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Triangle read by rows, T(n,k) = binomial(-k-n-1, -2*n-1)*E1(k+n, n), E1 the Eulerian numbers A173018, for n >= 0 and 0 <= k <= n.
1

%I #19 Dec 31 2018 04:13:58

%S 1,0,1,0,-4,11,0,15,-156,302,0,-56,1596,-9528,15619,0,210,-14400,

%T 193185,-882340,1310354,0,-792,122265,-3213760,30042672,-116857368,

%U 162512286,0,3003,-1005004,47887840,-802069632,6034981134,-21078701112,27971176092

%N Triangle read by rows, T(n,k) = binomial(-k-n-1, -2*n-1)*E1(k+n, n), E1 the Eulerian numbers A173018, for n >= 0 and 0 <= k <= n.

%e Triangle starts:

%e 1;

%e 0, 1;

%e 0, -4, 11;

%e 0, 15, -156, 302;

%e 0, -56, 1596, -9528, 15619;

%e 0, 210, -14400, 193185, -882340, 1310354;

%e 0, -792, 122265, -3213760, 30042672, -116857368, 162512286;

%p T := (n, k) -> binomial(-k-n-1, -2*n-1)*combinat:-eulerian1(k+n, n):

%p for n from 0 to 7 do seq(T(n,k), k=0..n) od;

%t E1[n_ /; n >= 0, 0] = 1; E1[n_, k_] /; k < 0 || k > n = 0;

%t E1[n_, k_] := E1[n, k] = (n - k) E1[n - 1, k - 1] + (k + 1) E1[n - 1, k];

%t T[n_, k_] := Binomial[-k - n - 1, -2 n - 1] E1[n + k, n];

%t Table[T[n, k], {n, 0, 7}, {k, 0, n}] // Flatten

%t (* _Jean-François Alcover_, Dec 30 2018 *)

%Y Row sums give A320337.

%Y Cf. A046739, A180056 (main diagonal), A271697, A001791.

%K sign,tabl

%O 0,5

%A _Peter Luschny_, Dec 18 2018