<s>[INST] Two interpretations of a sequence: an opportunity for combinatorics The sequence that is addressed here is resourced from the most useful site OEIS, listed as 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 $n$.

 [/INST] For what it's worth, here is a non-combinatorial proof.
That $\frac{1}{(1-x)^2}\prod_{k=1}^{\infty} \frac{1}{1-x^k}$ is the generating function for partitions of $n$ with three flavors of $1$'s is clear.
So I will focus on the sum of all the irredundant parts of partitions of $n$. Let me use $||\lambda||$ to denote this sum of irredundant parts (in contrast to $|\lambda|$ which is the usual size).
By basic combinatorial reasoning we have
$$\begin{align}\sum_{\lambda} q^{||\lambda||} x^{|\lambda|}  &= (1+qx+qx^2+\cdots)(1+q^2x^2+q^2x^4+\cdots)(1+q^3x^3+q^3x^6+\cdots)\cdots \\
&= \left(\frac{q}{1-x}+(1-q)\right)\left(\frac{q^2}{(1-x^2)}+(1-q^2)\right)\left(\frac{q^3}{(1-x^3)}+(1-q^3)\right)\cdots \\
&= \frac{1+x(q-1)}{(1-x)} \cdot \frac{1+x^2(q^2-1)}{(1-x^2)} \cdot \frac{1+x^3(q^3-1)}{(1-x^3)} \cdots \\
&= \frac{\prod_{k=1}(1+x^k(q^k-1))}{\prod_{k=1}^{\infty}(1-x^k)}.
\end{align} $$
Hence $\sum_{\lambda} ||\lambda|| \cdot x^{|\lambda|} $ is what we get from the above generating function by taking the derivative with respect to $q$ and setting $q:=1$. So let $f_k(q,x)=1+x^k(q^k-1)$. Being a little fast and loose with analytic issues, we can apply the "infinite product rule" to conclude
$$ 
\begin{align}
\sum_{\lambda} ||\lambda|| \cdot x^{|\lambda|} &= \frac{\sum_{k=1}^{\infty} \partial/\partial q f_k(x,q) \mid_{q=1} \cdot \prod_{i\neq k}f_i(x,q)\mid_{q=1}}{\prod_{k=1}^{\infty}(1-x^k)} \\
&=\frac{\sum_{k=1}^{\infty} (kx^kq^{k-1}) \mid_{q=1} \cdot \prod_{i\neq k}(1+x^i(q^i-1))\mid_{q=1}}{\prod_{k=1}^{\infty}(1-x^k)}\\
&= \frac{\sum_{k=1}^{\infty} kx^k}{\prod_{k=1}^{\infty}(1-x^k)} \\
&= \frac{x}{(1-x)^2} \cdot \prod_{k=1}^{\infty}\frac{1}{(1-x^k)},
\end{align}$$
exactly as claimed (with the extra power of $x$ reflecting the $n+1$ in $b(n+1)$).
</s>