A213558 Rectangular array: (row n) = b**c, where b(h) = h^3, c(h) = (n-1+h)^3, n>=1, h>=1, and ** = convolution.

1, 16, 8, 118, 91, 27, 560, 496, 280, 64, 2003, 1878, 1366, 637, 125, 5888, 5672, 4672, 2944, 1216, 216, 14988, 14645, 12917, 9542, 5446, 2071, 343, 34176, 33664, 30920, 25088, 17088, 9088, 3256, 512, 71445, 70716, 66620, 57359, 43535

Offset: 1

Author

Clark Kimberling, Jun 17 2012

Keywords

• nonn
• tabl
• easy

Comments

Principal diagonal: A213559

Antidiagonal sums: A213560

Row 1, (1,8,27,...)**(1,8,27,...): A145216

For a guide to related arrays, see A213500.

Examples

			Northwest corner (the array is read by falling antidiagonals):
1.....16.....118....560.....2003
8.....91.....496....1878....5672
27....280....1366...4672....12917
64....637....2944...9542....25088
125...1216...5446...17088...43535
		

Links

• Clark Kimberling, Antidiagonals n = 1..60, flattened
• Henri Muehle, Proper Mergings of Stars and Chains are Counted by Sums of Antidiagonals in Certain Convolution Arrays -- The Details, arXiv preprint arXiv:1301.1654, 2013.

Crossrefs

Cf. A213500.

Programs

Mathematica

b[n_] := n^3; c[n_] := n^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}]  (* A213558 *)
d = Table[t[n, n], {n, 1, 40}] (* A213559 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A213560 *)

Formula

T(n,k) = 8*T(n,k-1) - 28*T(n,k-2) + 56*T(n,k-3) - 70*T(n,k-4) + 56*T(n,k-5) - 28*T(n,k-6) + 8*T(n,k-7) - T(n,k-8).

G.f. for row n: f(x)/g(x), where f(x) = n^3 + ((n + 1)^3)*x + (-8*n^3 + 6*n^2 + 12*n + 8)*x^2 + (8*n^3 - 18*n^2 + 18)*x^3 - ((n - 2)^3)*x^4 - ((n + 1)^3)*x^5 and g(x) = (1 - x)^8.