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%I #3 Mar 30 2012 17:36:14
%S 1,2,3,4,1,5,3,6,7,7,13,1,8,22,4,9,34,12,10,50,28,1,11,70,58,5,12,95,
%T 108,18,13,125,188,50,1,14,161,308,121,6,15,203,483,261,25,16,252,728,
%U 520,80,1,17,308,1064,968,220,7,18,372,1512,1710,536,33,19,444,2100
%N Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 011 subwords (0<=k<=floor(n/3)). A Fibonacci binary word is a binary word having no 00 subword.
%C Also number of Fibonacci binary words of length n and having k 110 subwords. Row n has 1+floor(n/3) terms. Row sums are the Fibonacci numbers (A000045). T(n,0)=n+1. Sum(k*T(n,k), k>=0)=A023610(n-3).
%F G.f.=G(t,z)=(1+z)/(1-z-z^2+z^3-tz^3).
%e T(7,2)=4 because we have 1011011,0111011,0110110 and 0110111.
%e Triangle starts:
%e 1;
%e 2;
%e 3;
%e 4,1;
%e 5,3;
%e 6,7;
%e 7,13,1;
%e 8,22,4;
%e 9,34,12;
%e 10,50,28,1;
%p G:=(1+z)/(1-z-z^2+z^3-t*z^3): Gser:=simplify(series(G,z=0,23)): for n from 0 to 20 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 20 do seq(coeff(P[n],t,j),j=0..floor(n/3)) od; # yields sequence in triangular form
%Y Cf. A000045, A023610.
%K nonn,tabf
%O 0,2
%A _Emeric Deutsch_, May 12 2007
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