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%I #8 Jun 26 2019 18:04:10
%S 1,1,3,9,23,51,133,399,1321,4129,12457,38059,114177,349017,1056893,
%T 3188001,9592717,28778151,86390521,259279771,778196971,2334819901,
%U 7004459703,21010336329,63026931713,189077563341,567218511883
%N Number of (n+2)X(1+2) 0..2 arrays with each row and column divisible by 11, read as a base-3 number with top and left being the most significant digits.
%C Column 1 of A263333.
%H R. H. Hardin, <a href="/A263330/b263330.txt">Table of n, a(n) for n = 1..210</a>
%H Robert Israel, <a href="/A263330/a263330.pdf">Maple-assisted proof of empirical formula</a>
%F Empirical: a(n) = 3*a(n-1) +16*a(n-5) -48*a(n-6) -505*a(n-10) +1515*a(n-11) -5105*a(n-15) +15315*a(n-16) -10114*a(n-20) +30342*a(n-21) +15709*a(n-25) -47127*a(n-26).
%F Empirical formula confirmed: see link. - _Robert Israel_, Jun 26 2019
%e Some solutions for n=4
%e ..1..0..2....0..0..0....2..1..1....1..0..2....2..1..1....1..0..2....0..0..0
%e ..0..0..0....2..1..1....0..0..0....1..0..2....2..1..1....1..0..2....0..0..0
%e ..0..0..0....2..1..1....0..0..0....1..0..2....2..1..1....1..0..2....0..0..0
%e ..1..0..2....2..1..1....2..1..1....0..0..0....2..1..1....1..0..2....0..0..0
%e ..0..0..0....2..1..1....0..0..0....0..0..0....2..1..1....1..0..2....0..0..0
%e ..1..0..2....2..1..1....2..1..1....1..0..2....0..0..0....0..0..0....0..0..0
%p States:= [seq(seq(seq([x,y,z],z=0..10),y=0..10),x=0..10)]:
%p T:= Matrix(1331,1331,storage=sparse):
%p for i from 1 to 1331 do
%p for y in [[0,0,0],[1,0,2],[2,1,1]] do
%p z:= 3*States[i]+y mod 11;
%p j:= 121*z[1]+11*z[2]+z[3] + 1;
%p T[i,j]:= 1;
%p od od:
%p U[0]:= Vector([1,0$1330]):
%p for j from 1 to 40 do U[j]:= T . U[j-1] od:
%p seq(U[n+2][1],n=1..38); # _Robert Israel_, Jun 26 2019
%Y Cf. A263333.
%K nonn
%O 1,3
%A _R. H. Hardin_, Oct 15 2015
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