Now (sum from i=1 to k-1 of T(n-i,k-1)) is just the number of elements in {n-k+1,...,n-1) that are divisible by k-1. This is always equal to 1, because in k-1 consecutive integers, exactly one will be divisible by k-1.
On the other hand, (sum from i = 1 to k-1 of T(n-i,k)) is equal to either 0 or 1, because in k-1 consecutive integers, either 0 or 1 of them are divisible by k. We get 1 except if all are not divisible by k, which means the next integer /is/ divisible
Ok, by shifting the second columns recurrence index one step down, you can calculate the Möbius function as a matrix inverse from an arbitrary number sequence. But of course I told you this yesterday. @BalarkaSen
Interesting idea. I'd have to write it up later explicitly to understand what's he trying to say but I think the key is that $\sum_{d|n} \mu(d) = 0$ for $n > 1$ and $1$ otherwise. You'll have to start thinking about $A(1, 0, 0, \cdots )$. I can't imagine what shape it takes.
