A250162 Number of length n+1 0..3 arrays with the sum of adjacent differences multiplied by some arrangement of +-1 equal to zero.
4, 20, 96, 436, 1880, 7836, 32032, 129572, 521256, 2091052, 8376368, 33529908, 134168632, 536772668, 2147287104, 8589541444, 34358952008, 137437380684, 549752668240, 2199016964180, 8796080439384, 35184346923100, 140737438023776
Offset: 1
Keywords
Examples
Some solutions for n=6: ..1....3....3....2....3....2....2....2....2....1....1....2....1....0....2....1 ..1....0....2....3....3....0....2....2....2....1....0....0....1....2....0....0 ..0....3....1....3....3....2....3....3....2....3....3....0....1....2....3....2 ..3....1....0....0....0....3....1....1....2....2....3....2....3....3....0....1 ..0....2....3....3....3....2....3....2....1....0....1....3....3....1....2....2 ..1....1....0....1....3....0....2....0....0....1....2....2....0....3....0....2 ..3....3....1....2....3....2....2....0....2....3....1....0....1....2....2....3
Links
- R. H. Hardin, Table of n, a(n) for n = 1..59
Crossrefs
Column 3 of A250167.
Formula
Empirical: a(n) = 8*a(n-1) - 21*a(n-2) + 22*a(n-3) - 8*a(n-4).
Conjectures from Colin Barker, Nov 12 2018: (Start)
G.f.: 4*x*(1 - 3*x + 5*x^2) / ((1 - x)^2*(1 - 2*x)*(1 - 4*x)).
a(n) = 2*(2 - 3*2^n + 4^n + 2*n).
(End)
Comments