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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).

%I A247049 #20 Apr 17 2025 09:52:17
%S A247049 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,
%T A247049 34,55,55,55,89,89,89,144,144,144,233,233,233,377,377,377,610,610,610,
%U A247049 987,987,987,1597,1597,1597,2584,2584,2584,4181,4181,4181
%N 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).
%C A247049 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.
%C A247049 This is a 3-rowed array read upwards by columns. - _N. J. A. Sloane_, Sep 14 2014
%H A247049 Clark Kimberling, <a href="/A247049/b247049.txt">Table of n, a(n) for n = 0..1000</a>
%F A247049 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).
%F A247049 Conjectures from _Chai Wah Wu_, Apr 16 2025: (Start)
%F A247049 a(n) = a(n-3) + a(n-6) for n > 5.
%F A247049 G.f.: (-x^5 - x^4 + x^3 - 1)/(x^6 + x^3 - 1). (End)
%e A247049 First 10 columns:
%e A247049 0 .. 1 .. 1 .. 2 .. 3 .. 5 .. 8 .. 13 .. 21 .. 34
%e A247049 0 .. 1 .. 1 .. 2 .. 3 .. 5 .. 8 .. 13 .. 21 .. 34
%e A247049 1 .. 0 .. 1 .. 1 .. 2 .. 3 .. 5 .. 8 ... 13 .. 21
%e A247049 T(4,1) counts these 3 paths, given as vector sums applied to (0,0):
%e A247049 (1,2) + (1,-1) + (1,1) + (1,-1);
%e A247049 (1,1) + (1,-1) + (1,2) + (1,-1);
%e A247049 (1,2) + (1,-1) + (1,-1) + (1,1).
%e A247049 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).
%t A247049 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 *)
%t A247049 u = Flatten[Table[t[n, k], {n, 0, 20}, {k, 0, 2}]] (* sequence *)
%Y A247049 Cf.  A000045, A247050, A247051, A247309, A247310, A247311.
%K A247049 nonn,tabf,easy
%O A247049 0,11
%A A247049 _Clark Kimberling_, Sep 11 2014