Consider the number of integer partitions of $n$, usually denoted by $p(n)$ and generated by
$$\sum\_{n\geq0}p(n)x^n=\prod\_{k\geq1}\frac1{1-x^k}.$$


I have encountered an interesting enumeration.


Take all partitions of $n+2$ which do not use $1$ as a part. Then, count all different integers appearing in every such partition, call this number $u(n)$.


For example, say $n=4$ and hence $\{6,51,42,411,33,321,3111,222,2211,21111,111111\}$ are the partitions of $n+2=6$. Of these, $\{6,42,33,222\}$ avoid the integer $1$. The different integers here are $6,4,2,3,2$ for a total of $u(4)=5$.



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> **QUESTION.** Is it true that $u(n)=p(n)$ for all $n\geq1$?
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Number of partitions of $n$ and number of different integers in 1-avoiding partitions
Note that $u(n)=|\Theta(n+2)|$ where $\Theta(k)$ is the set of partitions of $k$ without 1's with one part being labelled (i.e., multisets of integers greater than 1 summing up to $k$ with one labelled element, like $\{{\bf 6}\},\{{\bf 4},2\},\{{\bf 2},4\},\{{\bf 3},3\}, \{{\bf 2},2,2\}$ for $k=6$).


Bijection from partitions of $n$ to $\Theta(n+2)$.


Take a partition $\lambda$ of $n$, let it have $s$ parts equal to 1. Remove them and add a labelled part $s+2$.


