A213771 Rectangular array: (row n) = b**c, where b(h) = 3*h-2, c(h) = n-1+h, n>=1, h>=1, and ** = convolution.

1, 6, 2, 18, 11, 3, 40, 30, 16, 4, 75, 62, 42, 21, 5, 126, 110, 84, 54, 26, 6, 196, 177, 145, 106, 66, 31, 7, 288, 266, 228, 180, 128, 78, 36, 8, 405, 380, 336, 279, 215, 150, 90, 41, 9, 550, 522, 472, 406, 330, 250, 172, 102, 46

Offset

1,2

Comments

Principal diagonal: A213772

Antidiagonal sums: A132117

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

Row 2, (1,4,7,10,...)**(2,3,4,5,...): A162260

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

Row 4, (1,2,3,4,5,...)**(10,13,16,...): (k^3 + 10*k^2 - 3*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*(n + (n+1)*x - (n+2)*x^2) and g(x) = (1 - x)^4.

Example

Northwest corner (the array is read by falling antidiagonals):
1....6....18...40....75....126
2....11...30...62....110...177
3....16...42...84....145...228
4....21...54...106...180...279
5....26...66...128...215...330

Mathematica

b[n_]:=3n-2; c[n_]:=n;
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}] (* A213771 *)
Table[t[n, n], {n, 1, 40}] (* A213772 *)
s[n_]:=Sum[t[i, n+1-i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A132117 *)

Crossrefs

Cf. A212500

Sequence in context: A136398 A228692 A368521 * A266711 A248269 A081778

Adjacent sequences: A213768 A213769 A213770 * A213772 A213773 A213774

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Jul 04 2012

Status

approved