A213833 Rectangular array: (row n) = b**c, where b(h) = 3*h-2, c(h) = 2*n-3+2*h, n>=1, h>=1, and ** = convolution.

1, 7, 3, 24, 17, 5, 58, 48, 27, 7, 115, 102, 72, 37, 9, 201, 185, 146, 96, 47, 11, 322, 303, 255, 190, 120, 57, 13, 484, 462, 405, 325, 234, 144, 67, 15, 693, 668, 602, 507, 395, 278, 168, 77, 17, 955, 927, 852, 742, 609, 465

Offset

1,2

Comments

Principal diagonal: A103748.

Antidiagonal sums: A213834.

Row 1, (1,3,5,7,...)**(1,3,5,7,...): A081436.

Row 2, (1,3,5,7,...)**(3,5,7,9,...): A144640.

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

For a guide to related arrays, see A212500.

Links

Clark Kimberling, Antidiagonals n = 1..12, 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) + (2*n+1)*x - (4*n-6)*x^2) and g(x) = (1-x)^4.

Example

Northwest corner (the array is read by falling antidiagonals):
1....7....24....58....115
3....17...48....102...185
5....27...72....146...255
7....37...96....190...325
9....47...120...234...395
11...57...144...278...465

Mathematica

b[n_]:=3n-2; 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}] (* A213833 *)
Table[t[n, n], {n, 1, 40}] (* A130748 *)
s[n_]:=Sum[t[i, n+1-i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A213834 *)

Crossrefs

Cf. A212500.

Sequence in context: A098231 A104716 A209313 * A282806 A283378 A104727

Adjacent sequences: A213830 A213831 A213832 * A213834 A213835 A213836

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Jul 04 2012

Status

approved