

A129721


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.


2



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, 32, 48, 38, 19, 6, 1, 64, 80, 56, 25, 7, 1, 64, 112, 104, 63, 26, 7, 1, 128, 192, 160, 88, 33, 8, 1, 128, 256, 272, 192, 96, 34, 8, 1, 256, 448, 432, 280, 129, 42, 9, 1, 256, 576, 688, 552
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OFFSET

0,2


COMMENTS

Row n has 1+floor(n/2) terms. Row sums are the Fibonacci numbers (A000045). T(2n+1,k)=T(2n,k)+T(2n1,k) (n>=1). T(2n,k)=A129719(2n,k). Sum(k*T(n,k), 0<=k<=floor(n/2))=A129722(n).


LINKS

Table of n, a(n) for n=0..75.


FORMULA

G.f.=G(t,z)=(1+2ztz^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)ztsz^3]/[1(1+t+s)z^2+tsz^4].


EXAMPLE

T(6,2)=4 because we have 111010, 101110, 101011 and 011010.
Triangle starts:
1;
2;
2,1;
4,1;
4,3,1;
8,4,1;
8,8,4,1;


MAPLE

G:=(1+2*zt*z^3)/(12*z^2t*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


CROSSREFS

Cf. A000045, A129719, A129722.
Sequence in context: A061298 A276468 A002126 * A268193 A238606 A054995
Adjacent sequences: A129718 A129719 A129720 * A129722 A129723 A129724


KEYWORD

nonn,tabf


AUTHOR

Emeric Deutsch, May 13 2007


STATUS

approved



