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Search: id:A129718
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| A129718 |
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Triangle read by rows: T(n,k) is the number of Fibonacci binary words of length n and having k runs of 1's (n>=0, 0<=k<=floor((n+1)/2))). A Fibonacci binary word is a binary word having no 00 subword. A run of 1's is a maximal subword of the form 11..1. |
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+0
2
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| 1, 1, 1, 0, 3, 0, 4, 1, 0, 4, 4, 0, 4, 8, 1, 0, 4, 12, 5, 0, 4, 16, 13, 1, 0, 4, 20, 25, 6, 0, 4, 24, 41, 19, 1, 0, 4, 28, 61, 44, 7, 0, 4, 32, 85, 85, 26, 1, 0, 4, 36, 113, 146, 70, 8, 0, 4, 40, 145, 231, 155, 34, 1, 0, 4, 44, 181, 344, 301, 104, 9, 0, 4, 48, 221, 489, 532, 259, 43, 1
(list; graph; listen)
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OFFSET
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0,5
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COMMENT
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Row n has 1+floor((n+1)/2) terms. Row sums are the Fibonacci numbers (A000045). T(n,k)=A129717(n,k-1) (since in each word: number of runs of 1's =1 + the number of 101's). Sum(k*T(n,k),k=0..floor((n+1)/2))=A055244(n) (n>=1).
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FORMULA
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G.f.=G(t,z)=(1+z)(1-z+tz)/(1-z-tz^2). T(n,k)=binom(n-k,k-1)+2binom(n-k-1,k-1)+binom(n-k-2,k-1) for n>=4 and 0<=k<floor((n+1)/2).
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EXAMPLE
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T(6,3)=5 because we have 110101, 101101, 101010, 101011 and 010101.
Triangle starts:
1;
1,1;
0,3;
0,4,1;
0,4,4;
0,4,8,1;
0,4,12,5;
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MAPLE
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G:=(1+z)*(1-z+t*z)/(1-z-t*z^2): 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 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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Cf. A000045, A129717, A055244.
Sequence in context: A027636 A060034 A035544 this_sequence A127375 A138376 A077140
Adjacent sequences: A129715 A129716 A129717 this_sequence A129719 A129720 A129721
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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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