A157181 A new general triangle sequence based on the Eulerian form in three parts ( subtraction):m=3; t0(n,k)=If[n*k == 0, 1, Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}]] t(n,k,m)=If[n == 0, 1, ( m*(n - k) + 1)*t0(n - 1 + 1, k - 1) + (m*k + 1)*t0(n - 1 + 1, k) - m*k*(n - k)*t0(n - 2 + 1, k - 1)].

1, 1, 1, 1, 5, 1, 1, 17, 17, 1, 1, 45, 106, 45, 1, 1, 105, 524, 524, 105, 1, 1, 229, 2231, 4258, 2231, 229, 1, 1, 481, 8547, 28771, 28771, 8547, 481, 1, 1, 989, 30424, 171283, 290126, 171283, 30424, 989, 1, 1, 2009, 102926, 928070, 2505074, 2505074, 928070

Offset

0,5

Comments

Row sums are:

{1, 2, 7, 36, 198, 1260, 9180, 75600, 695520, 7076160, 78926400,...}.

The m=0 of the general sequence is A008518.

Links

Table of n, a(n) for n=0..51.

Formula

m=3;

t0(n,k)=If[n*k == 0, 1, Sum[(-1)^j Binomial[n + 1, j](k + 1 - j)^n, {j, 0, k + 1}]];

t(n,k,m)=If[n == 0, 1, ( m*(n - k) + 1)*t0(n - 1 + 1, k - 1) +

(m*k + 1)*t0(n - 1 + 1, k) +

m*k*(n - k)*t0(n - 2 + 1, k - 1)].

Example

{1},
{1, 1},
{1, 5, 1},
{1, 17, 17, 1},
{1, 45, 106, 45, 1},
{1, 105, 524, 524, 105, 1},
{1, 229, 2231, 4258, 2231, 229, 1},
{1, 481, 8547, 28771, 28771, 8547, 481, 1},
{1, 989, 30424, 171283, 290126, 171283, 30424, 989, 1},
{1, 2009, 102926, 928070, 2505074, 2505074, 928070, 102926, 2009, 1},
{1, 4053, 336109, 4684096, 19330402, 30217078, 19330402, 4684096, 336109, 4053, 1}

Mathematica

Clear[t, n, k, m];
t[n_, k_, m_] = (m*(n - k) + 1)*Binomial[n - 1, k - 1] + (m*k + 1)*Binomial[n - 1, k] - m*k*(n - k)*Binomial[n - 2, k - 1];
Table[t[n, k, m], {m, 0, 10}, {n, 0, 10}, {k, 0, n}];
Table[Flatten[Table[Table[t[n, k, m], {k, 0, n}], {n, 0, 10}]], {m, 0, 10}]
Table[Table[Sum[t[n, k, m], {k, 0, n}], {n, 0, 10}], {m, 0, 10}];

Crossrefs

A008518

Sequence in context: A174159 A074060 A157637 * A347974 A029847 A154334

Adjacent sequences: A157178 A157179 A157180 * A157182 A157183 A157184

Keyword

nonn,tabl

Author

Roger L. Bagula and Gary W. Adamson, Feb 24 2009

Status

approved