%I #24 Feb 08 2019 12:09:48
%S 1,3,3,6,6,10,10,15,17,23,27,34,40,48,56,65,75,87,99,114,128,146,162,
%T 183,201,225,247,274,300,330,360,393,427,463,501,542,584,630,676,727,
%U 777,833,887,948,1008,1074,1140,1211,1283,1359,1437,1518,1602,1690,1780,1875
%N Number of nondecreasing strings of numbers (x(i), i=1..n) in -2..2 with sum x(i)^3 equal to 0.
%C Column 2 of A188277.
%H Robert Israel, <a href="/A188270/b188270.txt">Table of n, a(n) for n = 1..10000</a> (first 200 terms from R. H. Hardin)
%H Robert Israel, <a href="/A188270/a188270_1.pdf">Maple-assisted proof of empirical g.f.</a>
%H <a href="/index/Rec#order_13">Index entries for linear recurrences with constant coefficients</a>, signature (2,0,-2,1,0,0,0,0,1,-2,0,2,-1).
%F Empirical: a(n) = 2*a(n-1) - 2*a(n-3) + a(n-4) + a(n-9) - 2*a(n-10) + 2*a(n-12) - a(n-13).
%F Empirical g.f.: x*(1 + x - 3*x^2 + 2*x^3 - x^4 + x^5 - x^6 + x^7 + x^8 - 2*x^9 + 2*x^11 - x^12) / ((1 - x)^4*(1 + x)*(1 + x + x^2)*(1 + x^3 + x^6)). - _Colin Barker_, Apr 27 2018
%F From _Robert Israel_, Feb 07 2019: (Start)
%F Empirical g.f. verified: see link.
%F a(n) - (n^3/108 + (5/72)*n^2 + (19/36)*n) is periodic with period 18. (End)
%e All solutions for n=8 k=2:
%e .-1...-2...-2....0...-2...-2...-1...-2...-2...-2...-1...-2...-2...-1...-2
%e ..0....0...-2....0...-1...-1...-1...-2...-2...-2...-1...-2...-1...-1...-2
%e ..0....0...-2....0....0...-1...-1...-2....0...-2....0...-1...-1...-1...-1
%e ..0....0...-1....0....0....0...-1...-2....0....0....0...-1...-1....0....0
%e ..0....0....1....0....0....0....1....2....0....0....0....1....1....0....0
%e ..0....0....2....0....0....1....1....2....0....2....0....1....1....1....1
%e ..0....0....2....0....1....1....1....2....2....2....1....2....1....1....2
%e ..1....2....2....0....2....2....1....2....2....2....1....2....2....1....2
%p S:= series((y^8 - y^7 + y^6 - y^5 + y^4 - y^3 + y^2 - y + 1)/(
%p y^13 - 2*y^12 + 2*y^10 - y^9 - y^4 + 2*y^3 - 2*y + 1), y, 101):
%p seq(coeff(S,y,n),n=1..100); # _Robert Israel_, Feb 07 2019
%t f[z_] := (z^2 - z + 1)*(z^6 - z^3 + 1)/((z - 1)^4*(z + 1)*(z^2 + z + 1)*(z^6 + z^3 + 1)); CoefficientList[Series[f[z], {z, 0, 100}], z] (* By the result of _Robert Israel_,_Peter Luschny_, Feb 07 2019 *)
%Y Cf. A188277.
%K nonn,easy
%O 1,2
%A _R. H. Hardin_, Mar 26 2011