# How to get this terrible summation/product to run in Mathematica?

I've come across this formula and have no idea where to even start. (My assumption here is that $m,n$ are known and input into the expression to arrive at an answer.)

$$f(m,n) = \sum_{\substack{0 \leq a_{1} \leq m \\ 0 \leq a_{2} \leq m \\ ... \\ 0 \leq a_{\lfloor \frac{n}{2} \rfloor} \leq m}} \lbrack (m-a_{1}+2)2^{a_{1}-2} \prod ^{\lfloor \frac{n}{2} \rfloor -1} _{i=1} (m-a_{i+1}+2)(a_{i}+1)2^{|a_{i+1}-a_{i}|-2} \rbrack$$

I don't even know where to begin on coding this. I'm stuck at multiple summation indices and haven't yet found a help thread that made sense to me. I'd appreciate any and all help.

• How exactly is this expected to work, since the a in the sum cover subscript to floor[n/2]-1, yet in the product a with subscript up to floor[n/2] is used? A link to the source would be nice. – ciao Jun 22 '16 at 4:31
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• My mistake, @ciao, it should be a subscript floor[n/2] in the sum. It's fixed now. – SAWblade Jun 23 '16 at 3:47
• But, still no reference on where you encountered this sum? Anyway: Table[Sum[(m - K[1] + 2) 2^(K[1] - 2) Product[(m - K[i + 1] + 2) (K[i] + 1) 2^(Abs[K[i + 1] - K[i]] - 2), {i, Quotient[n, 2] - 1}], Evaluate[Sequence @@ Table[{K[k], 0, m}, {k, Quotient[n, 2]}]]], {m, 2, 5}, {n, 2, 5}] – J. M.'s ennui Jun 23 '16 at 4:05
• @J.M. - yes, I'm more curious about part 1 - looks to be something combinatorial, but rings no bells. – ciao Jun 23 '16 at 4:40