constant function under convolution with 3 * 
$p$ is prime
Can someone show the intermediate steps, I don't understand the $1$st step even with the definition of a convolution in front of me.
Thank you :)
 A: I knew the straight formula $\underbrace{(1*1*...*1)}_{i}(p^k)=\binom{k+i-1}{k}$ but not the intermediates. $(1*1)(p^i)=\sigma_0(p^i)=(i+1)$ and $1(p^{k-i})$ is just $1$.
The first step is just the sum over divisors expressed another way:
$(1*1*1)(p^k)=\sum\limits_{d\mid p^k}(1*1)(d)\cdot (1)(\frac{p^k}{d})$ where divisors of $p^k$ are the $p^i$ 
A: 
We obtain
  \begin{align*}
\color{blue}{(\mathbf{1}\star\mathbf{1}\star\mathbf{1})\left(p^k\right)}
&=\sum_{d|p^k}\left(\mathbf{1}\star\mathbf{1}\right)(d)(\mathbf{1})\left(\frac{p^k}{d}\right)\tag{1}\\
&=\sum_{d|p^k}\left(\mathbf{1}\star\mathbf{1}\right)(d)\tag{2}\\
&=\sum_{j=0}^k(\mathbf{1}\star \mathbf{1})\left(p^j\right)\tag{3}\\
&=\sum_{j=0}^k\sum_{d|p^j}\left(\mathbf{1}\right)(d)\left(\mathbf{1}\right)\left(\frac{p^j}{d}\right)\tag{4}\\
&=\sum_{j=0}^k\sum_{i=0}^j\left(\mathbf{1}\right)\left(p^i\right)\left(\mathbf{1}\right)\left({p^{j-i}}\right)\tag{5}\\
&=\sum_{j=0}^k\sum_{i=0}^j1\tag{6}\\
&=\sum_{j=0}^k(j+1)\\
&=\sum_{j=1}^{k+1}j\\
&\,\,\color{blue}{=\frac{1}{2}(k+1)(k+2)}
\end{align*}
  and the claim follows.

Comment:


*

*In (1) and (4) we do the convolution according to the definition $(f\star g)(n)=\sum_{d|n}f(d)g\left(\frac{n}{d}\right)$.

*In (2) and (6) we apply the definition of the unit function $\mathbf{1}(n)=1, n\geq 1$.

*In (3) and (5) we note the positive divisors $d$ of $p^r$, $p$ prime are $p^j,0\leq j\leq r$.
