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A129719
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Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 0's in odd positions (0 <= k <= ceiling(n/2)). A Fibonacci binary word is a binary word having no 00 subword.
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3
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1, 1, 1, 2, 1, 2, 2, 1, 4, 3, 1, 4, 5, 3, 1, 8, 8, 4, 1, 8, 12, 9, 4, 1, 16, 20, 13, 5, 1, 16, 28, 25, 14, 5, 1, 32, 48, 38, 19, 6, 1, 32, 64, 66, 44, 20, 6, 1, 64, 112, 104, 63, 26, 7, 1, 64, 144, 168, 129, 70, 27, 7, 1, 128, 256, 272, 192, 96, 34, 8, 1, 128, 320, 416, 360, 225, 104, 35
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OFFSET
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0,4
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COMMENTS
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Row n has 1+ceiling(n/2) terms.
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LINKS
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FORMULA
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G.f.: G(t,z) = (1+z)(1+tz-tz^2)/(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).
Row sums are the Fibonacci numbers (A000045).
T(2n,k) = T(2n-1,k) + T(2n-2,k) (n >= 1).
Sum_{k=0..ceiling(n/2)} k*T(n,k) = A129720(n).
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EXAMPLE
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T(6,2)=4 because we have 110101, 011101, 010110 and 010111.
Triangle starts:
1;
1, 1;
2, 1;
2, 2, 1;
4, 3, 1;
4, 5, 3, 1;
8, 8, 4, 1;
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MAPLE
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G:=(1+z)*(1+t*z-t*z^2)/(1-(2+t)*z^2+t*z^4): Gser:=simplify(series(G, z=0, 20)): for n from 0 to 17 do P[n]:=sort(coeff(Gser, z, n)) od: for n from 0 to 17 do seq(coeff(P[n], t, j), j=0..ceil(n/2)) od; # yields sequence in triangular form
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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