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A118424
Triangle read by rows: T(n,k) is the number of binary sequences of length n containing k subsequences 001 (n,k>=0).
3
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
OFFSET
0,2
COMMENTS
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).
LINKS
FORMULA
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.
EXAMPLE
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;
MAPLE
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
MATHEMATICA
nn=15; Map[Select[#, #>0&]&, CoefficientList[Series[1/(1-2z-(u-1)z^3), {z, 0, nn}], {z, u}]]//Grid (* Geoffrey Critzer, Dec 03 2013 *)
CROSSREFS
KEYWORD
nonn,tabf
AUTHOR
Emeric Deutsch, Apr 27 2006
STATUS
approved