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%I #11 Nov 13 2017 06:28:28
%S 4,29,140,741,3853,19956,104096,541775,2819027,14671776,76356631,
%T 397392690,2068192977,10763659993,56018290276,291541126302,
%U 1517293977155,7896590852333,41096944195828,213884557644749,1113138825391146
%N Number of nX3 0..1 arrays with each 1 horizontally or vertically adjacent to 1 or 2 1s.
%C Column 3 of A295040.
%H R. H. Hardin, <a href="/A295035/b295035.txt">Table of n, a(n) for n = 1..210</a>
%H Robert Israel, <a href="/A295035/a295035_1.pdf">Maple-assisted proof of formula</a>
%F Empirical: a(n) = 4*a(n-1) +2*a(n-2) +17*a(n-3) +18*a(n-4) +53*a(n-5) -23*a(n-6) -38*a(n-7) +52*a(n-8) -19*a(n-9) -14*a(n-10) +16*a(n-11) -11*a(n-12) +a(n-14) -a(n-15)
%F Empirical formula confirmed by _Robert Israel_, Nov 12 2017 (see link).
%e Some solutions for n=7
%e ..1..1..0. .1..0..0. .0..0..0. .0..0..1. .0..1..0. .1..1..0. .0..0..0
%e ..0..0..0. .1..0..0. .1..1..1. .1..0..1. .0..1..0. .1..0..1. .0..0..1
%e ..1..1..0. .1..1..0. .0..0..1. .1..1..0. .0..1..0. .0..1..1. .0..1..1
%e ..0..0..0. .0..0..1. .0..0..0. .0..0..0. .0..1..0. .0..1..0. .1..1..0
%e ..0..0..0. .0..0..1. .1..1..0. .1..1..1. .1..0..1. .0..1..1. .1..0..1
%e ..1..1..0. .1..1..0. .0..1..0. .0..0..0. .1..0..1. .1..0..1. .0..0..1
%e ..1..0..0. .0..0..0. .0..0..0. .0..0..0. .1..1..1. .1..0..1. .0..0..1
%p q:= proc(a,b) local r,s,t,M,i;
%p s:= floor((a-1)/8);
%p if s <> (b-1) mod 8 then return 0 fi;
%p s:= convert(s+8,base,2);
%p r:= convert(8+floor((b-1)/8),base,2);
%p t:= convert(8+ ((a-1) mod 8),base,2);
%p M:= Vector(3);
%p if s[1] = 1 and s[2] = 1 then M[1]:= 1; M[2]:= 1 fi;
%p if s[2]=1 and s[3]=1 then M[2]:= M[2]+1; M[3]:= 1 fi;
%p for i from 1 to 3 do if s[i]=1 then
%p M[i]:= M[i]+r[i]+t[i];
%p if M[i] = 0 or M[i]>2 then return 0 fi;
%p fi od;
%p 1
%p end proc:
%p T:= Matrix(64,64, q);
%p u:= Vector[row](64):
%p v:= Vector(64):
%p for i from 0 to 7 do u[8*i+1]:= 1; v[i+1]:= 1;
%p od:
%p seq(u . T^n . v, n = 1 .. 100); # _Robert Israel_, Nov 12 2017
%Y Cf. A295040.
%K nonn
%O 1,1
%A _R. H. Hardin_, Nov 12 2017
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