Summing up Unique Products A problem I have been working on recently requires sums similar to:
$6+5+4+3+2+1$
$3\cdot3+3\cdot2+3\cdot1+2\cdot2+2\cdot1+1\cdot1$
$7\cdot7\cdot7+7\cdot7\cdot6+...+7\cdot7\cdot1+7\cdot6\cdot6+...+2\cdot1\cdot1+1\cdot1\cdot1$
What I want to know is if there is an equation that could calculate these types of sums. I do have a way to do it with a recursive function:
$$S(a,b)=\sum_{i=1}^a{i\cdot S(i,b-1)}$$
where
$$S(x,0)=1$$ for any x
Going back to the previous examples:
$S(6,1)=6+5+4+3+2+1$
$S(3,2)=3\cdot3+3\cdot2+3\cdot1+2\cdot2+2\cdot1+1\cdot1$
However using this can be a little tedious at times, even with a computer.
EDIT:
If it helps any, what I am using this function for is for finding:
$$a! \cdot S(a,b)$$

I didn't manage to find a very satisfying result so far, maybe someone else can help. Note that $$S[a,1]=\frac{1}{2}a(a+1)$$
From the recursion formula, we can find the next few sums. Plugging this into Mathematica, shows we find a sequence of the form $$ S[a,b] = \sum_{i=1}^{a} \frac{(-1)^i i^{a+b}}{i!(a-i)!} $$
This can also be simplified to $$ S[a,b] =-\left(x \frac{\mathrm{d}}{\mathrm{d}x} \right)^{a+b-1} \frac{(-1)^{a-1}}{(a-1)!}\sum_{i=0}^{a-1} (-1)^{i+1} x^{i+1} {a-1 \choose i} \bigg|_{x=1} $$
Using the binomial expansion, we find
$$ S[a,b] =-\frac{(-1)^{a-1}}{(a-1)!} \left(x \frac{\mathrm{d}}{\mathrm{d}x} \right)^{a+b-1} x(1-x)^{a-1} \bigg|_{x=1} $$
which in turn can be simplified to
$$ S[a,b] = \frac{1}{(a-1)!} \left( \frac{\mathrm{d}}{\mathrm{d}t}\right)^{a+b-1}\left(e^t(e^t-1)^{a-1}\right) \bigg|_{t=0} $$
I believe this formula is correct, but I do not have a proof of the Mathematica step yet. You could try to prove it by induction.
Checking this formula for $$S[2,1] = \frac{\mathrm{d}}{\mathrm{d}t} \frac{\mathrm{d}}{\mathrm{d}t} (e^t(e^t-1)) \bigg|_{t=0} = \left(e^t(e^t-1) + 3e^{2t}\right)\bigg|_{t=0} = 3 $$
It also works for $S[3,2]$, for example, but it quickly becomes cumbersome to check.
