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

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.

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.

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.">>

Proofs.