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%I #14 Jun 03 2017 06:29:04
%S 1,11,31,71,187,491,1327,3623,10003,27827,77911,219263,619747,1758131,
%T 5003239,14276831,40836163,117049091,336123463,966831503,2785160227,
%U 8034032147,23203096519,67087148063,194165268355,562477731491
%N Number of base 11 circular n-digit numbers with adjacent digits differing by 1 or less.
%C [Empirical] a(base,n)=a(base-1,n)+A002426(n+1) for base>=1.int(n/2)+1
%C a(n) = T(n, 11) where T(n, k) = Sum_{j=1..k} (1+2*cos(j*Pi/(k+1)))^n. These are the number of smooth cyclic words of length n over the alphabet {1,2,..,11}. See theorem 3.3 in Knopfmacher and others, reference in A124696. - _Peter Luschny_, Aug 13 2012
%F G.f.: (1 - 45*x^2 + 150*x^3 - 18*x^4 - 504*x^5 + 490*x^6 + 300*x^7 - 420*x^8 - 32*x^9 + 72*x^10) / ((1 - x)*(1 - 2*x)*(1 - 2*x - x^2)*(1 - 2*x - 2*x^2)*(1 - 4*x + 2*x^2 + 4*x^3 - 2*x^4)) (conjectured). - _Colin Barker_, Jun 03 2017
%o (S/R) stvar $[N]:(0..M-1) init $[]:=0 asgn $[]->{*} kill +[i in 0..N-1](($[i]`-$[(i+1)mod N]`>1)+($[(i+1)mod N]`-$[i]`>1))
%K nonn,base
%O 0,2
%A _R. H. Hardin_, Dec 28 2006