A247049 Rectangular array read upwards by columns: T = T(n,k) = number of paths from (0,0) to (n,k), where 0 >= k <= 2, consisting of segments given by the vectors (1,1), (1,2), (1,-1).

1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 5, 5, 5, 8, 8, 8, 13, 13, 13, 21, 21, 21, 34, 34, 34, 55, 55, 55, 89, 89, 89, 144, 144, 144, 233, 233, 233, 377, 377, 377, 610, 610, 610, 987, 987, 987, 1597, 1597, 1597, 2584, 2584, 2584, 4181, 4181, 4181

Offset: 0

Author

Clark Kimberling, Sep 11 2014

Keywords

• nonn
• tabf
• easy

Comments

Also, T(n,k) = number of strings s(0)..s(n) of integers such that s(0) = 0, s(n) = k, and if i > 0, then s(i) is in {0,1,2} and s(i) - s(i-1) is in {1,2,-1}. Every row of T is a Fibonacci sequence (A000045), as is the sequence of column sums.

This is a 3-rowed array read upwards by columns. - N. J. A. Sloane, Sep 14 2014

Examples

			First 10 columns:
0 .. 1 .. 1 .. 2 .. 3 .. 5 .. 8 .. 13 .. 21 .. 34
0 .. 1 .. 1 .. 2 .. 3 .. 5 .. 8 .. 13 .. 21 .. 34
1 .. 0 .. 1 .. 1 .. 2 .. 3 .. 5 .. 8 ... 13 .. 21
T(4,1) counts these 3 paths, given as vector sums applied to (0,0):
(1,2) + (1,-1) + (1,1) + (1,-1);
(1,1) + (1,-1) + (1,2) + (1,-1);
(1,2) + (1,-1) + (1,-1) + (1,1).
Partial sums of second components in each vector sum give the 3 integer strings described in Comments:  (0,2,1,2,1), (0,1,0,2,1), (0,2,1,0,1).
		

Links

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

Crossrefs

Cf. A000045, A247050, A247051, A247309, A247310, A247311.

Programs

Mathematica

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

Formula

Let F = A000045, the Fibonacci numbers. Then (row 0, the bottom row) = F(n-1) for n >= 0; (row 1, the middle row) = F(n) for n >=0; (row 2, the top row) = (row 1).

Conjectures from Chai Wah Wu, Apr 16 2025: (Start)

a(n) = a(n-3) + a(n-6) for n > 5.

G.f.: (-x^5 - x^4 + x^3 - 1)/(x^6 + x^3 - 1). (End)