What is the canonical name (or a good one) for the following self-composition of functions? In trying to answer (my own) recent question I'm looking for the name for the following type of self-composition of a function.


*

*Let's define an infinite set of fixed coefficients $\{a_0,a_1,...\}$            

*Let us assume the domain for the function in question $ \mathbb N $     

*Let $g(n)$ be some simple function on $n$, say for example $g(n)=n$      


Then let us define the self-composing of $f(n)$ with iteration-index denoted as $f^{\circ h}(n)$  by the following scheme:        
$$ \begin{array} {} f^{\circ 0}(n) &= g(n) \\
f^{\circ h}(0)&=0 \\
   f^{\circ h+1}(n) &= a_0 f^{\circ h}(n) + a_1 f^{\circ h}(n-1) + ... + a_{n-1} f^{\circ h}(1)\\ 
\end{array}$$
Example: if $a_0=a_1=...=a_k=...=1 $ and $g(n)=n$ then we have for $f^{\circ 1}(n)$ the sum-of-natural-numbers up to $n$, for $f^{\circ 2}(n)$ the sum-of-sums and so on.    

Q: Is there a canonical name for this type of self-composition?         

It's surely not "iteration"; I looked for "convolution" but this seems to be different - I'm without an idea what other term I could look for.
 A: This can be written in terms of convolutions: For two functions $u,v\colon\mathbb{Z}\rightarrow\mathbb{R}$ you can define the convolution product formally as
$$
u*v(k)=\sum_{i\in\mathbb{Z}}u(i)v(k-i).
$$
In general this series does not converge in any sense, but assuming that $u(j)=v(j)=0$ for $j<0$, it is actually a finite sum for every $k\in\mathbb{Z}$ and for $k<0$ we even have $u*v(k)=0$. This means that  the convolution restricts to product on $R:=\{u\colon\mathbb{Z}_{\ge0}\rightarrow\mathbb{R}\}$, in the sense that it makes $(R,+,*)$ a commutative ring. (Which is ismorphic to $\mathbb{R}[[x]]$, the ring of formal power series.)
Now in your case you have some $a\in R$ defined by $a(j)=a_j$ and start with an arbitrary $g\in R$. Then $f^{\circ 1}(k)=a*g(k)$ for $k\ge 1$ and $f^{\circ0}(0)=0.$ But if $v\in R$ satisfies $v(0)=0$ also $a*v(0)=0$, hence from there on it is just convolution: $f^{\circ 2}=a*f^{\circ 1}$ and so on. If you aready start with a function $g\in R$ with $g(0)=0$, then
$$
f^{\circ h}=(a*\dots*a)*g=:a^{*h}*g.
$$
