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

0,5

Comments

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

Links

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

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)

Example

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;

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 *)

Prog

(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

Crossrefs

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

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

Sequence in context: A142596 A176063 A350060 * A152936 A152969 A138076

Adjacent sequences: A155464 A155465 A155466 * A155468 A155469 A155470

Keyword

nonn,tabl

Author

Roger L. Bagula, Jan 22 2009

Extensions

Edited by G. C. Greubel, Apr 01 2022

Status

approved