A130762 A fold back triangular sequence for A003991: symmetrical folding and addition of.

1, 4, 6, 4, 8, 12, 10, 16, 9, 12, 20, 24, 14, 24, 30, 16, 16, 28, 36, 40, 18, 32, 42, 48, 25, 20, 36, 48, 56, 60, 22, 40, 54, 64, 70, 36, 24, 44, 60, 72, 80, 84, 26, 48, 66, 80, 90, 96, 49, 28, 52, 72, 88, 100, 108, 112, 30, 56, 78, 96, 110, 120, 126, 64, 32, 60, 84, 104, 120

Offset: 1

Author

Roger L. Bagula, Jul 13 2007

Keywords

• nonn
• tabf
• uned

Comments

Row sum is still A000292. These sequences are related to heights of Cartan A_n groups. I spent half the night trying to get this algorithm to work and just barely got this to do what I was doing with ease by hand. The other fold back sequences were analogs for this.

Examples

			{1},
{4},
{6, 4},
{8, 12},
{10, 16, 9},
{12, 20, 24},
{14, 24, 30, 16},
{16, 28, 36, 40},
{18, 32, 42, 48, 25},
{20, 36, 48, 56, 60},
{22, 40, 54, 64, 70, 36},
{24, 44, 60, 72, 80, 84}
		

Crossrefs

Cf. A000292, A003991.

Programs

Mathematica

(* first A003991*) a = Table[Table[(n - i)*(1 + i), {i, 0, n - 1}], {n, 1, 20}]; (* then fold back from that*) Table[Table[If[ Mod[n, 2] == 1, a[[n]][[m]] + a[[n]][[Length[a[[n]]] - m]] - n, If[m - Floor[ n/2] == 0, (a[[n]][[m]] + a[[ n]][[Length[a[[n]]] - m]] - n)/ 2, a[[n]][[m]] + a[[n]][[Length[a[[n]]] - m]] - n]], {m, 1, Floor[n/ 2]}], {n, 1, Length[a]}]; Flatten[%]

Formula

a0(n,m) = (n-m)*(1+m) Doubling of all elements a(n,m)=2*a0(n,m)-> m->Floor[n/2] except for the uneven middle element on odd sequences in n.