A155467 Triangle T(n, k) = Eulerian(n+1, k)*Binomial(n+1, k)/(k+1), read by rows.

1, 1, 1, 1, 6, 1, 1, 22, 22, 1, 1, 65, 220, 65, 1, 1, 171, 1510, 1510, 171, 1, 1, 420, 8337, 21140, 8337, 420, 1, 1, 988, 40068, 218666, 218666, 40068, 988, 1, 1, 2259, 175296, 1852914, 3935988, 1852914, 175296, 2259, 1, 1, 5065, 717600, 13655760, 55034868, 55034868, 13655760, 717600, 5065, 1

Offset: 0

Author

Roger L. Bagula, Jan 22 2009

Keywords

• nonn
• tabl

Comments

The sequence substitutes Eulerian numbers for the binomial in a triangle of Narayana numbers A001263,

Examples

			Triangle begins as:
  1;
  1,    1;
  1,    6,      1;
  1,   22,     22,        1;
  1,   65,    220,       65,        1;
  1,  171,   1510,     1510,      171,        1;
  1,  420,   8337,    21140,     8337,      420,        1;
  1,  988,  40068,   218666,   218666,    40068,      988,      1;
  1, 2259, 175296,  1852914,  3935988,  1852914,   175296,   2259,    1;
  1, 5065, 717600, 13655760, 55034868, 55034868, 13655760, 717600, 5065, 1;
		

Links

• G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Crossrefs

Cf. A001263 (m=0), this sequence (m=1), A155491 (m=3), A155493 (m=4).

Cf. A001263, A008292, A099765 (row sums).

Programs

Mathematica

(* First program *)
Needs["Combinatorica`"]
T[n_, k_]:= Eulerian[n+1, k]*Binomial[n+1, k]/(k+1);
Table[T[n, k], {n,0,10}, {k,0,n}]//Flatten (* Roger L. Bagula, Apr 14 2010 *)
(* Second program *)
t[n_, k_, m_]:= t[n,k,m]= If[k==1 || k==n, 1, (m*n-m*k+1)*t[n-1,k-1,m] + (m*k -(m -1))*t[n-1,k,m]];
T[n_, k_, m_]:= Binomial[n+1,k]*t[n+1,k+1,m]/(k+1);
Table[T[n,k,1], {n,0,12}, {k,0,n}]//Flatten (* G. C. Greubel, Apr 01 2022 *)

Sage

@CachedFunction
def t(n,k,m):
    if (k==1 or k==n): return 1
    else: return (m*(n-k)+1)*t(n-1,k-1,m) + (m*k-m+1)*t(n-1,k,m)
def T(n,k,m): return binomial(n+1,k)*t(n+1,k+1,m)/(k+1)
flatten([[T(n,k,1) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Apr 01 2022

Formula

T(n, k) = binomial(n+1, k)*t(n, k, m)/(k+1), where t(n,k,m) = (m*(n-k)+1)*t(n-1,k-1,m) + (m*k-m+1)*t(n-1,k,m), t(n,1,m) = t(n,n,m) = 1, and m = 1.

Sum_{k=0..n} T(n, k) = A099765(n+2).

T(n, k) = Eulerian(n+1, k)*Binomial(n+1, k)/(k+1). - Roger L. Bagula, Apr 14 2010

From G. C. Greubel, Apr 01 2022: (Start)

T(n, k) = binomial(n+1, k)*A008292(n+1, k+1)/(k+1).

T(n, n-k) = T(n, k).

T(n, 1) = A003469(n). (End)

Extensions

Edited by G. C. Greubel, Apr 01 2022