A213836 Rectangular array: (row n) = b**c, where b(h) = 4*h-3, c(h) = n-1+h, n>=1, h>=1, and ** = convolution.

1, 7, 2, 22, 13, 3, 50, 37, 19, 4, 95, 78, 52, 25, 5, 161, 140, 106, 67, 31, 6, 252, 227, 185, 134, 82, 37, 7, 372, 343, 293, 230, 162, 97, 43, 8, 525, 492, 434, 359, 275, 190, 112, 49, 9, 715, 678, 612, 525, 425, 320, 218, 127

Offset

1,2

Comments

Principal diagonal: A213837.

Antidiagonal sums: A071238.

Row 1, (1,5,9,13,...)**(1,2,3,4,...): A002412.

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

Row 3, (1,5,9,13,...)**(3,4,5,6,...): (4*k^3 + 27*k^2 - 13*k)/6.

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

Example

Northwest corner (the array is read by falling antidiagonals):
1...7....22...50....95
2...13...37...78....140
3...19...52...106...185
4...25...67...134...230
5...31...82...162...275
6...37...97...190...320

Mathematica

b[n_]:=4n-3; 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}] (* A213836 *)
Table[t[n, n], {n, 1, 40}] (* A213837 *)
s[n_]:=Sum[t[i, n+1-i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A071238 *)

Crossrefs

Cf. A212500.

Sequence in context: A282362 A248282 A371214 * A340801 A338284 A269168

Adjacent sequences: A213833 A213834 A213835 * A213837 A213838 A213839

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Jul 04 2012

Status

approved