A213750 Rectangular array: (row n) = b**c, where b(h) = h, c(h) = 2*(n-1+h)-1, n>=1, h>=1, and ** = convolution.

1, 5, 3, 14, 11, 5, 30, 26, 17, 7, 55, 50, 38, 23, 9, 91, 85, 70, 50, 29, 11, 140, 133, 115, 90, 62, 35, 13, 204, 196, 175, 145, 110, 74, 41, 15, 285, 276, 252, 217, 175, 130, 86, 47, 17, 385, 375, 348, 308, 259, 205, 150, 98, 53, 19, 506, 495, 465, 420

Offset: 1

Author

Clark Kimberling, Jun 20 2012

Keywords

• nonn
• tabl
• easy

Comments

Principal diagonal: A007585

Antidiagonal sums: A002417

row 1, (1,2,3,4,5,...)**(1,3,5,7,9,...): A000330

row 2, (1,2,3,4,5,...)**(3,5,7,9,...): A051925

row 3, (1,2,3,4,5,...)**(5,7,9,11,...): (2*k^3 + 15*k^2 + 13*k)/6

row 4, (1,2,3,4,5,...)**(7,9,11,13,...): (2*k^3 + 21*k^2 + 19*k)/6

For a guide to related arrays, see A213500.

Examples

			Northwest corner (the array is read by falling antidiagonals):
1....5....14...30....55....91
3....11...26...50....85....133
5....17...38...70....115...175
7....23...50...90....145...217
9....29...62...110...175...259
11...35...74...130...205...301
		

Crossrefs

Cf. A213500.

Programs

Mathematica

b[n_] := n; c[n_] := 2 n - 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}]  (* A213750 *)
d = Table[t[n, n], {n, 1, 40}] (* A007585 *)
s1 = Table[s[n], {n, 1, 50}] (* A002417 *)
FindLinearRecurrence[s1]
FindGeneratingFunction[s1, x]

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) = (2*n - 1) - (2*n - 3)*x and g(x) = (1 - x )^4.