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Can't interpret a sum in Wolfram Alpha

Posted 1 year 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.

10 Replies
Posted 1 year ago

I tried several different ways with things like this:

Sum(Binomial(n-1,n-i)*Sum(Binomial(k+i,i)*Binomial(n-1,n-k),{k,0,n}),{i,0,n})

Unfortunately, none of those succeeded.

I finally used brute force calculation for n=0,n=1,...n=8 to get

{1, 2, 14, 106, 838, 6802, 56190, 470010, 3968310}

And I asked OEIS for all the information they had on this.

OEIS Link

which gives a collection of interesting places where this has been seen, information about the sequence, formulas to calculate it, etc.

Please check all this very carefully to find any of my mistakes before you even think about depending on this.

POSTED BY: Bill Nelson

@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"...

g[n_] := Sum[Binomial[n - 1, n - i]*
Sum[Binomial[k + i, i]*Binomial[n - 1, n - k], {k, 0, n}], {i, 0, 
n}];
Table[g[n], {n, 1, 15}]
(* {2, 14, 106, 838, 6802, 56190, 470010, 3968310, 33747490, 288654574, \
2480593546, 21400729382, 185239360178, 1607913963614, 13991107041306}*)

A051708 or Closed Form solution for the series works for: $n\in \mathbb{Z}$ and $n\geq 1$:

 W[n_] :=((-1)^n 4^(-2 + 
   n) ((-1 + n) (3 + n) JacobiP[-2 + n, -2 n, 3, -(1/2)] - 4 (7 + n (4 + n)) JacobiP[-1 + n, -1 - 2 n, 2, -(1/2)]))/(3 n (1 + n));
 Table[W[n], {n, 1, 15}]

 (*{2, 14, 106, 838, 6802, 56190, 470010, 3968310, 33747490, 288654574, \
 2480593546, 21400729382, 185239360178, 1607913963614, 13991107041306}*)

We can check(Code is Slow):

  g[n_] := 
     Sum[Binomial[n - 1, n - i]*
       Sum[Binomial[k + i, i]*Binomial[n - 1, n - k], {k, 0, n}], {i, 0, 
       n}];

  g1[n_] := ((-1)^
    n 4^(-2 + 
     n) ((-1 + n) (3 + n) JacobiP[-2 + n, -2 n, 3, -(1/2)] - 
      4 (7 + n (4 + n)) JacobiP[-1 + n, -1 - 2 n, 2, -(1/2)]))/(
   3 n (1 + n));
   Table[g[n] - g1[n], {n, 1, 200}]

   (*{0,0, and 0...}*)

Regards M.I.

POSTED BY: Mariusz Iwaniuk

Thanks! So as have understood your reply W[n] and W2[m, n_] are close approximations ?

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

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

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.">>

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

Proofs.

POSTED BY: Mariusz Iwaniuk

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

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