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Can't interpret a sum in Wolfram Alpha

Alexander Povolotsky
Posted 2 years ago

Wolframalpha is not capable to interpret the following syntax: 

Sum((C(n-1,n-i)*Sum(C(k+i,i)*C(n-1,n-k)),k=0..n)),i =0..n)

Could someone please help to (fix if necessary the above syntax) and obtain results for above.
POSTED BY: Alexander Povolotsky
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I am happy to report that the problem is solved in Mathematica by Christoph Koutschan by using his add-on package "HolonomicFunctions" (https://www3.risc.jku.at/research/combinat/software/ergosum/RISC/HolonomicFunctions.html.

The problem is solved by computing recurrences for both expressions and showing that they coincide.

This relates to https://oeis.org/A051708

Wolframalpha confirms that the recurrence obtained by Christoph Koutschan for both expressions

f[1]=2, f[2]=14, 9nf[n] + (-14 - 10n)f[1 + n] + (2 + n)*f[2 + n] == 0

gives n | f(n)

1 | 2

2 | 14

3 | 106

4 | 838

5 | 6802

6 | 56190

7 | 470010

...

which are correct values for https://oeis.org/A051708

So as the conclusion it follows that both:

the formula obtained byMariusz Iwaniuk (-1)^n((-4n^2 - 16n - 28)JacobiP[-1 + n, -1 - 2n, 2, -1/2] + JacobiP[-2 + n, -2n, 3, -1/2](3 + n)(-1 + n))4^n/(48(1 + n)*n)

and the recurrence obtained by Christoph Koutschan

$$ f[1]=2, f[2]=14, 9*n*f[n] + (-14 - 10*n)*f[1 + n] + (2 + n)*f[2 + n] == 0$$

rightfully belong to https://oeis.org/A051708

Thanks , Best Regards, Alexander R Povolotsky

Attachments:
Povolotsky.nb
POSTED BY: Alexander Povolotsky

Proofs.
POSTED BY: Mariusz Iwaniuk

I received this message from Joerg Arndt.

I am skeptical about the machine-generated (and IMO overly complicated) formula. There are several formulas here which are all human-readable!

Looks like my submissions was rejected in both cases :(
POSTED BY: Mariusz Iwaniuk

Dear Mariusz, Is your submission for your formula in A035002 ready to be proposed for review?

<< OEIS Server: This sequence has not been edited or commented on for a week yet is not proposed for review. If it is ready for review, please visit https://oeis.org/draft/A035002 and click the button that reads "These changes are ready for review by an OEIS Editor.">>
POSTED BY: Alexander Povolotsky

Dear Mariusz - thanks!
I hope that you will do me a great favor by submitting your closed form solutions to both A035002 and A051708 correspondingly.

Cheers,
Alex
POSTED BY: Alexander Povolotsky

No, W[n] and W2[m,n] are not close approximations. W[n] and W2[m,n] are exact solution.

For $n\in \mathbb{Z}$ and $m\in \mathbb{Z}$ this values are removable singularity points that why I use approximate values to get a solution.

If you need a exact value then you need to use a Limit.

Table[Limit[W[n], n -> j], {j, 2, 5}]
(* {14, 106, 838, 6802} *)
Limit[W2[m, n], {m -> 5, n -> 9}](*nested limit *)
(*365696*)


FunctionSingularities[W[n], n]

(*(n >= -(1/2) && Sin[(-(1/2) - n) \[Pi]] == 0) || 
 n == 0 || (n >= 2 && Sin[(2 - n) \[Pi]] == 0) || (n <= -2 && 
   Sin[(2 + n) \[Pi]] == 0) || (n <= 1/2 && 
   Sin[(-(1/2) + n) \[Pi]] == 0) || Gamma[2 - n] == 0 || 
 Gamma[2 + n] == 0 || (Sin[(1 - 2 n) \[Pi]] == 0 && Re[n] >= 1/2) || 
 Sin[2 n \[Pi]] == 
  0 || (Re[n] >= 1 && Sin[(2 - 2 n) \[Pi]] == 0) || (Re[n] <= -(1/2) &&
    Sin[(1 + 2 n) \[Pi]] == 0) || (Sin[2 (1 + n) \[Pi]] == 0 && 
   1 + Re[n] <= 0)*)

Regards M.I.
POSTED BY: Mariusz Iwaniuk

Thanks! So as have understood your reply W[n] and W2[m, n_] are close approximations ?
POSTED BY: Alexander Povolotsky
POSTED BY: Mariusz Iwaniuk

@Bill Nelson - Your information is absolutely correct. Thank you!

Please note that CROSSREF section of A051708 says: "Main diagonal of the square array given in A035002"

That will be A035002(n,n) while general formula for A035002 (see FORMULA section there) is:

A035002(n,m) = Sum_{i=0..m} C(m-1,m-i)Sum_{k=0..n} C(k+i,i)C(n-1,n-k)

So the formula which I have asked about here (which is A035002(n,n)) comes out by the substituting "m" with "n"...
POSTED BY: Alexander Povolotsky
Bill Nelson
Bill Nelson
Posted 2 years ago
POSTED BY: Bill Nelson
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