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0

After trying a little bit more, I was able to produce a working version of my compiled function: FC = Compile[{{part1, _Integer, 1}, {part2, _Integer, 1}, {a, _Integer}}, Module[{i1, j1, i2, j2, nnslist}, i1 = part1[[1]]; j1 = part1[[2]]; i2 = part2[[1]]; j2 = part2[[2]]; nnslist = nns[i1, j1]; p[[i1, j1, a]]*Sum[p[[i2, j2, c]], ...


3

Something like this ought to work for you: With[{m = 4}, Sum[F @@ Array[K, m], Evaluate[Sequence @@ Prepend[Table[{K[j + 1], 1, K[j] - 1}, {j, m - 1}], {K[1], 1, n - 1}]]]] which yields $$\sum_{\tt K[1]=1}^{\tt -1+n} \sum_{\tt K[2]=1}^{\tt -1+K[1]} \sum_{\tt K[3]=1}^{\tt -1+K[2]} \sum_{\tt ...


2

In the process of addressing question 85900, I noticed that the question above can be solved compactly as follows. The solution draws upon insights from the answers by Hector and ybeltukov. Series[2^n/Sum[2^i*Binomial[n - i - 1, 2*n/3 - 1], {i, 0, n/3}], {n, ∞, 0}] // Normal // FullSimplify[#, n > 0] & (* 2^(5 n/3) 3^-n Sqrt[n ...


4

The suggested answer to a similar question asked here just explains how adding a zero causes a formal overall shift by an integer in the summation range, which causes the Dirichlet regularization to become ambiguous. Indeed, it is not a problem with Mathematica but with the regularization technique. Luckily, this blog entry gives a much sturdier ...


5

For n even, the Sum in the question can be performed explicitly, Evaluate[Unevaluated[n*Sum[Binomial[2 n - 4 i, n - 2 i]*Binomial[n, 2 i]* Binomial[4 i, 2 i], {i, 0, Floor[n/2]}]/2^(3 n - 1)] /. n -> 2 m /. Floor[(2*m)/2] -> m] // Simplify (* 4^(1 - 3 m) m Binomial[4 m, 2 m] HypergeometricPFQ[ {1/4, 3/4, 1/2 - m, 1/2 - m, ...


1

Clear[expr] expr[n_Integer] := n*Sum[Binomial[2 n - 4 i, n - 2 i]*Binomial[n, 2 i]* Binomial[4 i, 2 i], {i, 0, Floor[n/2]}]/2^(3 n - 1); expr2 = expr[10000] // N 0.63662 RootApproximant[expr2*Pi]/Pi 2/Pi


2

expr1 = Sum[50/10000*1/4*1/4*n*97/100*Exp[-h1*1/4*n], {n, 1, 4}] - Sum[1/2*1/4*n*97/100*(Exp[-h1*1/4*(n - 1)] - Exp[-h1*1/4*n]), {n, 1, 4}]; expr2 = Sum[77/10000*1/4*1/4*n*97/100*Exp[-h1*1/4*n], {n, 1, 4}] + Sum[77/10000*1/4*1/4*n*94/100*Exp[-h2*1/4*n], {n, 5, 8}] - Sum[1/2*1/4*n*97/100*(Exp[-h1*1/4*(n - 1)] - Exp[-h1*1/4*n]), {n, 1, 4}] + ...


1

You wrote: Both left hand and right hand side of second expression is the sum of eight quantities So I guessed your Mathematica expression is wrong. You forgot a Plus sign here So, after correcting: Solve[.... == ... .. && ... + ... == ... + ...] // N {{h1 -> 0.00998752, h2 -> 0.017552}, {h1 -> 0.00998752, h2 -> 5.94171}} ...


1

As I noted in a comment, there are no Reals solutions to your equations. There are, however, 17 Complex solutions, obtained by deleting Reals from Solve. One such solution is, {h1 -> ConditionalExpression[-4 (2 I π C[1] - Log[401/400]), C[1] ∈ Integers], h2 -> ConditionalExpression[-4 (2 I π C[2] + Log[Root[958425069093800 - ...


1

I feel I've worked on a problem like this before. Maybe there is a similar question on the site, but I can't find it. It reminds me a little of Why does Mathematica simplify $x/x\to1$?, in which the symbolic result is generically true. Analysis The method used in the sum is Method -> "IteratedSummation" so we can examine the major steps: Sum[ ...


2

Here's one way: Total[Subscript[p, #] & /@ lst f @@@ lst] or Total[Subscript[p, #] & /@ lst f /@ lst] But you're probably making a mistake defining your p's this way, especially if you later need to do something with them. If you are willing to define p as a function, then Total[p @@@ lst f @@@ lst] would work.



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