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%I #4 Nov 30 2014 18:07:48
%S 512,1728,5832,19683,46656,110592,262144,512000,1000000,1953125,
%T 3375000,5832000,10077696,16003008,25412184,40353607,60236288,
%U 89915392,134217728,191102976,272097792,387420489,531441000,729000000,1000000000,1331000000
%N Number of (n+2)X(1+2) 0..1 arrays with nondecreasing sum of every three consecutive values in every row and column
%C Column 1 of A251192
%H R. H. Hardin, <a href="/A251187/b251187.txt">Table of n, a(n) for n = 1..210</a>
%F Empirical: a(n) = 2*a(n-1) -a(n-2) +8*a(n-3) -16*a(n-4) +8*a(n-5) -28*a(n-6) +56*a(n-7) -28*a(n-8) +56*a(n-9) -112*a(n-10) +56*a(n-11) -70*a(n-12) +140*a(n-13) -70*a(n-14) +56*a(n-15) -112*a(n-16) +56*a(n-17) -28*a(n-18) +56*a(n-19) -28*a(n-20) +8*a(n-21) -16*a(n-22) +8*a(n-23) -a(n-24) +2*a(n-25) -a(n-26)
%F Empirical for n mod 3 = 0: a(n) = (1/19683)*n^9 + (5/2187)*n^8 + (11/243)*n^7 + (377/729)*n^6 + (304/81)*n^5 + (484/27)*n^4 + (1520/27)*n^3 + 112*n^2 + 128*n + 64
%F Empirical for n mod 3 = 1: a(n) = (1/19683)*n^9 + (5/2187)*n^8 + (100/2187)*n^7 + (3500/6561)*n^6 + (8750/2187)*n^5 + (43750/2187)*n^4 + (437500/6561)*n^3 + (312500/2187)*n^2 + (390625/2187)*n + (1953125/19683)
%F Empirical for n mod 3 = 2: a(n) = (1/19683)*n^9 + (5/2187)*n^8 + (11/243)*n^7 + (3397/6561)*n^6 + (8248/2187)*n^5 + (13232/729)*n^4 + (378880/6561)*n^3 + (256256/2187)*n^2 + (100352/729)*n + (1404928/19683)
%e Some solutions for n=6
%e ..0..0..0....0..0..0....0..0..0....1..0..0....0..0..0....0..0..0....1..0..0
%e ..0..0..0....0..0..1....0..1..1....0..1..0....0..0..0....1..1..0....1..0..0
%e ..1..0..1....0..1..0....0..1..1....0..0..0....1..0..0....0..1..1....1..0..1
%e ..1..0..0....0..1..1....1..1..1....1..0..0....0..0..1....0..0..0....1..0..0
%e ..1..0..1....0..0..1....1..1..1....0..1..1....0..1..0....1..1..0....1..0..0
%e ..1..0..1....0..1..0....0..1..1....0..0..0....1..0..0....0..1..1....1..0..1
%e ..1..1..0....0..1..1....1..1..1....1..1..0....1..1..1....0..0..1....1..1..1
%e ..1..0..1....1..1..1....1..1..1....0..1..1....0..1..1....1..1..0....1..1..1
%K nonn
%O 1,1
%A _R. H. Hardin_, Nov 30 2014
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