A207724 Number of n X 3 0..1 arrays avoiding 0 0 0 and 0 1 0 horizontally and 0 1 1 and 1 0 1 vertically.
6, 36, 78, 189, 490, 1113, 2449, 5474, 12036, 26100, 56595, 122472, 264061, 568695, 1224190, 2633000, 5660226, 12165489, 26141850, 56165805, 120662377, 259206930, 556801480, 1196027864, 2569059663, 5518244160, 11852866905, 25459111647
Offset: 1
Examples
Some solutions for n=4: 1 1 0 1 1 1 1 0 0 1 0 0 0 1 1 0 1 1 0 0 1 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 1 1 1 1 0 0 1 1 0 1 0 0 1 0 0 1 1 1 0 1 1 0 0 1 1 1 0 1 0 0 1 1 1 1 0 0 0 0 1 0 1 1 0 0 1
Links
- R. H. Hardin, Table of n, a(n) for n = 1..210
- Robert Israel, Maple-assisted proof of formula
- Index entries for linear recurrences with constant coefficients, signature (3,-2,3,-6,0,0,3,1,0,-1).
Crossrefs
Cf. A207729.
Programs
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Maple
Configs:= select(A -> A[1..3] <> [0,0,0] and A[4..6] <> [0,0,0] and A[1..3] <> [0,1,0] and A[4..6] <> [0,1,0], [seq(convert(x,base,2)[1..6],x=2^6..2^7-1)]): compat:= proc(i,j) local k,col; if Configs[i][4..6] <> Configs[j][1..3] then return 0 fi; for k from 1 to 3 do col:= [Configs[i][k],Configs[i][k+3],Configs[j][k+3]]; if col = [0,1,1] or col = [1,0,1] then return 0 fi; od; 1 end proc: T:= Matrix(36,36,compat): u:= Vector[row](36, 1): v:= Vector(36,1): 6,seq(u . T^(n-2) . v,n=2..50); # Robert Israel, Mar 05 2018 -
Mathematica
LinearRecurrence[{3, -2, 3, -6, 0, 0, 3, 1, 0, -1}, {6, 36, 78, 189, 490, 1113, 2449, 5474, 12036, 26100}, 30] (* Jean-François Alcover, May 15 2023, after Robert Israel's confirmed formula *)
Formula
Empirical: a(n) = 3*a(n-1) - 2*a(n-2) + 3*a(n-3) - 6*a(n-4) + 3*a(n-7) + a(n-8) - a(n-10).
Formula confirmed by Robert Israel, Mar 05 2018 (see link).
G.f.: x*(6 + 18*x - 18*x^2 + 9*x^3 + 7*x^4 + 3*x^5 - 9*x^6 - x^7 - x^8 + x^9) / ((1 - x)*(1 + x^2 - x^3)*(1 - x - x^3)*(1 - x - 2*x^2 - x^3)). - Colin Barker, Mar 05 2018
Comments