A213761 Rectangular array: (row n) = b**c, where b(h) = h, c(h) = 3*n-5+3*h, n>=1, h>=1, and ** = convolution.

1, 6, 4, 18, 15, 7, 40, 36, 24, 10, 75, 70, 54, 33, 13, 126, 120, 100, 72, 42, 16, 196, 189, 165, 130, 90, 51, 19, 288, 280, 252, 210, 160, 108, 60, 22, 405, 396, 364, 315, 255, 190, 126, 69, 25, 550, 540, 504, 448, 378, 300

Offset

1,2

Comments

Principal diagonal: A172073.

Antidiagonal sums: A002419.

Row 1, (1,2,3,4,5,...)**(1,4,7,10,13,...): A002411.

Row 2, (1,2,3,4,5,...)**(4,7,10,13,16,...): A077414.

Row 3, (1,2,3,4,5,...)**(7,10,13,16,...): (k^3 + 7*k^2 + 6*k)/2.

Row 4, (1,2,3,4,5,...)**(10,13,16,...): (k^3 + 10*k^2 + 9*k)/2.

For a guide to related arrays, see A212500.

Links

Clark Kimberling, Antidiagonals n = 1..45, 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*(3*n - 2 - (3*n - 5)*x) and g(x) = (1 - x)^4.

Example

Northwest corner (the array is read by falling antidiagonals):
1....6....18...40....75....126
4....15...36...70....120...189
7....24...54...100...165...252
10...33...72...130...210...315
13...42...90...160...255...378

Mathematica

b[n_]:=n; c[n_]:=3n-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}] (* A213761 *)
Table[t[n, n], {n, 1, 40}] (* A172073 *)
s[n_]:=Sum[t[i, n+1-i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A002419 *)

Crossrefs

Cf. A212500.

Sequence in context: A171089 A368257 A180495 * A160248 A356044 A317858

Adjacent sequences: A213758 A213759 A213760 * A213762 A213763 A213764

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Jul 04 2012

Status

approved