%I #8 Jan 18 2022 21:46:34
%S 1,1,1,3,4,1,15,21,7,1,105,148,52,10,1,945,1333,472,96,13,1,10395,
%T 14664,5197,1066,153,16,1,135135,190633,67567,13873,2009,223,19,1,
%U 2027025,2859496,1013512,208116,30170,3380,306,22,1
%N Triangle T(n,k) read by rows with T(n,0) = (2*n)! / (2^n * n!) for n >= 0 and T(n,k) = (Sum_{i=k..n} binomial(i-1,k-1) * 2^i * i! / (2*i)!) * (2*n)! / (2^n * n!) for 0 < k <= n.
%F T(n,n) = 1.
%F T(n,k) = binomial(n-1,k-1) + (2*n - 1) * T(n-1,k) for 0 < k < n.
%F Conjecture: M(n,k) = (-1)^(n-k) * T(n,k) is matrix inverse of A350512.
%e Triangle T(n,k) for 0 <= k <= n starts:
%e n\k : 0 1 2 3 4 5 6 7 8
%e =================================================================
%e 0 : 1
%e 1 : 1 1
%e 2 : 3 4 1
%e 3 : 15 21 7 1
%e 4 : 105 148 52 10 1
%e 5 : 945 1333 472 96 13 1
%e 6 : 10395 14664 5197 1066 153 16 1
%e 7 : 135135 190633 67567 13873 2009 223 19 1
%e 8 : 2027025 2859496 1013512 208116 30170 3380 306 22 1
%e etc.
%t Flatten[Table[If[k==0,(2n)!/(2^n n!),Sum[Binomial[i-1,k-1]2^i i!/(2i)!,{i,k,n}](2n)!/(2^n n!)],{n,0,8},{k,0,n}]] (* _Stefano Spezia_, Jan 06 2022 *)
%Y Cf. A001147 (column 0), A286286 (column 1), A249349 (column 2).
%Y Cf. A000007 (alternating row sums).
%Y Cf. A350512.
%K nonn,easy,tabl
%O 0,4
%A _Werner Schulte_, Jan 05 2022