A247323 Number of paths from (0,0) to (n,0), with vertices (i,k) satisfying 0 <= k <= 3, consisting of segments given by the vectors (1,1), (1,2), (1,-1).
1, 0, 1, 1, 2, 5, 7, 18, 29, 63, 116, 229, 445, 856, 1677, 3229, 6298, 12185, 23675, 45922, 89097, 172931, 335460, 651065, 1263145, 2451184, 4756105, 9228777, 17907538, 34747357, 67424063, 130828370, 253859365, 492585879, 955810772, 1854647997, 3598744709
Offset: 0
Examples
a(5) counts these 5 paths, each represented by a vector sum applied to (0,0): (1,2) + (1,1) + (1,-1) + (1,-1) + (1,-1); (1,1) + (1,2) + (1,-1) + (1,-1) + (1,-1); (1,2) + (1,-1) + (1,1) + (1,-1) + (1,-1); (1,1) + (1,-1) + (1,2) + (1,-1) + (1,-1); (1,2) + (1,-1) + (1,-1) + (1,1) + (1,-1).
Links
- Clark Kimberling, Table of n, a(n) for n = 0..1000
Programs
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Mathematica
z = 25; t[0, 0] = 1; t[0, 1] = 0; t[0, 2] = 0; t[0, 3] = 0; t[1, 3] = 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] + t[n - 1, 3]; t[n_, 3] := t[n, 3] = t[n - 1, 1] + t[n - 1, 2]; Table[t[n, 0], {n, 0, z}]; (* A247323 *)
Formula
Empirically, a(n) = 3*a(n-2) + 2*a(n-3) - a(n-4) and g.f. = (1 + 2*x^2 - x^3)/(1 - 3 x^2 - 2 x^3 + x^4).
Comments