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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.

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    $\begingroup$ 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. $\endgroup$ – ciao Jun 22 '16 at 4:31
  • $\begingroup$ Welcome to Mathematica.SE! I suggest the following: 1) As you receive help, try to give it too, by answering questions in your area of expertise. 2) Take the tour! 3) When you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge. Also, please remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign! $\endgroup$ – Michael E2 Jun 22 '16 at 4:46
  • $\begingroup$ My mistake, @ciao, it should be a subscript floor[n/2] in the sum. It's fixed now. $\endgroup$ – SAWblade Jun 23 '16 at 3:47
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    $\begingroup$ 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}] $\endgroup$ – J. M.'s ennui Jun 23 '16 at 4:05
  • $\begingroup$ @J.M. - yes, I'm more curious about part 1 - looks to be something combinatorial, but rings no bells. $\endgroup$ – ciao Jun 23 '16 at 4:40

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