cp's OEIS Frontend

A183897 Number of nondecreasing arrangements of n+3 numbers in 0..2 with each number being the sum mod 3 of three others.

%I A183897 #14 Apr 05 2018 09:52:28
%S A183897 1,7,17,25,34,44,55,67,80,94,109,125,142,160,179,199,220,242,265,289,
%T A183897 314,340,367,395,424,454,485,517,550,584,619,655,692,730,769,809,850,
%U A183897 892,935,979,1024,1070,1117,1165,1214,1264,1315,1367,1420,1474,1529,1585,1642
%N A183897 Number of nondecreasing arrangements of n+3 numbers in 0..2 with each number being the sum mod 3 of three others.
%C A183897 Column 2 of A183904.
%H A183897 R. H. Hardin, <a href="/A183897/b183897.txt">Table of n, a(n) for n = 1..62</a>
%H A183897 Charles Cratty, Samuel Erickson, Frehiwet Negass, Lara Pudwell, <a href="http://www.valpo.edu/mathematics-statistics/files/2015/07/Pattern-Avoidance-in-Double-Lists.pdf">Pattern Avoidance in Double Lists</a>, preprint, 2015.
%F A183897 Empirical: a(n) = (1/2)*n^2 + (9/2)*n - 1 for n>2.
%F A183897 Conjectures from _Colin Barker_, Apr 05 2018: (Start)
%F A183897 G.f.: x*(1 + x - x^2)*(1 + 3*x - 3*x^2) / (1 - x)^3.
%F A183897 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>5.
%F A183897 (End)
%e A183897 Some solutions for n=4:
%e A183897 ..0....0....0....0....1....0....0....0....1....0....0....0....0....0....0....0
%e A183897 ..1....1....0....0....1....0....0....0....1....0....0....0....0....0....1....0
%e A183897 ..1....1....2....0....1....0....1....0....1....0....1....0....0....1....1....1
%e A183897 ..2....1....2....0....2....1....1....0....1....1....1....0....0....1....1....1
%e A183897 ..2....1....2....1....2....1....1....2....2....2....2....0....1....1....1....1
%e A183897 ..2....2....2....1....2....1....1....2....2....2....2....1....2....1....1....2
%e A183897 ..2....2....2....1....2....1....1....2....2....2....2....1....2....2....2....2
%Y A183897 Cf. A183904.
%K A183897 nonn
%O A183897 1,2
%A A183897 _R. H. Hardin_, Jan 07 2011