A118645 Number of binary strings of length n such that there exist three consecutive digits where at least two of them are 1's.

0, 0, 1, 4, 10, 23, 51, 109, 228, 471, 964, 1960, 3967, 8003, 16107, 32362, 64941, 130200, 260866, 522415, 1045831, 2093129, 4188408, 8379967, 16764552, 33535872, 67081663, 134177863, 268377031, 536785286, 1073616333, 2147299732

Offset: 0

Author

Tanya Khovanova, May 10 2006

Keywords

• nonn
• easy

Comments

We set a(2) = 1 by convention; there is one string of length 2 which has two consecutive 1's, namely 11. This also makes various formulas simpler.

For n>=3, a(n) = 2^n - the sum of all terms in the (n-3)rd power of the 4 X 4 matrix [[1 1 0 0] [0 0 1 0] [0 0 0 1] [1 1 0 0]] because this matrix represents the transitions from the state where the last three bits are 000, 001, 010, 100 to the state after the next bit, always avoiding two 1's out of the last three bits. - Joshua Zucker, Aug 04 2006

Complementary to A048625 which starts 4,6,9,13,19,28,41,60,88,129,189. For n >= 3, a(n) + A048625(n-3) = 2^n. A048625 is a subsequence of A000930, A068921 and A078012. All of them satisfy the recurrence a(n) = a(n-1) + a(n-3). - Tanya Khovanova, Aug 22 2006

Examples

			a(4) = 10 because we have: 0011, 0101, 0110, 0111, 1010, 1011, 1100, 1101, 1110, 1111. - _Geoffrey Critzer_, Jan 19 2014
		

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-2,1,-2).

Programs

Mathematica

nn=31;r=Solve[{s==1+x s+x b,a==x s,b==x a,c==x a+x b+2x c},{s,a,b,c}]; CoefficientList[Series[c/.r,{x,0,nn}],x] (* Geoffrey Critzer, Jan 19 2014 *)
LinearRecurrence[{3,-2,1,-2},{0,0,1,4},40] (* Harvey P. Dale, Dec 15 2014 *)

PARI

x='x+O('x^50); concat([0,0], Vec((x^3 + x^2)/(2*x^4 - x^3 + 2*x^2 - 3*x + 1))) \\ G. C. Greubel, May 02 2017

Formula

a(n) = 3*2^(n-3) + a(n-1) + a(n-3) for n >= 3. - Tanya Khovanova, Aug 22 2006

From R. J. Mathar, Oct 03 2011: (Start)

G.f.: (x^3 + x^2)/(2*x^4 - x^3 + 2*x^2 - 3*x + 1).

G.f.: x^2 * (x+1)/((2*x-1)*(x^3+x-1)).

a(n) = 2^n - A000930(n+2). (End)

Extensions

More terms from Joshua Zucker, Aug 04 2006

Edited by Franklin T. Adams-Watters, Sep 30 2011