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%I #3 Mar 30 2012 17:36:14
%S 1,2,2,1,4,1,4,3,1,8,4,1,8,8,4,1,16,12,5,1,16,20,13,5,1,32,32,18,6,1,
%T 32,48,38,19,6,1,64,80,56,25,7,1,64,112,104,63,26,7,1,128,192,160,88,
%U 33,8,1,128,256,272,192,96,34,8,1,256,448,432,280,129,42,9,1,256,576,688,552
%N Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 0's in even positions (0<=k<=floor(n/2)). A Fibonacci binary word is a binary word having no 00 subword.
%C Row n has 1+floor(n/2) terms. Row sums are the Fibonacci numbers (A000045). T(2n+1,k)=T(2n,k)+T(2n-1,k) (n>=1). T(2n,k)=A129719(2n,k). Sum(k*T(n,k), 0<=k<=floor(n/2))=A129722(n).
%F G.f.=G(t,z)=(1+2z-tz^3)/[1-(2+t)z^2+tz^4]. The trivariate generating function H(t,s,z), where t marks number of 0's in odd position and s marks number of 0's in even position, is given by H(t,s,z)=[1+(1+t)z-tsz^3]/[1-(1+t+s)z^2+tsz^4].
%e T(6,2)=4 because we have 111010, 101110, 101011 and 011010.
%e Triangle starts:
%e 1;
%e 2;
%e 2,1;
%e 4,1;
%e 4,3,1;
%e 8,4,1;
%e 8,8,4,1;
%p G:=(1+2*z-t*z^3)/(1-2*z^2-t*z^2+t*z^4): Gser:=simplify(series(G,z=0,21)): for n from 0 to 18 do P[n]:=sort(coeff(Gser,z,n)) od: for n from 0 to 18 do seq(coeff(P[n],t,j),j=0..floor(n/2)) od; # yields sequence in triangular form
%Y Cf. A000045, A129719, A129722.
%K nonn,tabf
%O 0,2
%A _Emeric Deutsch_, May 13 2007
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