A271697 Triangle read by rows, T(n,k) = Sum_{j=0..n} C(-j-1,-n-1)*E1(j,k), E1 the Eulerian numbers A173018, for n>=0 and 0<=k<=n.

1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 0, 1, 7, 1, 0, 0, 1, 21, 21, 1, 0, 0, 1, 51, 161, 51, 1, 0, 0, 1, 113, 813, 813, 113, 1, 0, 0, 1, 239, 3361, 7631, 3361, 239, 1, 0, 0, 1, 493, 12421, 53833, 53833, 12421, 493, 1, 0, 0, 1, 1003, 42865, 320107, 607009, 320107, 42865, 1003, 1, 0

Offset: 0

Author

Peter Luschny, Apr 12 2016

Keywords

• nonn
• tabl

Examples

			Triangle starts:
  1;
  0, 0;
  0, 1,   0;
  0, 1,   1,   0;
  0, 1,   7,   1,   0;
  0, 1,  21,  21,   1,   0;
  0, 1,  51, 161,  51,   1, 0;
  0, 1, 113, 813, 813, 113, 1, 0;
  ...
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..1325 (rows 0..50)
• Peter Luschny, Extensions of the binomial

Crossrefs

Variant: A046739 (main entry for this triangle).

Cf. A000166 (row sums), A122045 (Euler numbers are the alternating row sums), A070313 (col. 2) and (diag. n,n-2).

Cf. A173018.

T(2n,n) gives A320337.

Programs

Maple

A271697 := (n,k) -> add(binomial(-j-1,-n-1)*combinat:-eulerian1(j,k), j=0..n):
seq(seq(A271697(n, k), k=0..n), n=0..11);

Mathematica

<= 0, 0] = 1;
E1[n_, k_] /; k < 0 || k > n = 0;
E1[n_, k_] := E1[n, k] = (n-k) E1[n-1, k-1] + (k+1) E1[n-1, k];
T[n_, k_] := Sum[Binomial[-j-1, -n-1] E1[j, k], {j, 0, n}];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 29 2020 *)

PARI

T(n)={my(x='x+O('x^(n+1)), v=Vec(serlaplace((y-1)/(y*exp(x)-exp(x*y))))); vector(#v,n,Vecrev(v[n],n))}
{ my(A=T(10)); for(i=1, #A, print(A[i])) } \\ Andrew Howroyd, Nov 13 2024

Formula

T(n,k) = T(n,n-k). - Alois P. Heinz, Oct 29 2020