A118892 -id:A118892 - cp's OEIS frontend

A049864 a(n) = Sum_{k=0,1,2,...,n-4,n-2,n-1} a(k); a(n-3) is not a summand, with a(0)=a(1)=a(2)=1.

1, 1, 1, 2, 4, 8, 15, 28, 52, 97, 181, 338, 631, 1178, 2199, 4105, 7663, 14305, 26704, 49850, 93058, 173717, 324288, 605368, 1130077, 2109583, 3938086, 7351463, 13723420, 25618337, 47823297, 89274637, 166654357, 311103754, 580756168, 1084132616, 2023815835

Offset: 0

Clark Kimberling

Number of binary sequences of length n-2 with no subsequence 0110. E.g., a(7)=28 because among the 32 (=2^5) binary sequences of length 5 only 01100,01101,00110 and 10110 contain the subsequence 0110. - Emeric Deutsch, May 04 2006
This is a_3(n) in the Doroslovacki reference. - Max Alekseyev, Jun 26 2007
Column 0 of A118890. - Emeric Deutsch, May 04 2006

R. Doroslovacki, Binary sequences without 011...110 (k-1 1's) for fixed k, Mat. Vesnik 46 (1994), no. 3-4, 93-98.
Index entries for linear recurrences with constant coefficients, signature (2,0,-1,1).

Cf. A005251, A049858, A118890, A118891, A118892.

(With a different offset:) a[0]:=1:a[1]:=2:a[2]:=4:a[3]:=8: for n from 4 to 35 do a[n]:=2*a[n-1]-a[n-3]+a[n-4] od: seq(a[n],n=0..35); # Emeric Deutsch, May 04 2006

LinearRecurrence[{2,0,-1,1},{1,1,1,2},40] (* Harvey P. Dale, Sep 24 2013 *)

a(n) = 2*a(n-1) - a(n-3) + a(n-4).

G.f.: (1+z)*(1-z)^2/(1 - 2z + z^3 - z^4). - Emeric Deutsch, May 04 2006

Edited by N. J. A. Sloane, Nov 16 2007, at the suggestion of Max Alekseyev

A118890 Triangle read by rows: T(n,k) is the number of binary sequences of length n containing k subsequences 0110 (n,k >= 0).

Original entry on oeis.org

1, 2, 4, 8, 15, 1, 28, 4, 52, 12, 97, 30, 1, 181, 70, 5, 338, 156, 18, 631, 339, 53, 1, 1178, 722, 142, 6, 2199, 1515, 357, 25, 4105, 3140, 862, 84, 1, 7663, 6444, 2018, 252, 7, 14305, 13116, 4614, 700, 33, 26704, 26513, 10348, 1846, 124, 1, 49850, 53280, 22844

Offset: 0

Emeric Deutsch, May 04 2006

Row n has ceiling(n/3) terms (n>=1).
Sum of entries in row n is 2^n (A000079).
T(n,0) = A049864(n).
T(n,1) = A118892(n).
Sum_{n>=0} k*T(n,k) = (n-3)*2^(n-4) (A001787).

			T(8,2) = 5 because we have 01100110, 01101100, 01101101, 00110110 and 10110110.
Triangle starts:
    1;
    2;
    4;
    8;
   15,   1;
   28,   4;
   52,  12;
   97,  30,  1;
  181,  70,  5;
  338, 156, 18;
  631, 339, 53, 1;

Alois P. Heinz, Rows n = 0..250, flattened

Cf. A000079, A049864, A118892, A001787.

G:=(1+(1-t)*z^3)/(1-2*z+(1-t)*(1-z)*z^3): Gser:=simplify(series(G,z=0,24)): P[0]:=1: for n from 1 to 18 do P[n]:=sort(coeff(Gser,z^n)) od: 1; for n from 1 to 18 do seq(coeff(P[n],t,j),j=0..ceil(n/3)-1) od; # yields sequence in triangular form

nn=18;c=x^3;Map[Select[#,#>0&]&,CoefficientList[Series[1/(1-2x - (y-1)x^4/ (1-(y-1)c)),{x,0,nn}],{x,y}]]//Flatten (* Geoffrey Critzer, Dec 25 2013 *)

G.f.: G(t,z) = (1+(1-t)z^3)/(1 - 2z + (1-t)(1-z)z^3).

cp's OEIS Frontend

A049864 a(n) = Sum_{k=0,1,2,...,n-4,n-2,n-1} a(k); a(n-3) is not a summand, with a(0)=a(1)=a(2)=1.

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