A152971 A vector sequence with set row sum function: row(n)=(2*n)!/n! and linear build up and decline function: f(n,m)=Floor[(m/n)*row(n)].

1, 1, 1, 1, 10, 1, 1, 59, 59, 1, 1, 210, 1258, 210, 1, 1, 3024, 12095, 12095, 3024, 1, 1, 55440, 110880, 332638, 110880, 55440, 1, 1, 1235520, 2471040, 4942079, 4942079, 2471040, 1235520, 1, 1, 32432400, 64864800, 97297200, 129729598, 97297200

Offset: 0

Author

Roger L. Bagula, Dec 16 2008

Keywords

• nonn
• tabl
• uned

Comments

row sums (2*n)!/n!:

{1, 2, 12, 120, 1680, 30240, 665280, 17297280, 518918400, 17643225600, 67044257280,...}

Examples

			{1},
{1, 1},
{1, 10, 1},
{1, 59, 59, 1},
{1, 210, 1258, 210, 1},
{1, 3024, 12095, 12095, 3024, 1},
{1, 55440, 110880, 332638, 110880, 55440, 1},
{1, 1235520, 2471040, 4942079, 4942079, 2471040, 1235520, 1},
{1, 32432400, 64864800, 97297200, 129729598, 97297200, 64864800, 32432400, 1},
{1, 980179200, 1960358400, 2940537600, 2940537599, 2940537599, 2940537600, 1960358400, 980179200, 1},
{1, 33522128640, 67044257280, 100566385920, 134088514560, -2, 134088514560, 100566385920, 67044257280, 33522128640, 1}
		

Crossrefs

A142459

Programs

Mathematica

Clear[v, n, row, f]; row[n_] = (2*n)!/n!;
f[n_, m_] = Floor[(m/n)*row[n]/2]; v[0] = {1}; v[1] = {1, 1};
v[n_] := v[n] = If[Mod[n, 2] == 0, Join[{1}, Table[ f[n, m], {m, 1, Floor[ n/2] - 1}], {row[n] - 2*Sum[ f[n, m], {m, 1, Floor[n/2] - 1}] - 2}, Table[ f[n, m], {m, Floor[n/ 2] - 1, 1, -1}], { 1}],
Join[{1}, Table[ f[n, m], {m, 1, Floor[n/2] - 1}], {row[n]/2 - Sum[ f[n, m], { m, 1, Floor[n/2] - 1}] - 1, row[n]/ 2 - Sum[ f[n, m], {m, 1, Floor[ n/2] - 1}] - 1}, Table[ f[n, m], {m, Floor[n/ 2] - 1, 1, -1}], {1}]];
Table[v[n], {n, 0, 10}]; Flatten[%]

Formula

row(n)=(2*n)!/n!: f(n,m)=Floor[(m/n)*row(n)].