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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).
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%I #14 Sep 14 2014 14:26:57

%S 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,

%T 33,34,33,56,55,56,88,89,88,145,144,145,232,233,232,378,377,378,609,

%U 610,609,988,987,988,1596,1597,1596,2585,2584,2585,4180,4181

%N 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).

%C Also, T(n,k) = number of strings s(0)..s(n) of integers such that s(0) = 2, 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}. The column sums form the Fibonacci sequence (A000045).

%C This is a 3-rowed array read upwards by columns. - _N. J. A. Sloane_, Sep 14 2014

%H Clark Kimberling, <a href="/A247051/b247051.txt">Table of n, a(n) for n = 0..1000</a>

%F 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.

%e First 10 columns:

%e 1 .. 0 .. 1 .. 1 .. 2 .. 3 .. 5 .. 8 .. 13 .. 21

%e 0 .. 1 .. 0 .. 2 .. 1 .. 4 .. 4 .. 9 .. 12 .. 22

%e 0 .. 0 .. 1 .. 0 .. 2 .. 1 .. 4 .. 4 .. 9 ... 12

%e 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).

%e 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).

%t 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 *)

%t Flatten[Table[t[n, k], {n, 0, 20}, {k, 0, 2}]] (* sequence *)

%Y Cf. A000045, A247049, A247050.

%K nonn,tabf,easy

%O 0,11

%A _Clark Kimberling_, Sep 11 2014