cp's OEIS Frontend

A200182 Number of -n..n arrays x(0..3) of 4 elements with zero sum and no two consecutive declines, no adjacent equal elements, and no element more than one greater than the previous (random base sawtooth pattern).

%I A200182 #26 May 07 2015 12:28:23
%S A200182 3,6,11,14,19,26,31,38,47,54,63,74,83,94,107,118,131,146,159,174,191,
%T A200182 206,223,242,259,278,299,318,339,362,383,406,431,454,479,506,531,558,
%U A200182 587,614,643,674,703,734,767,798,831,866,899,934,971,1006,1043,1082,1119,1158
%N A200182 Number of -n..n arrays x(0..3) of 4 elements with zero sum and no two consecutive declines, no adjacent equal elements, and no element more than one greater than the previous (random base sawtooth pattern).
%H A200182 R. H. Hardin, <a href="/A200182/b200182.txt">Table of n, a(n) for n = 1..200</a>
%F A200182 Empirical: a(n) = 2*a(n-1) -a(n-2) +a(n-3) -2*a(n-4) +a(n-5)
%F A200182      a(3*k-2) = ((3*k+1)^2)/3 - 7/3.
%F A200182      a(3*k-1) = ((3*k+2)^2)/3 - 7/3.
%F A200182      a(3*k)   = ((3*k+3)^2)/3 -  1  = 3*(k+1)^2 - 1.
%F A200182      a(3*k+1) = ((3*k+4)^2)/3 - 7/3.
%F A200182      a(3*k+2) = ((3*k+5)^2)/3 - 7/3 ... and so on.
%F A200182 The terms a(3*k-1) and a(3*k+1) seem to be terms of A241199: numbers n such that 4 consecutive terms of binomial(n,k) satisfy a quadratic relation for 0 <= k <= n/2. - _Avi Friedlich_, Apr 28 2015
%F A200182 Empirical g.f.: -x*(2*x^4-5*x^3+2*x^2+3) / ((x-1)^3*(x^2+x+1)). - _Colin Barker_, Apr 28 2015
%e A200182 Some solutions for n=6:
%e A200182 ..3....4....2....6....5....2....0....6....1....0....0....5....6....1....4....3
%e A200182 .-2....0....1...-2....6....3...-1...-1....2....1....1....0...-3....0...-1....1
%e A200182 .-1....1....2...-1...-6...-3....0....0....3....2...-1....1...-2....1....0....2
%e A200182 ..0...-5...-5...-3...-5...-2....1...-5...-6...-3....0...-6...-1...-2...-3...-6
%Y A200182 Row 4 of A200181.
%Y A200182 A014206 is a related sequence.
%K A200182 nonn
%O A200182 1,1
%A A200182 _R. H. Hardin_, Nov 13 2011