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Let be a recurrence for polynomials $R_{m,j}$

R[n_, k_] := 0
R[n_, k_] := (2 k + 1)*Binomial[2 k, k]*
   Sum[R[n, j]*Binomial[j, 2 k + 1]*(-1)^(j - 1)/(j - k)*
   BernoulliB[2 j - 2 k], {j, 2 k + 1, n}] /; 2 k + 1 <= n
R[n_, k_] := (2 n + 1)*Binomial[2 n, n] /; k == n; 
T[n_, k_] := Numerator[R[n, k]];

By the following command

Column[Table[T[n, k], {n, 0, 15}, {k, 0, n}], Center]

It produces the following set of numbers

{
 {{1}},
 {{1, 6}},
 {{1, 0, 30}},
 {{1, -14, 0, 140}},
 {{1, -120, 0, 0, 630}},
 {{1, -1386, 660, 0, 0, 2772}},
 {{1, -21840, 18018, 0, 0, 0, 12012}},
 {{1, -450054, 491400, -60060, 0, 0, 0, 51480}}
}

How to reverse it mirrorwise? I.e first column suppoe to be {1,6,30,140...} and continue similarly for each line/column

Note that initial function should be changed

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  • $\begingroup$ To be clear, what do you mean ty "Note that initial function should be changed"? $\endgroup$ – Somos Apr 7 at 2:17
  • $\begingroup$ @Somos Assume we want to have for each $m$, the $A_{m,0}=(2 n + 1)*\binom{2n}{n}$. To do that we just have to change variable $j$ in the recursion to $m-j$, so now for $A_{m,0}$ we get $A_{m,0}=(2 n + 1)*\binom{2n}{n}$. But this is not so simply to fix recurrence for $0<j \leq m$ $\endgroup$ – Petro Kolosov Apr 7 at 6:11
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use

Column[Reverse /@ Table[T[n, k], {n, 0, 15}, {k, 0, n}], Center]
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  • $\begingroup$ Note that initial function should be changed $\endgroup$ – Petro Kolosov Apr 6 at 17:54
  • $\begingroup$ @PetroKolosov I just added the Reverse/@ thing. You can do whatever you like with those functions $\endgroup$ – J42161217 Apr 6 at 17:57

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