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Search: id:A129708
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| A129708 |
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Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k 010 subwords (n>=0, 0<=k<=floor((n-1)/2)). A Fibonacci binary word is a binary word having no 00 subword. |
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+0
1
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| 1, 2, 3, 4, 1, 6, 2, 9, 3, 1, 13, 6, 2, 19, 11, 3, 1, 28, 18, 7, 2, 41, 30, 14, 3, 1, 60, 50, 24, 8, 2, 88, 81, 43, 17, 3, 1, 129, 130, 77, 30, 9, 2, 189, 208, 132, 57, 20, 3, 1, 277, 330, 224, 108, 36, 10, 2, 406, 520, 379, 193, 72, 23, 3, 1, 595, 816, 633, 342, 143, 42, 11, 2, 872
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OFFSET
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0,2
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COMMENT
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Row n has floor((n+1)/2) terms (n>=1). Row sums are the Fibonacci numbers (A000045). T(n,0)=A000930(n+2). Sum(k*T(n,k), k>=0)=A001629(n-1).
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FORMULA
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G.f.=G(t,z)=(1+z+z^2-tz^2)/(1-z-tz^2+tz^3-z^3). Row generating polynomials P[n] are given by P[n](t)=Q[n](t,1), where Q[0]=1, Q[1]=1+x, Q[n](t,x)=Q[n-1](t,1)+xQ[n-2](t,t) for n>=2.
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EXAMPLE
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T(7,2)=3 because we have 1101010, 1010101 and 0101011.
Triangle starts:
1;
2;
3;
4,1;
6,2;
9,3,1;
13,6,2;
19,11,3,1;
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MAPLE
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Q[0]:=1: Q[1]:=1+x: for n from 2 to 30 do Q[n]:=expand(subs(x=1, Q[n-1])+x*subs(x=t, Q[n-2])) od: for n from 0 to 18 do P[n]:=subs(x=1, Q[n]) od; 1; for n from 1 to 18 do seq(coeff(P[n], t, j), j=0..floor((n-1)/2)) od; # yields sequence in triangular form
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CROSSREFS
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Cf. A000045, A000930, A001629.
Sequence in context: A071439 A100941 A082119 this_sequence A071518 A065338 A001438
Adjacent sequences: A129705 A129706 A129707 this_sequence A129709 A129710 A129711
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KEYWORD
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nonn,tabf
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AUTHOR
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Emeric Deutsch (deutsch(AT)duke.poly.edu), May 12 2007
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