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%I #21 Mar 31 2024 19:03:46
%S 1,6,16,44,122,340,950,2658,7442,20844,58392,163594,458356,1284250,
%T 3598338,10082246,28249720,79153804,221783810,621424108,1741191198,
%U 4878708658,13669836930,38302030548,107319902744,300703682402
%N Number of base 6 n-digit numbers with adjacent digits differing by one or less.
%C Empirical: a(base,n) = a(base-1,n) + 3^(n-1) for base >= n; a(base,n) = a(base-1,n) + 3^(n-1) - 2 when base = n-1.
%C Leading 0's are allowed. - _Robert Israel_, Aug 12 2019
%H Robert Israel, <a href="/A126360/b126360.txt">Table of n, a(n) for n = 0..2231</a>
%H Arnold Knopfmacher, Toufik Mansour, Augustine Munagi, and Helmut Prodinger, <a href="http://arxiv.org/abs/0809.0551">Smooth words and Chebyshev polynomials</a>, arXiv:0809.0551v1 [math.CO], 2008.
%F From _Colin Barker_, Nov 26 2012: (Start)
%F Conjecture: a(n) = 4*a(n-1) - 3*a(n-2) - a(n-3) for n > 3.
%F G.f.: -(x^3 + 5*x^2 - 2*x - 1)/(x^3 + 3*x^2 - 4*x + 1). (End)
%F From _Robert Israel_, Aug 12 2019: (Start)
%F a(n) = e^T A^(n-1) e for n >= 1, where A is the 6 X 6 matrix with 1 on the main diagonal and first super- and subdiagonals, 0 elsewhere, and e the column vector (1,1,1,1,1,1).
%F Barker's conjecture follows from the fact that (A^3 - 4*A^2 + 3*A + 1) e = 0. (End)
%p A:=LinearAlgebra:-ToeplitzMatrix([1,1,0,0,0,0],symmetric):
%p e:= Vector(6,1):
%p 1, seq(e^%T . A^n . e, n=0..30); # _Robert Israel_, Aug 12 2019
%o (S/R) stvar $[N]:(0..M-1) init $[]:=0 asgn $[]->{*} kill +[i in 0..N-2](($[i]`-$[i+1]`>1)+($[i+1]`-$[i]`>1))
%K nonn,base
%O 0,2
%A _R. H. Hardin_, Dec 26 2006
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