A213822 Rectangular array: (row n) = b**c, where b(h) = 3*h-1, c(h) = 3*n-4+3*h, n>=1, h>=1, and ** = convolution.

4, 20, 10, 57, 41, 16, 124, 102, 62, 22, 230, 202, 147, 83, 28, 384, 350, 280, 192, 104, 34, 595, 555, 470, 358, 237, 125, 40, 872, 826, 726, 590, 436, 282, 146, 46, 1224, 1172, 1057, 897, 710, 514, 327, 167, 52, 1660, 1602

Offset

1,1

Comments

Principal diagonal: A213823.

Antidiagonal sums: A213824.

Row 1, (2,5,8,11,...)**(2,5,8,11,...): (3*k^3 + 3*k^2 + 2*k)/2.

Row 2, (2,5,8,11,...)**(5,8,11,14,...): (3*k^3 + 12*k^2 + 5*k)/2.

Row 3, (2,5,8,11,...)**(8,11,14,17,...): (3*k^3 + 21*k^2 + 8*k)/2.

For a guide to related arrays, see A212500.

Links

Clark Kimberling, Antidiagonals n = 1..80, 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-2) - (3*n-7)*x - (3*n-4)*x^2) and g(x) = (1-x)^4.

Example

Northwest corner (the array is read by falling antidiagonals):
4....20....57....124...230
10...41....102...202...350
16...62....147...280...470
22...83....192...358...590
28...104...237...436...710

Mathematica

b[n_]:=3n-1; c[n_]:=3n-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}] (* A213822 *)
Table[t[n, n], {n, 1, 40}] (* A213823 *)
s[n_]:=Sum[t[i, n+1-i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A213824 *)

Crossrefs

Cf. A212500.

Sequence in context: A118392 A263964 A180855 * A182456 A196380 A227997

Adjacent sequences: A213819 A213820 A213821 * A213823 A213824 A213825

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Jul 04 2012

Status

approved