Is there a way to find all number series whose formulae of general term contain progressions? Let $\{c_{m,n}\}_{m,n\in\mathbb{N}}$ be known complex numbers. My question is, how to find all number series  $\{a_{n}\}_{n\in\mathbb{N}}$ such that
$$a_n=\sum_{m=0}^\infty c_{m,n}a_{m+n},~\forall n\in\mathbb{N}?$$
For some special cases, the question is easy. For example, if
$$c_{m,n}=\begin{cases}0,&m=0,\\
1.&m>0,\end{cases}$$
then
$$a_n=\sum_{m=1}^\infty a_{m+n},~\forall n\in\mathbb{N},$$
so
$$a_n=\sum_{m=1}^\infty a_{m+n}=a_{n+1}+\sum_{m=1}^\infty a_{m+n+1}=a_{n+1}+a_{n+1}=2a_{n+1},~\forall n\in\mathbb{N}$$
and we can easily conclude that
$$a_{n}=\frac{a_0}{2^n},~\forall n\in\mathbb{N}.$$
In general the simplification like above does not work, e.g., maybe the following problem is hard to solve:
$$a_n=\sum_{m=0}^\infty \binom{m+n}{m}a_{m+n},~\forall n\in\mathbb{N}.$$
So is there any way to tackle all such problems? Or like Diophantine equations, only  special cases can be tackled?
 A: (Some basic functional analysis is assumed in this answer).
Formally, you are asking for the fixed points (i.e. eigenvalue-1 eigenvectors) of the infinite upper triangular matrix
$$
T:=\begin{bmatrix}c_{01}&c_{11}&c_{21}&c_{31}&\cdots\\0&c_{02}&c_{12}&c_{22}&\cdots\\0&0&c_{03}&c_{13}&\cdots\\0&0&0&c_{04}&\cdots\\\vdots&\vdots&\vdots&\vdots&\ddots \end{bmatrix}
$$
on the vector space $\mathbb{C}^{\aleph_0}$ of complex sequences. Equivalently, you are asking for a description of $\ker(T-I)$, where $I$ is the identity operator on $\mathbb{C}^{\aleph_0}$. However, $\mathbb{C}^{\aleph_0}$ is purely a vector space (with no choice topology and admissible basis) and can be unwieldy. Desirably, a norm $\|\cdot\|$ on suitable subspaces $\mathcal{X}\subsetneqq\mathbb{C}^{\aleph_0}$ that naturally admit as Schauder basis the vectors $\{e_k\}_{k\ge1}$, where $e_k=(\delta_{jk})_{j\ge1}$ and $\delta_{jk}$ is Kronecker delta symbol, will be much more convenient. Preferably, $T$ should be bounded (i.e. continuous) in $(\mathcal{X},\|\cdot\|)$. For instance, you may consider the sequence spaces $(\ell^p,\|\cdot\|_p)$, with the $p$-norm where $1\le p<\infty$, or generally the null-sequence space $(c_0,\|\cdot\|_\infty)$ with the supremum norm (or, more generally, the convergent sequence space $(c,\|\cdot\|_\infty)$). In particular, $\ker(T-I)=0$ if $T-I$ is invertible; for example if (the operator norm) $\|T\|_{\rm op}<1$ or, generally, if (the spectral radius) $r(T):=\lim_{n\to\infty}\|T^n\|_ {\rm op}^{1/n}<1$. Otherwise, we observe that
$${\rm diag}(T):=\{c_{0n}\}_{n\ge1}\subseteq\sigma_p(T):=\{{\rm eigenvalues~of~}T\}\,.$$
Hence, $\ker(T-I)\ne0$ if $1\in{\rm diag}(T)$; however, if $1\notin{\rm diag}(T)$, then $\dim\ker(T-I)\le1$.
To wit, as in your special case example, when $1\notin{\rm diag}(T)$, there’d essentially be a unique solution to your problem (up to scalar multiplication), which can be trivial (i.e. the null sequence). However, in your problem involving the binomial coefficients, ${\rm diag}(T)=\{1,1,1,\ldots\}$ as $c_{0n}=\binom{0+n}{0}=1$, which implies that $\ker(T-I)$ would have a solution in some suitable Banach sequence subspaces of $\mathbb{C}^{\aleph_0}$, but the solution space could be infinite-dimensional.
Hopefully, I will search to add a reference on determining the unique solution, when it is non-trivial (i.e. when ${\rm diag}(T)\not\ni1\in\sigma_p(T)$), or solutions in the general case when $1\in{\rm diag}(T)$.
