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%I #40 Sep 26 2021 14:10:47
%S 4,13,32,71,150,309,628,1267,2546,5105,10224,20463,40942,81901,163820,
%T 327659,655338,1310697,2621416,5242855,10485734,20971493,41943012,
%U 83886051,167772130,335544289,671088608,1342177247,2684354526,5368709085
%N Number of 3 X n 0..1 arrays with new values 0..1 introduced in row major order and no element equal to more than one of its immediate leftward or upward or right-upward antidiagonal neighbors.
%C Row 3 of A208637. Possibly row 4 of the convolution array A213568. - _Clark Kimberling_, Jun 20 2012
%C From _Noah Carey_, Aug 31 2021: (Start)
%C Conjecture: a(n) is equal to half the sum along the edges of (centered, height 2, width n, starting at line n+1) rectangles in Pascal's triangle, as shown here for n=3 (not including the terms inside the rectangles):
%C 1
%C 1 1
%C 1 2 1 a(3) = (4+6+4 + 15+20+15)/2
%C 1 3 3 1
%C 1 4---6---4 1
%C 1 5 | | 5 1
%C 1 6 15--20--15 6 1
%C 1 7 21 35 35 20 7 1 (End)
%H R. H. Hardin, <a href="/A208638/b208638.txt">Table of n, a(n) for n = 1..210</a>
%F Empirical: a(n) = 4*a(n-1) - 5*a(n-2) + 2*a(n-3).
%F Conjectures from _Colin Barker_, Mar 07 2018: (Start)
%F G.f.: x*(4 - 3*x) / ((1 - x)^2*(1 - 2*x)).
%F a(n) = 5*2^n - n - 5.
%F (End)
%e Some solutions for n=4:
%e 0 1 0 1 0 0 1 0 0 1 0 0 0 0 0 1 0 0 0 0
%e 0 1 0 0 1 0 1 0 0 1 1 1 1 1 0 0 1 1 1 0
%e 1 0 1 0 1 0 1 0 1 0 0 1 0 1 1 1 0 0 1 1
%Y Cf. A208637.
%K nonn
%O 1,1
%A _R. H. Hardin_, Feb 29 2012
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