What is the name for a matrix which is generated by a recursive sum whose form equals a recursive product when replacing the sums with products?

Question

In this answer to the question "Do these series converge to logarithms?" it is shown by George Lowther that each Dirichlet series involving the pattern of divisors converge to $\log(n)$ in row $n$. Jaume Oliver Lafont pointed out that this was originally(?) discovered by Lehmer in 1975, and several people on this forum have rediscovered this result.

The infinite table $T$ has the definition:

$$T(n,k) =\left\{ \begin{matrix} \text{ if }\; n|k \\ \text{ else } \end{matrix} \right|\begin{matrix} -(n-1) \\ +1 \end{matrix}$$

Starting:

$$\displaystyle T = \left( \begin{array}{ccccccc} +0&+0&+0&+0&+0&+0&+0&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&+1&+1&-3&+1&+1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&+1&+1&+1&+1&-5&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$

which has the said property $\displaystyle \log(n)=\sum\limits_{k=1}^{\infty}\frac{T(n,k)}{k}$$\;$

The sum recurrence for the (transposed) matrix $T$ is: $$T(n,1)=0$$ $$T(n,k)=\text{If } n\geq k \text{ then } \sum _{i=1}^{k-1} T(n-i,k-1)-\sum _{i=1}^{k-1} T(n-i,k) \text{ else } 1$$

The product recurrence for the (transposed) matrix $T$ is: $$T(n,1)=0$$

$$T(n,k)=\text{If } n\geq k \text{ then } \prod _{i=1}^{k-1} T(n-i,k-1)-\prod _{i=1}^{k-1} T(n-i,k) \text{ else } 1$$

As you see they are equivalent in the sense that only the sum symbols have been replaced with product symbols: $$\sum \rightarrow \prod$$

Question: What is the name for such a equivalence in form, of (recurrence) formulas?

I have been told that (at least) the sum recurrence is algebraic, but I don't know exactly what that means, other than that each column depends by recursion on the previous column.

Mathematica program to demonstrate the equivalence of the sum and product recurrences:

(*Mathematica start*)
"Sum recurrence:"
Clear[t, T, n, k, i];
nn = 8;
t[n_, 1] = If[n >= 1, 0, 0];
t[n_, k_] := 
 t[n, k] = 
  If[n >= k, (Sum[t[n - i, k - 1], {i, 1, k - 1}] - 
     Sum[t[n - i, k], {i, 1, k - 1}]), 1]
TableForm[T = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}]]
"Product recurrence:"
Clear[t, T, n, k, i];
nn = 8;
t[n_, 1] = If[n >= 1, 0, 0];
t[n_, k_] := 
 t[n, k] = 
  If[n >= k, (Product[t[n - i, k - 1], {i, 1, k - 1}] - 
     Product[t[n - i, k], {i, 1, k - 1}]), 1]
TableForm[T = Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}]]
(*Mathematica end*)