A247309 Rectangular array read upwards by columns: T = T(n,k) = number of paths from (0,1) to (n,k), where 0 <= k <= 2, consisting of segments given by the vectors (1,1), (1,0), (1,-1), (1,-2).

1, 0, 0, 1, 1, 1, 2, 3, 3, 5, 8, 8, 13, 21, 21, 34, 55, 55, 89, 144, 144, 233, 377, 377, 610, 987, 987, 1597, 2584, 2584, 4181, 6765, 6765, 10946, 17711, 17711, 28657, 46368, 46368, 75025, 121393, 121393, 196418, 317811, 317811, 514229, 832040, 832040

Offset: 0

Author

Clark Kimberling, Sep 12 2014

Keywords

• nonn
• tabf
• easy

Comments

Every member of T is a Fibonacci number, and the sum of the numbers in column n is A000045(2n+2).

Examples

			First 10 columns:
0 .. 1 .. 3 .. 8 .. 21 .. 55 .. 144 .. 377 .. 987 ... 2584
0 .. 1 .. 3 .. 8 .. 21 .. 55 .. 144 .. 377 .. 987 ... 2584
1 .. 1 .. 2 .. 5 .. 13 .. 34 .. 89 ... 233 .. 610 ... 1597
T(2,2) counts these 3 paths, given as vector sums applied to (0,0):
(1,2) + (1,0); (1,1) + (1,1); (1,0) + (1,2).
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A247049, A247310, A000045.

Programs

Mathematica

t[0, 0] = 1; t[0, 1] = 0; t[0, 2] = 0; t[1, 2] = 1;
t[n_, 0] := t[n, 0] = t[n - 1, 0] + t[n - 1, 1];
t[n_, 1] := t[n, 1] = t[n - 1, 0] + t[n - 1, 1] + t[n - 1, 2];
t[n_, 2] := t[n, 2] = t[n - 1, 0] + t[n - 1, 1] + t[n - 1, 2]
TableForm[Reverse[Transpose[Table[t[n, k], {n, 0, 12}, {k, 0, 2}]]]] (*  array *)
Flatten[Table[t[n, k], {n, 0, 20}, {k, 0, 2}]] (*  A247309 *)

Formula

Let F = A000045 (Fibonacci numbers); then

(row 0, the bottom row) = (F(2n)), n >= 0;

(row 1, the middle row) = (F(2n)), n >= 0;

(row 2, the top row) = (F(2n-1)), n >= 0.

(n-th column sum) = (F(2n+2)), n >= 0.