A213847 Rectangular array: (row n) = b**c, where b(h) = 4*h-1, c(h) = 2*n-3+2*h, n>=1, h>=1, and ** = convolution.

3, 16, 9, 47, 36, 15, 104, 89, 56, 21, 195, 176, 131, 76, 27, 328, 305, 248, 173, 96, 33, 511, 484, 415, 320, 215, 116, 39, 752, 721, 640, 525, 392, 257, 136, 45, 1059, 1024, 931, 796, 635, 464, 299, 156, 51, 1440, 1401

Offset

1,1

Comments

Principal diagonal: A213848.

Antidiagonal sums: A180324.

Row 1, (3,7,11,15,...)**(1,3,5,7,...): A172482.

Row 2, (3,7,11,15,...)**(3,5,7,9,...): (4*k^3 + 15*k^2 + 8*k)/3.

Row 3, (3,7,11,15,...)**(5,7,9,13,...): (4*k^3 + 27*k^2 + 14*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*(6*n-3 + 4*(n-2)x - (2*n-3)*x^2) and g(x) = (1-x)^4.

Example

Northwest corner (the array is read by falling antidiagonals):
3....16...47....104...195...328
9....36...89....176...305...484
15...56...131...248...415...640
21...76...173...320...525...796

Mathematica

b[n_]:=4n-1; 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}] (* A213847 *)
Table[t[n, n], {n, 1, 40}] (* A213848 *)
s[n_]:=Sum[t[i, n+1-i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A180324 *)

Crossrefs

Cf. A212500.

Sequence in context: A286021 A286649 A320543 * A195883 A272329 A076623

Adjacent sequences: A213844 A213845 A213846 * A213848 A213849 A213850

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Jul 05 2012

Status

approved