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Number of base 8 n-digit numbers with adjacent digits differing by one or less.
8

%I #19 Jan 13 2022 05:07:58

%S 1,8,22,62,176,502,1436,4116,11814,33942,97582,280676,807574,2324116,

%T 6689624,19257202,55439298,159611886,459545688,1323132230,3809653732,

%U 10969153364,31583803574,90940708414,261850874726,753964626300

%N Number of base 8 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.

%H Robert Israel, <a href="/A126362/b126362.txt">Table of n, a(n) for n = 0..2174</a>

%H Jim Bumgardner, <a href="http://jbum.com/papers/componium_variations.pdf">Variations of the Componium</a>, 2013.

%F From _Colin Barker_, Nov 26 2012: (Start)

%F Conjecture: a(n) = 5*a(n-1) - 6*a(n-2) - a(n-3) + 2*a(n-4) for n > 4.

%F G.f.: -(4*x^4 + x^3 - 12*x^2 + 3*x + 1)/((2*x - 1)*(x^3 - 3*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 8 X 8 matrix with 1 on the main diagonal and first super- and subdiagonals, 0 elsewhere, and e the column vector (1,1,1,1,1,1,1,1). Barker's conjecture follows from the fact that (A^4 - 5*A^3 + 6*A^2 + A - 2*I)*e = 0. (End)

%p f:= gfun:-rectoproc({a(n)=5*a(n-1)-6*a(n-2)-a(n-3)+2*a(n-4),a(0)=1,a(1)=8,a(2)=22,a(3)=62,a(4)=176},a(n),remember):

%p map(f, [$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