Recurrence with Minimum
Let us prove that for $m \geq 1$,
$$
T(n,m) = \max(1, 2\lfloor \tfrac{m-1}{(n-1)!} \rfloor - 1).
$$
(This is the same as your formula.)

If $m \leq 2(n-1)!$ then by definition $T(n,m) = 1$, and indeed $\lfloor \tfrac{m-1}{(n-1)!} \rfloor \leq 1$, showing that our formula also gives the value $1$. If $m &gt; 2(n-1)!$, then 
$$
T(n,m) = \min_{\substack{a,b \geq 1 \\ ab \leq (n-1)!}} T(n-1,a) + T(n-1,b) + T(n-ab).
$$
Since $ab \leq (n-1)!$ also $a,b \leq (n-1)!$, and so $T(n-1,a) = T(n-1,b) = 1$. Therefore
$$
T(n,m) = 2 + \min_{\substack{a,b \geq 1 \\ ab \leq (n-1)!}} T(n,m-ab) = 2 + \min_{1 \leq c \leq (n-1)!} T(n,m-c).
$$
We can prove inductively that $T(n,m)$ is monotone nondecreasing, and so
$$
T(n,m) = 2 + T(n,m-(n-1)!).
$$
From here it's not too hard to prove the formula by induction.