally, Lemma \ref{lem:recuplanar} implies that the coefficient of $t$ in the polynomial $P\left(\xy {\ar@{-}(0,-3)*{\bullet};(0,1)};{(0,3)*{\slab{T}}};
				\endxy\right)(t)$ is precisely $E_T$: this follows from the fact that $P(\widehat{T_v})(t)$ is either $1$ if $\widehat{T_v}=\xy {\ar@{-}(0,-2);(0,2)}\endxy$ or it has no constant term if $\widehat{T_v}\neq \xy {\ar@{-}(0,-2);(0,2)}\endxy\,$.}, thus $E_{T_{2m+1}}=\frac{m!m!}{(2m+1)!}$, as desired. On the other hand, the above computation of $P(T_{2m+2})(t)$	shows that only odd powers of $t$ appear in it, and since the odd Bernoulli numbers with the exception of $B_1$ vanish, the Eulerian coefficient $E_{T_{2m+2}}=\left<\left. \frac{D}{1-e^{-D}}\right|P(T_{2m+2})(t)\right>$ is given by
			\[ E_{T_{2m+2}}= \frac{1}{2}\times\frac{m! m!}{(2m+1)!}=\frac{m!(m+1)!}{(2m+2)!}. \]
		\end{proof}
		
		We conclude this section by giving a combinatorial interpretation of the polynomial $P(T)(t)$ associated to a planar (although, as we know, the polynomial is independent of the planar structure) rooted tree $T\in\mathcal{T}_{\mathrm{p}}$, in the spirit of Theorem \ref{th: combinatorial} from Subsection \ref{subsection: log in sT}.
		
		\begin{definition}\label{def:alternating decoration}
			We call a decreasing decoration $l$ of $T$ (Definition \ref{def:decreasing decoration}) {\em alternating}
			if for all edges of $T$ the labels of the two endpoints have opposite parity.
			
			We denote the set of decreasing, alternating decorations of $T$, which associate the number $i$ to the root, by $\mathscr{D}_\mathrm{alt}(T,i)$.
		\end{definition}
		
		\begin{example}
			The following two decreasing decorations are alternating,
			$$
			\xy     {(-6,1)*{_0}};
			{(0,-7)*{_3}};
			{(6,1)*{_2}};
			{(6,6)*{_1}};
			{\ar@{-}(0,-4)*{\bullet};(-4,1)*{\circ}};
			{\ar@{-}(0,-4)*{\circ};(4,1)*{\circ}};
			{\ar@{-}(4,1)*{\circ};(4,6)*{\circ}};
			\endxy \qquad \qquad \qquad
			\xy
			{(0,-8)*{_7}};
			{(-8,-2)*{_4}};
			{(-12,4)*{_3}};
			{(-6,7)*{_3}};
			{(0,4)*{_1}};
			{(8,3)*{_3}};
			{(8,-2)*{_6}};
			{(3,11)*{_2}};
			{(9,11)*{_0}};		
			{\ar@{-}(0,-6)*{\bullet};(-6,-2)*{\circ}};
			{\ar@{-}(0,-6)*{\bullet};(6,-2)*{\circ}};
			{\ar@{-}(-6,-2)*{\circ};(-6,4)*{\circ}};
			{\ar@{-}(-6,-2)*{\circ};(-10,4)*{\circ}};
			{\ar@{-}(-6,-2)*{\circ};(-2,4)*{\circ}};
			{\ar@{-}(6,-2)*{\circ};(6,3)*{\circ}}