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I don't know what else this may be called. But if the coefficients $r_{n,m}$ are written as lower triangular matrix $R$ and the coefficients $a_n$ as column-vector $A$ then this is an eigenvector-problem at eigenvalue $\lambda =1$ :

$ \qquad R*A=A*\lambda = A*1$

(I should note, that the formulation of the problem indicates, that the diagonal of $R$ at least from the second row on seems to be zero here because the sum-index goes only to rowindex $n$ minus 1)
Note 2: in the OEIS as well as in its associated electronic journal I've seen the term "eigensequence" and "invariant sequence" with the same meaning.

Example:

$ \qquad \small \begin{array} {rrrrrr} & & & & & | & 1 \\ & & & & & | & 2 \\ & R*&A =&A & & | & 7 \\ & & & & & | & 33 \\ & & & & & | & 201 \\ - & - & - & - & - & - & - \\ 1 & . & . & . & . & | & 1 \\ 2 & . & . & . & . & | & 2 \\ 1 & 3 & . & . & . & | & 7 \\ 1 & 2 & 4 & . & . & | & 33 \\ 2 & 3 & 4 & 5 & . & | & 201 \end{array} $

Here we have, for a rowindex $n$ : $ \qquad \sum_{m=1}^{n-1} R_{n,m}*A_m = A_n$
show/hide this revision's text 2 added example; added 9 characters in body

I don't know what else this may be called. But if the coefficients $r_{n,m}$ are written as lower triangular matrix $R$ and the coefficients $a_n$ as column-vector $A$ then this is an eigenvector-problem at eigenvalue $\lambda =1$ :

$ \qquad R*A=A*\lambda = A*1$

(I should note, that the formulation of the problem indicates, that the diagonal of $R$ at least from the second row on seems to be zero here because the sum-index goes only to rowindex $n$ minus 1)

Example:

$ \qquad \small \begin{array} {rrrrrr} & & & & & | & 1 \\ & & & & & | & 2 \\ & R*&A =&A & & | & 7 \\ & & & & & | & 33 \\ & & & & & | & 201 \\ - & - & - & - & - & - & - \\ 1 & . & . & . & . & | & 1 \\ 2 & . & . & . & . & | & 2 \\ 1 & 3 & . & . & . & | & 7 \\ 1 & 2 & 4 & . & . & | & 33 \\ 2 & 3 & 4 & 5 & . & | & 201 \end{array} $

Here we have, for a rowindex $n$ : $ \qquad \sum_{m=1}^{n-1} R_{n,m}*A_m = A_n$
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I don't know what else this may be called. But if the coefficients $r_{n,m}$ are written as lower triangular matrix $R$ and the coefficients $a_n$ as column-vector $A$ then this is an eigenvector-problem at eigenvalue $\lambda =1$ :

$ \qquad R*A=A*\lambda = A*1$

(I should note, that the formulation of the problem indicates, that the diagonal of $R$ at least from the second row on seems to be zero here because the sum-index goes only to rowindex $n$ minus 1)