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).
3, 6, 11, 14, 19, 26, 31, 38, 47, 54, 63, 74, 83, 94, 107, 118, 131, 146, 159, 174, 191, 206, 223, 242, 259, 278, 299, 318, 339, 362, 383, 406, 431, 454, 479, 506, 531, 558, 587, 614, 643, 674, 703, 734, 767, 798, 831, 866, 899, 934, 971, 1006, 1043, 1082, 1119, 1158
Offset: 1
Keywords
Examples
Some solutions for n=6: ..3....4....2....6....5....2....0....6....1....0....0....5....6....1....4....3 .-2....0....1...-2....6....3...-1...-1....2....1....1....0...-3....0...-1....1 .-1....1....2...-1...-6...-3....0....0....3....2...-1....1...-2....1....0....2 ..0...-5...-5...-3...-5...-2....1...-5...-6...-3....0...-6...-1...-2...-3...-6
Links
- R. H. Hardin, Table of n, a(n) for n = 1..200
Formula
Empirical: a(n) = 2*a(n-1) -a(n-2) +a(n-3) -2*a(n-4) +a(n-5)
a(3*k-2) = ((3*k+1)^2)/3 - 7/3.
a(3*k-1) = ((3*k+2)^2)/3 - 7/3.
a(3*k) = ((3*k+3)^2)/3 - 1 = 3*(k+1)^2 - 1.
a(3*k+1) = ((3*k+4)^2)/3 - 7/3.
a(3*k+2) = ((3*k+5)^2)/3 - 7/3 ... and so on.
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
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
Comments