A152972 A vector sequence with set row sum function: row(n)=-Product[3*k - 1, {k, 0, n}] and linear build up and decline function: f(n,m)=Floor[(m/n)*row(n)].
1, 1, 1, 1, 8, 1, 1, 39, 39, 1, 1, 110, 658, 110, 1, 1, 1232, 4927, 4927, 1232, 1, 1, 17453, 34906, 104720, 34906, 17453, 1, 1, 299200, 598400, 1196799, 1196799, 598400, 299200, 1, 1, 6021400, 12042800, 18064200, 24085598, 18064200, 12042800
Offset: 0
Examples
{1}, {1, 1}, {1, 8, 1}, {1, 39, 39, 1}, {1, 110, 658, 110, 1}, {1, 1232, 4927, 4927, 1232, 1}, {1, 17453, 34906, 104720, 34906, 17453, 1}, {1, 299200, 598400, 1196799, 1196799, 598400, 299200, 1}, {1, 6021400, 12042800, 18064200, 24085598, 18064200, 12042800, 6021400, 1}, {1, 139161244, 278322488, 417483733, 417483734, 417483734, 417483733, 278322488, 139161244, 1}, {1, 3632108480, 7264216960, 10896325440, 14528433920, -2, 14528433920, 10896325440, 7264216960, 3632108480, 1}
Programs
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Mathematica
Clear[v, n, row, f]; row[n_] = -Product[3*k - 1, {k, 0, 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[FullSimplify[v[n]], {n, 0, 10}]; Flatten[%]
Formula
row(n)=(2*n)!/n!: f(n,m)=Floor[(m/n)*row(n)].
Comments