A213777 Rectangular array: (row n) = b**c, where b(h) = F(h), c(h) = F(h+1), F=A000045 (Fibonacci numbers), n>=1, h>=1, and ** = convolution.

1, 3, 2, 7, 5, 3, 15, 12, 8, 5, 30, 25, 19, 13, 8, 58, 50, 40, 31, 21, 13, 109, 96, 80, 65, 50, 34, 21, 201, 180, 154, 130, 105, 81, 55, 34, 365, 331, 289, 250, 210, 170, 131, 89, 55, 655, 600, 532, 469, 404, 340, 275, 212, 144, 89, 1164, 1075, 965, 863

Offset

1,2

Comments

Principal diagonal: A001870

Antidiagonal sums: A152881

row 1, (1,1,2,3,5,8,...)**(1,2,3,5,8,13,...): A023610(k-1)

row 2, (1,1,2,3,5,8,...)**(2,3,5,8,13,21,...): A067331(k-1)

row 3, (1,1,2,3,5,8,...)**(3,5,8,13,21,34,...)

For a guide to related arrays, see A213500.

Links

Clark Kimberling, Antidiagonals n=1..80, flattened

Formula

T(n,k) = 2*T(n,k-1) + T(n,k-2) - 2*T(n,k-3) - T(n,k-4).

G.f. for row n: f(x)/g(x), where f(x) = F(n-1) + F(n-2)*x and g(x) = (1 - x - x^2)^2.

T(n,k) = (k*Lucas(n+k+1) + Lucas(n)*Fibonacci(k))/5. - Ehren Metcalfe, Jul 10 2019

Example

Northwest corner (the array is read by falling antidiagonals):
1....3....7....15....30....58
2....5....12...25....50....96
3....8....19...40....80....154
5....13...31...65....130...250
8....21...50...105...210...404
13...34...81...170...340...654

Mathematica

b[n_] := Fibonacci[n]; c[n_] := Fibonacci[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}]  (* A213777 *)
Table[t[n, n], {n, 1, 40}] (* A001870 *)
s[n_] := Sum[t[i, n + 1 - i], {i, 1, n}]
Table[s[n], {n, 1, 50}] (* A152881 *)

Crossrefs

Cf. A213500.

Sequence in context: A128140 A213579 A137225 * A118834 A255547 A087468

Adjacent sequences: A213774 A213775 A213776 * A213778 A213779 A213780

Keyword

nonn,tabl,easy

Author

Clark Kimberling, Jun 21 2012

Status

approved