A183520 Number of n X 3 0..2 arrays with each element equal to either the sum mod 3 of its horizontal and vertical neighbors or the sum mod 3 of its diagonal and antidiagonal neighbors.

5, 31, 101, 543, 2233, 10003, 47685, 215451, 994397, 4603823, 21240401, 98257363, 454235165, 2100740935, 9717553917, 44943606231, 207898873245, 961691911899, 4448501263357, 20578124472715, 95191234404373, 440340646073843

Offset: 1

Author

R. H. Hardin, Jan 05 2011

Keywords

• nonn

Examples

			Some solutions for 5X3
..2..0..2....2..0..2....2..1..0....0..1..1....2..2..1....1..0..1....2..0..2
..0..2..0....1..2..2....2..2..2....2..1..2....2..2..2....2..1..2....0..2..1
..2..1..2....0..2..0....1..1..2....0..1..0....1..0..1....0..0..0....2..1..0
..2..0..2....0..2..2....2..2..2....2..2..1....2..2..2....2..2..1....2..0..1
..0..1..0....2..1..0....0..1..2....0..1..2....1..2..2....1..2..2....0..0..1
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..200
• Robert Israel, Linear recurrence of order 156
• Robert Israel, Maple-assisted derivation of recurrence

Crossrefs

Column 3 of A183526.

Programs

Maple

Configs:= map(t -> convert(t+3^6, base, 3)[1..6],[$0..3^6-1]):
q:= proc(a,b) local A,B;
   A:= Configs[a]; B:= Configs[b];
   if A[4..6]<> B[1..3] then return 0 fi;
   if A[4] <> A[1]+A[5]+B[4] mod 3  and A[4] <> A[2]+B[5] mod 3 then return 0 fi;
   if A[5] <> A[2]+A[4]+A[6]+B[5] mod 3 and A[5] <> A[1]+A[3]+B[4]+B[6] mod 3 then return 0 fi;
   if A[6] <> A[3]+A[5]+B[6] mod 3 and A[6] <> A[2]+B[5] mod 3 then return 0 fi;
   1
end proc:
T:= Matrix(3^6,3^6, q):
u:= Vector[row](3^6,proc(a) if Configs[a][1..3]=[0,0,0] then 1 else 0 fi end proc):
v:= Vector(3^6,proc(a) if Configs[a][4..6]=[0,0,0] then 1 else 0 fi end proc):
V[0]:= v:
for nn from 1 to 30 do V[nn]:= T . V[nn-1] od:
seq(u . V[n],n=1..30); # Robert Israel, Nov 17 2019

Formula

Linear recurrence of order 156: see links. - Robert Israel, Nov 17 2019