Return to Revisions

The result does not seem to be stated correctly. For instance, $\underline{\mathcal{P}_3}=\{ 3,21,1\}$, yet $p(0)+p(1)+p(2)=4$, where $p(n)=\#\mathcal{P}_n$. It is true that $$\sum_{\lambda\in\underline{\mathcal{P_n}}}\ell(\lambda)=p(0)+p(1)+\cdots p(n-1),$$ where $\ell(\lambda)$ is the number of parts of $\lambda$. This is equivalent to the case $k=1$ of Enumerative Combinatorics, vol. 1, second ed., Exercise 1.80, which has a simple combinatorial proof.