A213551 Rectangular array: (row n) = b**c, where b(h) = h*(h+1)/2, c(h) = b(n-1+h), n>=1, h>=1, and ** = convolution.

1, 6, 3, 21, 15, 6, 56, 46, 28, 10, 126, 111, 81, 45, 15, 252, 231, 186, 126, 66, 21, 462, 434, 371, 281, 181, 91, 28, 792, 756, 672, 546, 396, 246, 120, 36, 1287, 1242, 1134, 966, 756, 531, 321, 153, 45, 2002, 1947, 1812, 1596, 1316, 1001, 686, 406

Offset

1,2

Comments

Principal diagonal: A213552

Antidiagonal sums: A051923

Row 1, (1,3,6,...)**(1,3,6,...): A000389

Row 2, (1,3,6,...)**(3,6,10,...): (k^5 + 15*k^4 + 85*k^3 + 165*k^2 + 94*k)/120

Row 3, (1,3,6,...)**(6,10,15,...): (k^5 + 20*k^4 + 155*k^3 + 340*k^2 + 204*k)/120

For a guide to related arrays, see A213500.

Links

Clark Kimberling, Antidiagonals n = 1..60, flattened

Formula

T(n,k) = 6*T(n,k-1) - 15*T(n,k-2) + 20*T(n,k-3) - 15*T(n,k-4) + 6*T(n,k-5) - T(n,k-6).

G.f. for row n: f(x)/g(x), where f(x) = n*(n+1) - 2*((n-1)^2)*x + 2*(n-1)*x^2 and g(x) = 2*(1 - x)^2.

Example

Northwest corner (the array is read by falling antidiagonals):
1....6....21....56....126....252
3....15...46....111...231....434
6....28...81....186...371....672
10...45...126...281...546....966
15...66...181...396...756....1316
21...91...246...531...1001...1722

Mathematica

b[n_] := n (n + 1)/2; c[n_] := n (n + 1)/2
t[n_, k_] := Sum[b[k - i] c[n + i], {i, 0, k - 1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n - k + 1, k], {n, 12}, {k, n, 1, -1}]]
r[n_] := Table[t[n, k], {k, 1, 60}]  (* A213551 *)
d = Table[t[n, n], {n, 1, 40}] (* A213552 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A051923 *)

Crossrefs

Cf. A213500.

Sequence in context: A281851 A282217 A213756 * A213753 A213747 A286203

Adjacent sequences: A213548 A213549 A213550 * A213552 A213553 A213554

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Jun 17 2012

Status

approved