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Search: id:A119462
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| A119462 |
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Triangle read by rows: T(n,k) is the number of circular binary words of length n having k occurrences of 01 (0<=k<=floor(n/2)). |
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+0
2
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| 1, 2, 2, 2, 2, 6, 2, 12, 2, 2, 20, 10, 2, 30, 30, 2, 2, 42, 70, 14, 2, 56, 140, 56, 2, 2, 72, 252, 168, 18, 2, 90, 420, 420, 90, 2, 2, 110, 660, 924, 330, 22, 2, 132, 990, 1848, 990, 132, 2, 2, 156, 1430, 3432, 2574, 572, 26, 2, 182, 2002, 6006, 6006, 2002, 182, 2, 2, 210, 2730
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OFFSET
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0,2
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COMMENT
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Row n contains 1+floor(n/2) terms. Sum of entries in row n is 2^n (A000079). T(n,0)=2 for n>=1. T(n,1)=2*binomial(n,2)=A002378(n-1). T(n,2)=2*binomial(n,4)=A034827(n). T(n,k)=2*A034239(n-1,k) for n>=1. Sum(k*T(n,k),k=0..floor(n/2))=A057711(n).
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REFERENCES
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L. Carlitz and R. Scoville, Zero-one sequences and Fibonacci numbers, Fibonacci Quarterly, 15 (1977), 246-254.
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FORMULA
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T(n,k)=2*binomial(n,2k) for n>=1; T(0,0)=1. T(n,k)=2T(n-1,k)-T(n-2,k)+T(n-2,k-1) for n>=3. G.f.=G(t,z)=(1-z^2+tz^2)/(1-2z+z^2-tz^2).
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EXAMPLE
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T(3,1)=6 because we have 001,010,011,100,101 and 110.
Triangle starts:
1;
2;
2,2;
2,6;
2,12,2;
2,20,10;
2,30,30,2;
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MAPLE
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T:=proc(n, k) if n=0 and k=0 then 1 else 2*binomial(n, 2*k) fi end: for n from 0 to 15 do seq(T(n, k), k=0..floor(n/2)) od; # yields sequence in triangular form
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CROSSREFS
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Cf. A000079, A002378, A034827, A034239, A057711.
Sequence in context: A163368 A151948 A080400 this_sequence A096625 A103222 A061033
Adjacent sequences: A119459 A119460 A119461 this_sequence A119463 A119464 A119465
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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 21 2006
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