A247051 Rectangular array read upwards by columns: T = T(n,k) = number of paths from (0,2) to (n,k), where 0 <= k <= 2, consisting of segments given by the vectors (1,1), (1,2), (1,-1).
0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1, 4, 3, 4, 4, 5, 4, 9, 8, 9, 12, 13, 12, 22, 21, 22, 33, 34, 33, 56, 55, 56, 88, 89, 88, 145, 144, 145, 232, 233, 232, 378, 377, 378, 609, 610, 609, 988, 987, 988, 1596, 1597, 1596, 2585, 2584, 2585, 4180, 4181
Offset: 0
Examples
First 10 columns: 1 .. 0 .. 1 .. 1 .. 2 .. 3 .. 5 .. 8 .. 13 .. 21 0 .. 1 .. 0 .. 2 .. 1 .. 4 .. 4 .. 9 .. 12 .. 22 0 .. 0 .. 1 .. 0 .. 2 .. 1 .. 4 .. 4 .. 9 ... 12 T(3,1) counts these 2 paths, given as vector sums applied to (0,2): (1,-1) + (1,1) + (1,-1) and (1,-1) + (1,-1) + (1,1). Partial sums of second components in each vector sum give the 2 integer strings described in Comments: (2,1,2,1), (2,1,0,1).
Links
- Clark Kimberling, Table of n, a(n) for n = 0..1000
Programs
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Mathematica
t[0, 0] = 0; t[0, 1] = 0; t[0, 2] = 1; 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 *) 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-2) + (-1)^n for n >= 1; (row 1, the middle row) = F(n-1) - (-1)^n for n >=0; (row 2, the top row) = F(n+1) for n >= 1.
Comments