A126474 Number of arrays in [1..6]^n with adjacent elements differing by three or less.

1, 6, 30, 154, 788, 4034, 20650, 105708, 541122, 2770018, 14179796, 72586754, 371573530, 1902094812, 9736874082, 49843318162, 255149275268, 1306115946338, 6686042370634, 34226029248972, 175203956722818

Offset: 0

Author

R. H. Hardin, Dec 27 2006

Keywords

• nonn
• base

Comments

[Empirical] a(base,n)=a(base-1,n)+7^(n-1) for base>=3n-2; a(base,n)=a(base-1,n)+7^(n-1)-2 when base=3n-3

Original name: Number of base 6 n-digit numbers with adjacent digits differing by three or less.

Examples

			For n=2 the a(2)=30 solutions are [1, 1], [1, 2], [1, 3], [1, 4], [2, 1], [2, 2], [2, 3], [2, 4], [2, 5], [3, 1], [3, 2], [3, 3], [3, 4], [3, 5], [3, 6], [4, 1], [4, 2], [4, 3], [4, 4], [4, 5], [4, 6], [5, 2], [5, 3], [5, 4], [5, 5], [5, 6], [6, 3], [6, 4], [6, 5], [6, 6]. - _Robert Israel_, Jan 23 2018
		

Links

• Robert Israel, Table of n, a(n) for n = 0..1408

Crossrefs

Cf. Base 6 differing by two or less A126393, one or less A126360.

Programs

Maple

f:= gfun:-rectoproc({a(n) = 5*a(n-1) + a(n-2) - 2*a(n-3),a(0)=1,a(1)=6,a(2)=30},a(n),remember):
map(f, [$0..30]); # Robert Israel, Jan 23 2018

Formula

Conjectures from Colin Barker, Jan 20 2017: (Start)

a(n) = 5*a(n-1) + a(n-2) - 2*a(n-3) for n>2.

G.f.: (1 + x - x^2) / (1 - 5*x - x^2 + 2*x^3).

(End)

From Robert Israel, Jan 23 2018: (Start)

a(n) = e^T M^(n-1) e where e = [1,1,1,1,1,1]^T and M is the 6 X 6 matrix with entries M(i,j) = 1 if |i-j|<=3, 0 otherwise.

The fact that (M^3-5*M^2-M+2I) e = 0 implies Colin Barker's recursion, and the G.f. follows. (End)