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A157179
A new general triangle sequence based on the Eulerian form in three parts ( subtraction):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)].
0
1, 1, 1, 1, 3, 1, 1, 9, 9, 1, 1, 23, 50, 23, 1, 1, 53, 236, 236, 53, 1, 1, 115, 983, 1822, 983, 115, 1, 1, 241, 3723, 11995, 11995, 3723, 241, 1, 1, 495, 13168, 70369, 117534, 70369, 13168, 495, 1, 1, 1005, 44382, 377918, 997974, 997974, 377918, 44382, 1005, 1, 1
OFFSET
0,5
COMMENTS
Row sums are:
{1, 2, 5, 20, 98, 580, 4020, 31920, 285600, 2842560, 31147200,...}.
The m=0 of the general sequence is A008518.
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, 3, 1},
{1, 9, 9, 1},
{1, 23, 50, 23, 1},
{1, 53, 236, 236, 53, 1},
{1, 115, 983, 1822, 983, 115, 1},
{1, 241, 3723, 11995, 11995, 3723, 241, 1},
{1, 495, 13168, 70369, 117534, 70369, 13168, 495, 1},
{1, 1005, 44382, 377918, 997974, 997974, 377918, 44382, 1005, 1},
{1, 2027, 144605, 1896720, 7620498, 11819498, 7620498, 1896720, 144605, 2027, 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
KEYWORD
nonn,tabl
AUTHOR
STATUS
approved