

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, 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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,4


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 (n3)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(n3) = 2^n. A048625 is a subsequence of A000930, A068921 and A078012. All of them satisfy the recurrence a(n) = a(n1) + a(n3).  Tanya Khovanova, Aug 22 2006


LINKS

Table of n, a(n) for n=0..31.
Index entries for linear recurrences with constant coefficients, signature (3,2,1,2).


FORMULA

a(n) = 3*2^(n3) + a(n1) + a(n3) for n >= 3.  Tanya Khovanova, Aug 22 2006
G.f.: (x^3 + x^2)/(2*x^4  x^3 + 2*x^2  3*x + 1) = x^2 * (x+1)/((2*x1)*(x^3+x1)). a(n) = 2^nA000930(n+2).  R. J. Mathar, Oct 03 2011


EXAMPLE

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


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 *)


CROSSREFS

Sequence in context: A001980 A266376 A057750 * A200759 A159347 A137531
Adjacent sequences: A118642 A118643 A118644 * A118646 A118647 A118648


KEYWORD

nonn,easy


AUTHOR

Tanya Khovanova, May 10 2006


EXTENSIONS

More terms from Joshua Zucker, Aug 04 2006
Edited by Franklin T. AdamsWatters, Sep 30 2011


STATUS

approved



