A157012 Riordan's general Eulerian recursion: T(n,k) = (k+2)*T(n-1, k) + (n-k) * T(n-1, k-1), with T(n,0) = 1, T(n,n) = 0.

1, 1, 0, 1, 1, 0, 1, 5, 1, 0, 1, 18, 14, 1, 0, 1, 58, 110, 33, 1, 0, 1, 179, 672, 495, 72, 1, 0, 1, 543, 3583, 5163, 1917, 151, 1, 0, 1, 1636, 17590, 43730, 32154, 6808, 310, 1, 0, 1, 4916, 81812, 324190, 411574, 176272, 22904, 629, 1, 0

Offset: 0

Author

Roger L. Bagula, Feb 21 2009

Keywords

• nonn
• tabl

Comments

Row sums are:

{1, 1, 2, 7, 34, 203, 1420, 11359, 102230, 1022299,...}.

This recursion set doesn't seem to produce the Eulerian 2nd A008517.

The Mathematica code gives ten sequences of which the first few are in the OEIS (see Crossrefs section). - G. C. Greubel, Feb 22 2019

Examples

			Triangle begins with:
1.
1,    0.
1,    1,     0.
1,    5,     1,      0.
1,   18,    14,      1,      0.
1,   58,   110,     33,      1,      0.
1,  179,   672,    495,     72,      1,     0.
1,  543,  3583,   5163,   1917,    151,     1,   0.
1, 1636, 17590,  43730,  32154,   6808,   310,   1,   0.
1, 4916, 81812, 324190, 411574, 176272, 22904, 629,   1,   0.
		

References

• J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 214-215

Links

• G. C. Greubel, Rows n=0..100 of triangle, flattened

Crossrefs

Cf. A008517.

Cf. A157011 (m=0), A008292 (m=1), This Sequence (m=2), A157013 (m=3).

Programs

Mathematica

e[n_, 0, m_]:= 1;
e[n_, k_, m_]:= 0 /; k >= n;
e[n_, k_, m_]:= (k+m)*e[n-1, k, m] +(n-k+1-m)*e[n-1, k-1, m];
Table[Flatten[Table[Table[e[n, k, m], {k,0,n-1}], {n,1,10}]], {m,0,10}]
T[n_, 0]:= 1; T[n_, n_]:= 0; T[n_, k_]:= T[n, k] = (k+2)*T[n-1, k] +(n-k) *T[n-1, k-1]; Table[T[n, k], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Feb 22 2019 *)

PARI

{T(n, k) = if(k==0, 1, if(k==n, 0, (k+2)*T(n-1, k) + (n-k)* T(n-1, k-1)))};
for(n=0, 12, for(k=0, n, print1(T(n, k), ", "))) \\ G. C. Greubel, Feb 22 2019

Sage

def T(n, k):
    if (k==0): return 1
    elif (k==n): return 0
    else: return (k+2)*T(n-1, k) + (n-k)* T(n-1, k-1)
[[T(n, k) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Feb 22 2019

Formula

e(n,k,m) = (k+m)*e(n-1, k, m) + (n-k+1-m)*e(n-1, k-1, m) for m=2.

T(n,k) = (k+2)*T(n-1, k) + (n-k)*T(n-1, k-1), with T(n,0) = 1, T(n,n) = 0. - G. C. Greubel, Feb 22 2019