A157180 A new general triangle sequence based on the Eulerian form in three parts ( subtraction):m=2; 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, 4, 1, 1, 13, 13, 1, 1, 34, 78, 34, 1, 1, 79, 380, 380, 79, 1, 1, 172, 1607, 3040, 1607, 172, 1, 1, 361, 6135, 20383, 20383, 6135, 361, 1, 1, 742, 21796, 120826, 203830, 120826, 21796, 742, 1, 1, 1507, 73654, 652994, 1751524, 1751524, 652994, 73654
Offset: 0
Examples
{1},
{1, 1},
{1, 4, 1},
{1, 13, 13, 1},
{1, 34, 78, 34, 1},
{1, 79, 380, 380, 79, 1},
{1, 172, 1607, 3040, 1607, 172, 1},
{1, 361, 6135, 20383, 20383, 6135, 361, 1},
{1, 742, 21796, 120826, 203830, 120826, 21796, 742, 1},
{1, 1507, 73654, 652994, 1751524, 1751524, 652994, 73654, 1507, 1},
{1, 3040, 240357, 3290408, 13475450, 21018288, 13475450, 3290408, 240357, 3040, 1}
Crossrefs
Programs
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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}];
Formula
m=2;
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)].
Comments