Proofs.

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

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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.

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.

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.