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A200892 Number of 0..n arrays x(0..8) of 9 elements without any interior element greater than both neighbors.

%I A200892 #10 Oct 16 2017 12:23:26
%S A200892 200,4059,34350,181336,710976,2269938,6233356,15250675,34054592,
%T A200892 70608021,137674186,254905378,451556600,769941268,1269757336,
%U A200892 2033423669,3172578200,4835901375,7218440614,10572623996,15221164112,21572066022
%N A200892 Number of 0..n arrays x(0..8) of 9 elements without any interior element greater than both neighbors.
%C A200892 Row 7 of A200886.
%H A200892 R. H. Hardin, <a href="/A200892/b200892.txt">Table of n, a(n) for n = 1..210</a>
%F A200892 Empirical: a(n) = (2/2835)*n^9 + (131/630)*n^8 + (2803/945)*n^7 + (1349/90)*n^6 + (41449/1080)*n^5 + (20423/360)*n^4 + (1149293/22680)*n^3 + (22741/840)*n^2 + (2011/252)*n + 1.
%F A200892 Conjectures from _Colin Barker_, Oct 16 2017: (Start)
%F A200892 G.f.: x*(200 + 2059*x + 2760*x^2 - 3509*x^3 - 1714*x^4 + 288*x^5 + 208*x^6 - 45*x^7 + 10*x^8 - x^9) / (1 - x)^10.
%F A200892 a(n) = 10*a(n-1) - 45*a(n-2) + 120*a(n-3) - 210*a(n-4) + 252*a(n-5) - 210*a(n-6) + 120*a(n-7) - 45*a(n-8) + 10*a(n-9) - a(n-10) for n>10.
%F A200892 (End)
%e A200892 Some solutions for n=3
%e A200892 ..2....3....2....3....2....2....1....2....3....2....1....2....0....2....1....1
%e A200892 ..0....2....3....0....1....2....1....2....3....0....2....0....3....2....0....1
%e A200892 ..2....2....3....0....1....3....3....3....0....0....3....0....3....0....1....1
%e A200892 ..2....3....1....2....2....3....3....3....0....0....3....1....2....3....2....2
%e A200892 ..2....3....3....2....2....1....1....3....2....2....1....1....3....3....2....3
%e A200892 ..0....0....3....2....0....1....1....1....3....2....2....1....3....1....1....3
%e A200892 ..3....0....1....2....1....1....0....3....3....1....2....3....2....3....3....0
%e A200892 ..3....2....1....2....1....0....2....3....0....1....2....3....0....3....3....0
%e A200892 ..1....2....3....0....1....1....2....0....3....1....2....2....1....0....0....0
%K A200892 nonn
%O A200892 1,1
%A A200892 _R. H. Hardin_, Nov 23 2011