Let $\mathcal{U}_n$ denote the set of unrooted unlabelled trees with $n$ edges. For $T\in\mathcal{U}_n$, let $1^{u_1}2^{u_2}\cdots n^{u_n}$ be its degree distribution, that is, $u_i=\#$ of vertices having $\deg i$ so that $\sum_i u_i=n+1$ amd $\sum_i iu_i=2n$, and let $\ell(T)$ be the number of labelled trees whose underlying tree is $T$.

QUESTION. Is this true? $$\sum_{T\in\mathcal{U}_n}\, \ell(T)\cdot 1!^{u_1}2!^{u_2}\cdots n!^{u_n}=n!\cdot \frac1{2n+1}\binom{3n}n.$$

Remark. Hence, $n$-edge trees on the left correspond to $3n$-edge trees on the right.

Let $\mathcal{U}_n$ denote the set of unrooted unlabelled trees with $n$ edges. For $T\in\mathcal{U}_n$, let $1^{u_1}2^{u_2}\cdots n^{u_n}$ be its degree distribution, that is, $u_i=\#$ of vertices having $\deg i$ so that $\sum_i u_i=n+1$ amd $\sum_i iu_i=2n$, and let $\ell(T)$ be the number of labelled trees whose underlying tree is $T$.

QUESTION. Is this true? $$\sum_{T\in\mathcal{U}_n}\, \ell(T)\cdot 1!^{u_1}2!^{u_2}\cdots n!^{u_n}=n!\cdot \frac1{2n+1}\binom{3n}n.$$

Remark. Hence, $n$-edge trees on the left correspond to $3n$-edge trees on the right.

My thanks goes to Sam Hopkins, Ilya Bogdanov and Fedor Petrov for instructive and beautiful ideas for the variety of proofs. These help every reader.

Let $\mathcal{U}_n$ denote the set of unrooted unlabelled trees with $n$ edges. For $T\in\mathcal{U}_n$, let $1^{u_1}2^{u_2}\cdots n^{u_n}$ be its degree distribution, that is, $u_i=\#$ of vertices having $\deg i$ so that $\sum_i u_i=n+1$ amd $\sum_i iu_i=2n$, and let $\ell(T)$ be the number of labelled trees whose underlying tree is $T$.

QUESTION. Is this true? $$\sum_{T\in\mathcal{U}_n}\, \ell(T)\cdot 1!^{u_1}2!^{u_2}\cdots n!^{u)n}=n!\cdot \frac1{2n+1}\binom{3n}n.$$

Remark. Hence, $n$-edge trees on the left correspond to $3n$-edge trees on the right.

Let $\mathcal{U}_n$ denote the set of unrooted unlabelled trees with $n$ edges. For $T\in\mathcal{U}_n$, let $1^{u_1}2^{u_2}\cdots n^{u_n}$ be its degree distribution, that is, $u_i=\#$ of vertices having $\deg i$ so that $\sum_i u_i=n+1$ amd $\sum_i iu_i=2n$, and let $\ell(T)$ be the number of labelled trees whose underlying tree is $T$.

QUESTION. Is this true? $$\sum_{T\in\mathcal{U}_n}\, \ell(T)\cdot 1!^{u_1}2!^{u_2}\cdots n!^{u_n}=n!\cdot \frac1{2n+1}\binom{3n}n.$$

Remark. Hence, $n$-edge trees on the left correspond to $3n$-edge trees on the right.