

A118429


Triangle read by rows: T(n,k) is the number of binary sequences of length n containing k subsequences 010 (n,k >= 0).


4



1, 2, 4, 7, 1, 12, 4, 21, 10, 1, 37, 22, 5, 65, 47, 15, 1, 114, 98, 38, 6, 200, 199, 91, 21, 1, 351, 396, 210, 60, 7, 616, 777, 468, 158, 28, 1, 1081, 1508, 1014, 396, 89, 8, 1897, 2900, 2151, 952, 255, 36, 1, 3329, 5534, 4487, 2212, 687, 126, 9, 5842, 10492, 9229
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OFFSET

0,2


COMMENTS

Row n has ceiling(n/2) terms (n >= 1).
Sum of entries in row n is 2^n (A000079).
T(n,0) = A005251(n+3), T(n,1) = A118430(n).
Sum_{k=0..n1} k*T(n,k) = (n2)*2^(n3) (A001787).


LINKS

Alois P. Heinz, Rows n = 0..199, flattened


FORMULA

G.f.: G(t,z) = (1+(1t)z^2)/(1  2z + (1t)z^2  (1t)z^3).
Recurrence relation: T(n,k) = 2T(n1,k)  T(n2,k) + T(n3,k) + T(n2,k1)  T(n3,k1) for n >= 3.


EXAMPLE

T(6,2) = 5 because we have 010010, 010100, 010101, 001010 and 101010.
Triangle starts:
1;
2;
4;
7, 1;
12, 4;
21, 10, 1;
37, 22, 5;


MAPLE

G:=(1+(1t)*z^2)/(12*z+(1t)*z^2(1t)*z^3): Gser:=simplify(series(G, z=0, 18)): P[0]:=1: for n from 1 to 16 do P[n]:=sort(coeff(Gser, z^n)) od: 1; for n from 1 to 16 do seq(coeff(P[n], t, j), j=0..ceil(n/2)1) od; # yields sequence in triangular form


MATHEMATICA

nn=15; Map[Select[#, #>0&]&, CoefficientList[Series[1/(12z(u1)z^3/(1(u1)z^2)), {z, 0, nn}], {z, u}]]//Grid (* Geoffrey Critzer, Dec 03 2013 *)


CROSSREFS

Cf. A000079, A005251, A001787, A118430.
Sequence in context: A272001 A089349 A118424 * A110317 A098073 A118390
Adjacent sequences: A118426 A118427 A118428 * A118430 A118431 A118432


KEYWORD

nonn,tabf


AUTHOR

Emeric Deutsch, Apr 27 2006


STATUS

approved



