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Search: id:A119473
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| A119473 |
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Triangle read by rows: T(n,k) is number of binary words of length n and having k runs of 0's of odd length, 0<=k<=ceil(n/2). (A run of 0's is a subsequence of consecutive 0's of maximal length). |
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+0
1
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| 1, 1, 1, 2, 2, 3, 4, 1, 5, 8, 3, 8, 15, 8, 1, 13, 28, 19, 4, 21, 51, 42, 13, 1, 34, 92, 89, 36, 5, 55, 164, 182, 91, 19, 1, 89, 290, 363, 216, 60, 6, 144, 509, 709, 489, 170, 26, 1, 233, 888, 1362, 1068, 446, 92, 7, 377, 1541, 2580, 2266, 1105, 288, 34, 1, 610, 2662, 4830
(list; graph; listen)
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OFFSET
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0,4
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COMMENT
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Row n has 1+ceiling(n/2) terms. T(n,0)=Fibonacci(n+1)=A000045(n+1). T(n,1)=A029907(n). Sum(k*T(n,k),k>=0)=A059570(n).
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REFERENCES
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R. Grimaldi and S. Heubach, Binary strings without odd runs of zeros, Ars Combinatoria 75 (2005), 241-255.
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FORMULA
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G.f.=(1+tz)/(1-z-z^2-tz^2). G.f. of column k (k>=1) is z^(2k-1)*(1-z^2)/(1-z-z^2)^(k+1).
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EXAMPLE
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T(5,2)=8 because we have 00010, 01000, 01011, 01101, 01110, 10101, 10110 and 11010.
Triangle starts:
1;
1,1;
2,2;
3,4,1;
5,8,3;
8,15,8,1;
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MAPLE
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G:=(1+t*z)/(1-z-z^2-t*z^2): Gser:=simplify(series(G, z=0, 18)): P[0]:=1: for n from 1 to 14 do P[n]:=sort(coeff(Gser, z^n)) od: for n from 0 to 14 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, A029907, A059570.
Sequence in context: A104567 A087824 A008951 this_sequence A002122 A105689 A117632
Adjacent sequences: A119470 A119471 A119472 this_sequence A119474 A119475 A119476
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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 22 2006
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