A247301 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), (2,1), (1,-1).
1, 0, 0, 0, 1, 0, 1, 1, 1, 1, 2, 2, 2, 4, 3, 4, 6, 6, 6, 12, 10, 12, 20, 18, 20, 36, 32, 36, 64, 56, 64, 112, 100, 112, 200, 176, 200, 352, 312, 352, 624, 552, 624, 1104, 976, 1104, 1952, 1728, 1952, 3456, 3056, 3456, 6112, 5408, 6112, 10816, 9568, 10816
Offset: 0
Examples
First 10 columns: 0 .. 0 .. 1 .. 2 .. 3 .. 6 .. 10 .. 18 .. 32 .. 56 0 .. 1 .. 1 .. 2 .. 4 .. 6 .. 12 .. 20 .. 36 .. 64 1 .. 0 .. 1 .. 1 .. 2 .. 4 .. 6 ... 12 .. 20 .. 36 T(4,1) counts these 4 paths, given as vector sums applied to (0,0): (1,1) + (1,-1) + (2,1); (2,1) + (1,-1) + (1,-1); (2,1) + (1,1) + (1,-1); (1,1) + (2,1) + (1,-1).
Links
- Clark Kimberling, Table of n, a(n) for n = 0..1000
Programs
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Mathematica
t[0, 0] = 1; t[0, 1] = 0; t[0, 2] = 0; t[1, 0] = 0; t[1, 1] = 1; t[1, 2] = 0; t[2, 0] = 1; t[2, 1] = 1; t[2, 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, 0]; t[n_, 2] := t[n, 2] = t[n - 1, 1] + t[n - 2, 1]; TableForm[Reverse[Transpose[Table[t[n, k], {n, 0, 12}, {k, 0, 2}]]]] Flatten[Table[t[n, k], {n, 0, 20}, {k, 0, 2}]] (* A247301 *)
Formula
(row 0, the bottom row): r(n) = 2*r(n-2) + 2*r(n-3), with r(0) = 1, r(1) = 0, r(2) = 1, r(3) = 1;
(row 1, the middle row): r(n) = 2*r(n-2) + 2*r(n-3), with r(0) = 0, r(1) = 2, r(2) = 1, r(3) = 2;
(row 2, the top row): r(n) = 2*r(n-2) + 2*r(n-3), with r(0) = 0, r(1) = 0, r(2) = 1, r(3) = 2.
From Chai Wah Wu, Jan 24 2020: (Start)
a(n) = 2*a(n-6) + 2*a(n-9) for n > 14.
G.f.: (-x^14 - 2*x^11 + x^9 - x^8 - x^7 + x^6 - x^4 - 1)/(2*x^9 + 2*x^6 - 1). (End)