A213838 Rectangular array: (row n) = b**c, where b(h) = 4*h-3, c(h) = 2*n-3+2*h, n>=1, h>=1, and ** = convolution.

1, 8, 3, 29, 20, 5, 72, 59, 32, 7, 145, 128, 89, 44, 9, 256, 235, 184, 119, 56, 11, 413, 388, 325, 240, 149, 68, 13, 624, 595, 520, 415, 296, 179, 80, 15, 897, 864, 777, 652, 505, 352, 209, 92, 17, 1240, 1203, 1104, 959, 784

Offset

1,2

Comments

Principal diagonal: A213839.

Antidiagonal sums: A213840.

Row 1, (1,5,9,13,...)**(1,3,5,7,...): A100178.

Row 2, (1,5,9,13,...)**(3,5,7,9,...): (4*k^3 + 9*k^2 - 4*k)/3.

Row 3, (1,5,9,13,...)**(5,7,9,11,...): (4*k^3 + 21*k^2 - 10*k)/3.

For a guide to related arrays, see A212500.

Links

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

Formula

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

G.f. for row n: f(x)/g(x), where f(x) = x*(2*n-1 + 4*n*x - (6*n-9)*x^2) and g(x) = (1-x)^4.

Example

Northwest corner (the array is read by falling antidiagonals):
1....8....29....72....145
3....20...59....128...235
5....32...89....184...325
7....44...119...240...415
9....56...149...296...505
11...68...179...352...595

Mathematica

b[n_]:=4n-3; c[n_]:=2n-1;
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}] (* A213838 *)
Table[t[n, n], {n, 1, 40}] (* A213839 *)
s[n_]:=Sum[t[i, n+1-i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A213840 *)

Crossrefs

Cf. A212500.

Sequence in context: A137481 A228886 A182160 * A248294 A370564 A004734

Adjacent sequences: A213835 A213836 A213837 * A213839 A213840 A213841

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Jul 05 2012

Status

approved