A213841 Rectangular array: (row n) = b**c, where b(h) = 2*h-1, c(h) = 4*n-7+4*h, n>=1, h>=1, and ** = convolution.

1, 8, 5, 29, 24, 9, 72, 65, 40, 13, 145, 136, 101, 56, 17, 256, 245, 200, 137, 72, 21, 413, 400, 345, 264, 173, 88, 25, 624, 609, 544, 445, 328, 209, 104, 29, 897, 880, 805, 688, 545, 392, 245, 120, 33, 1240, 1221, 1136, 1001

Offset

1,2

Comments

Principal diagonal: A213842.

Antidiagonal sums: A213843.

Row 1, (1,5,9,13,...)**(1,3,5,7,...): A100178.

Row 2, (1,5,9,13,...)**(3,5,7,9,...): (4*k^3 + 9*k^2 + 2*k)/3.

Row 3, (1,5,9,13,...)**(5,7,9,11,...): (4*k^3 + 21*k^2 + 2*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*(4*n-3 + 4*x - (4*n-7)*x^2) and g(x) = (1-x)^4.

Example

Northwest corner (the array is read by falling antidiagonals):
1....8....29....72....145
5....24...65....136...245
9....40...101...200...345
13...56...137...264...445
17...72...173...328...545
21...88...209...392...645

Mathematica

b[n_]:=2n-1; c[n_]:=4n-3;
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}] (* A213841 *)
Table[t[n, n], {n, 1, 40}] (* A213842 *)
s[n_]:=Sum[t[i, n+1-i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A213843 *)

Crossrefs

Cf. A212500.

Sequence in context: A248292 A196123 A196120 * A271594 A272273 A272049

Adjacent sequences: A213838 A213839 A213840 * A213842 A213843 A213844

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Jul 05 2012

Status

approved