The sequence that is addressed here is resourced from the most useful site OEIS, [listed as A014153](https://oeis.org/A014153), with a generating function
$$\frac1{(1-x)^2}\prod\_{k=1}^{\infty}\frac1{1-x^k}.$$
In particular, look at these two interpretations mentioned there.


$a(n)=$ Number of partitions of $n$ with three kinds of $1$.


**Example.** $a(2)=7$ because we have $2, 1+1, 1+1', 1+1'', 1'+1', 1'+1'', 1''+1''$.


$b(n)=$ Sum of parts, counted without multiplicity, in all partitions of $n$.


**Example.** $b(3)=7$ because the partitions are $3, 2+1, 1+1+1$ and removing repetitions leaves $3, 2+1, 1$. Adding these shows $b(3)=7$.


I could not resist asking:



> 
> **QUESTION.** Can you provide a combinatorial proof for $a(n)=b(n+1)$ for all $n$.
> 
> 
> 



Two interpretations of a sequence: an opportunity for combinatorics
Here is a bijective proof that equates both $a(n)$ and $b(n+1)$ to the quantity
$$p(n)+2p(n-1)+\cdots+np(1)+(n+1)p(0) \tag1$$
where $p(n)$ is the number of partitions of $n$.


For $a(n)$: each partition with three types of $1$'s corresponds to a partition of $k$ together with $(n-k)$ parts from $\{1',1''\}$. This second gadget can be chosen in $(n-k+1)$ ways, so we get $a(n)=\sum\_{k=0}^n p(k)(n-k+1)$.


For $b(n+1)$: Let us count the number of times that a summand "$k$" appears, for $1\le k\le n+1$. It only appears in partitions that have at least one part equal to $k$. The number of such partitions is $p(n+1-k)$. So the total sum calculating $b(n+1)$ is $\sum\_{k=1}^{n+1}kp(n+1-k)$.


**Remark.** The origin of (1) is this: $\frac1{1-x}\prod\_{k\geq1}\frac1{1-x^k}=\sum\_{m\geq0}\sum\_{k=0}^mp(k)x^n$ and hence $\frac1{(1-x)^2}\prod\_{k\geq}\frac1{1-x^k}=\sum\_{n\geq0}\sum\_{m=0}^n\sum\_{k=0}^mp(k)x^n$. Furthermore,
$$\sum\_{m=0}^n\sum\_{k=0}^mp(k)=(1).$$


