A157177 A new general triangle sequence based on the Eulerian form in three parts:m=1; 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, 13, 13, 1, 1, 29, 82, 29, 1, 1, 61, 368, 368, 61, 1, 1, 125, 1399, 3010, 1399, 125, 1, 1, 253, 4863, 19243, 19243, 4863, 253, 1, 1, 509, 16048, 106099, 194846, 106099, 16048, 509, 1, 1, 1021, 51298, 532466, 1622734, 1622734, 532466, 51298

Offset

0,5

Comments

Row sums are:

{1, 2, 7, 28, 142, 860, 6060, 48720, 440160, 4415040, 48686400,...}.

The m=1 of the general sequence is A008518.

Links

Table of n, a(n) for n=0..52.

Formula

m=1;

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, 13, 13, 1},
{1, 29, 82, 29, 1},
{1, 61, 368, 368, 61, 1},
{1, 125, 1399, 3010, 1399, 125, 1},
{1, 253, 4863, 19243, 19243, 4863, 253, 1},
{1, 509, 16048, 106099, 194846, 106099, 16048, 509, 1},
{1, 1021, 51298, 532466, 1622734, 1622734, 532466, 51298, 1021, 1},
{1, 2045, 160669, 2510256, 11855730, 19628998, 11855730, 2510256, 160669, 2045, 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: A152654 A176487 A272644 * A298240 A299366 A299135

Adjacent sequences: A157174 A157175 A157176 * A157178 A157179 A157180

Keyword

nonn,tabl

Author

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

Status

approved