%I #11 Apr 05 2018 09:52:41
%S 16,292,2574,13344,59518,250124,1024174,4143944,16671934,66886900,
%T 267965390,1072744752,4292852990,17175392028,68710014830,274857954776,
%U 1099469650494,4397958390916,17592001449102,70368358249024
%N Number of arrangements of n+3 numbers in 0..3 with each number being the sum mod 4 of three others.
%C Column 3 of A183892.
%H R. H. Hardin, <a href="/A183886/b183886.txt">Table of n, a(n) for n = 1..62</a>
%F Empirical (for n>=2): 4^(n+3) - (n+3)*2^(n+4) - 2/3*(7*n^3+51*n^2+125*n+108). - _Vaclav Kotesovec_, Nov 27 2012.
%F Conjectures from _Colin Barker_, Apr 05 2018: (Start)
%F G.f.: 2*x*(8 + 50*x - x^2 - 1488*x^3 + 4469*x^4 - 5336*x^5 + 2996*x^6 - 656*x^7) / ((1 - x)^4*(1 - 2*x)^2*(1 - 4*x)).
%F a(n) = 12*a(n-1) - 58*a(n-2) + 148*a(n-3) - 217*a(n-4) + 184*a(n-5) - 84*a(n-6) + 16*a(n-7) for n>8.
%F (End)
%e Some solutions for n=2:
%e ..1....3....1....2....0....0....1....0....1....2....2....0....3....1....3....0
%e ..1....0....0....3....1....2....0....2....3....3....3....3....3....1....0....1
%e ..0....3....1....3....1....1....2....0....1....1....0....0....0....0....1....2
%e ..2....2....2....1....3....1....1....1....2....0....1....3....2....3....3....3
%e ..3....0....3....0....2....0....2....1....0....3....3....2....2....2....2....1
%Y Cf. A183892.
%K nonn
%O 1,1
%A _R. H. Hardin_, Jan 07 2011