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A118424
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Triangle read by rows: T(n,k) is the number of binary sequences of length n containing k subsequences 001 (n,k>=0).
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1
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1, 2, 4, 7, 1, 12, 4, 20, 12, 33, 30, 1, 54, 68, 6, 88, 144, 24, 143, 291, 77, 1, 232, 568, 216, 8, 376, 1080, 552, 40, 609, 2012, 1318, 156, 1, 986, 3688, 2988, 520, 10, 1596, 6672, 6504, 1552, 60, 2583, 11941, 13702, 4266, 275, 1, 4180, 21180, 28104, 11000
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OFFSET
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0,2
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COMMENTS
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Row n has 1+floor(n/3) terms. Sum of entries in row n is 2^n (A000079). T(n,0)=A000071(n+3)=fibonacci(n+3)-1. T(n,1)=A118425(n). Sum(k*T(n,k),k=0..n-1)=(n-2)*2^(n-3) (A001787).
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LINKS
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Table of n, a(n) for n=0..54.
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FORMULA
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G.f.=G(t,z)=1/[1-2z+(1-t)z^3]. Recurrence relation: T(n,k)=2T(n-1,k)-T(n-3,k)+T(n-3,k-1) for n>=3.
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EXAMPLE
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T(7,2)=6 because we have 0bb,1bb,b0b,b1b,bb0 and bb1, where b=001.
Triangle starts:
1;
2;
4;
7,1;
12,4;
20,12;
33,30,1;
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MAPLE
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G:=1/(1-2*z+(1-t)*z^3): Gser:=simplify(series(G, z=0, 20)): P[0]:=1: for n from 1 to 17 do P[n]:=coeff(Gser, z^n) od: for n from 0 to 17 do seq(coeff(P[n], t, j), j=0..floor(n/3)) od; # yields sequence in triangular form
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CROSSREFS
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Cf. A000071, A000079, A001787, A118425.
Sequence in context: A019748 A061501 A089349 * A118429 A110317 A098073
Adjacent sequences: A118421 A118422 A118423 * A118425 A118426 A118427
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KEYWORD
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nonn,tabf
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AUTHOR
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Emeric Deutsch, Apr 27 2006
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STATUS
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approved
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