A152938 A vector recursion designed around a factorial row sum : v(n)=if[odd,{1.n,n^2,...,(n+1)!/2-Sum[2^m,{m,0,n/2-1}],(n+1)!/2-Sum2^m,{m,0,n/2-1}],...n^2.n,1}],if[ even{1.n,n^2,...,(n+1)!-2Sum[2^m,{m,0,n/2-1}],...n^2.n,1}].

1, 1, 1, 1, 4, 1, 1, 11, 11, 1, 1, 4, 110, 4, 1, 1, 5, 354, 354, 5, 1, 1, 6, 36, 4954, 36, 6, 1, 1, 7, 49, 20103, 20103, 49, 7, 1, 1, 8, 64, 512, 361710, 512, 64, 8, 1, 1, 9, 81, 729, 1813580, 1813580, 729, 81, 9, 1, 1, 10, 100, 1000, 10000, 39894578, 10000, 1000, 100

Offset

0,5

Comments

Row sums are:

{1, 2, 6, 24, 120, 720, 5040, 40320, 362880, 3628800, 39916800,...}.

This designed symmetrical triangle is meant to be like the Eulerian numbers

in row sum ( the Stirling numbers of the first kind also have factorial row sums).

Links

Table of n, a(n) for n=0..63.

Formula

v(n)=if[odd,{1.n,n^2,...,(n+1)!/2-Sum[2^m,{m,0,n/2-1}],(n+1)!/2-Sum2^m,{m,0,n/2-1}],...n^2.n,1}],

if[ even{1.n,n^2,...,(n+1)!-2Sum[2^m,{m,0,n/2-1}],...n^2.n,1}].

Example

{1},
{1, 1},
{1, 4, 1},
{1, 11, 11, 1},
{1, 4, 110, 4, 1},
{1, 5, 354, 354, 5, 1},
{1, 6, 36, 4954, 36, 6, 1},
{1, 7, 49, 20103, 20103, 49, 7, 1},
{1, 8, 64, 512, 361710, 512, 64, 8, 1},
{1, 9, 81, 729, 1813580, 1813580, 729, 81, 9, 1},
{1, 10, 100, 1000, 10000, 39894578, 10000, 1000, 100, 10, 1}

Mathematica

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

Crossrefs

Sequence in context: A087903 A287532 A112500 * A154096 A146898 A152970

Adjacent sequences: A152935 A152936 A152937 * A152939 A152940 A152941

Keyword

nonn,tabl

Author

Roger L. Bagula, Dec 15 2008

Status

approved