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*)
