cp's OEIS Frontend

A183900 Number of nondecreasing arrangements of n+3 numbers in 0..5 with each number being the sum mod 6 of three others.

%I A183900 #8 Apr 05 2018 11:13:05
%S A183900 3,24,130,364,771,1386,2281,3534,5236,7492,10422,14162,18865,24702,
%T A183900 31863,40558,51018,63496,78268,95634,115919,139474,166677,197934,
%U A183900 233680,274380,320530,372658,431325,497126,570691,652686,743814,844816,956472,1079602
%N A183900 Number of nondecreasing arrangements of n+3 numbers in 0..5 with each number being the sum mod 6 of three others.
%C A183900 Column 5 of A183904.
%H A183900 R. H. Hardin, <a href="/A183900/b183900.txt">Table of n, a(n) for n = 1..57</a>
%F A183900 Empirical: a(n) = (1/120)*n^5 + (1/4)*n^4 + (71/24)*n^3 + (57/4)*n^2 - (262/15)*n - 50 for n>4.
%F A183900 Conjectures from _Colin Barker_, Apr 05 2018: (Start)
%F A183900 G.f.: x*(3 + 6*x + 31*x^2 - 116*x^3 + 102*x^4 - 38*x^5 + 59*x^6 - 78*x^7 + 38*x^8 - 6*x^9) / (1 - x)^6.
%F A183900 a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>10.
%F A183900 (End)
%e A183900 Some solutions for n=2:
%e A183900 ..0....1....0....0....0....0....0....0....0....0....0....3....1....3....0....1
%e A183900 ..2....3....0....1....0....0....0....1....2....1....0....3....3....3....2....1
%e A183900 ..4....5....2....3....2....2....0....1....4....2....4....5....3....3....2....1
%e A183900 ..4....5....4....4....2....2....0....2....5....3....4....5....5....3....2....3
%e A183900 ..4....5....4....5....4....2....0....4....5....4....4....5....5....3....4....3
%Y A183900 Cf. A183904.
%K A183900 nonn
%O A183900 1,1
%A A183900 _R. H. Hardin_, Jan 07 2011