A213819 Rectangular array: (row n) = b**c, where b(h) = h, c(h) = 3*n-4+3*h, n>=1, h>=1, and ** = convolution.

2, 9, 5, 24, 18, 8, 50, 42, 27, 11, 90, 80, 60, 36, 14, 147, 135, 110, 78, 45, 17, 224, 210, 180, 140, 96, 54, 20, 324, 308, 273, 225, 170, 114, 63, 23, 450, 432, 392, 336, 270, 200, 132, 72, 26, 605, 585, 540, 476, 399, 315

Offset

1,1

Comments

Principal diagonal: A213820.

Antidiagonal sums: A153978.

Row 1, (1,2,3,4,...)**(2,5,8,11,...): A006002.

Row 2, (1,2,3,4,...)**(5,8,11,14,...): is it the sequence A212343?.

Row 3, (1,2,3,4,...)**(8,11,14,17,...): (k^3 + 8*k^2 + 7*k)/2.

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(3*n-1 - (3*n-4)*x) and g(x) = (1-x)^4.

Example

Northwest corner (the array is read by falling antidiagonals):
2....9....24....50....90....147
5....18...42....80....135...210
8....27...60....110...180...273
11...36...78....140...225...336
14...45...96....170...270...399
17...54...114...200...315...462

Mathematica

b[n_]:=n; 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}] (* A213819 *)
Table[t[n, n], {n, 1, 40}] (* A213820 *)
d/2 (* A002414 *)
s[n_]:=Sum[t[i, n+1-i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A153978 *)
s1/2 (* A001296 *)

Crossrefs

Cf. A212500

Sequence in context: A021342 A069857 A076841 * A361013 A193088 A162916

Adjacent sequences: A213816 A213817 A213818 * A213820 A213821 A213822

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Jul 04 2012

Status

approved