Timeline for Recurrence relation for number of bit strings of length n that contain two consecutive 1s

14 events

when toggle format	what		by	license	comment
Feb 10, 2022 at 13:27	comment	added	R. J. Mathar		This is in oeis.org/A008466, where the recurrence is in a formula.
S Oct 21, 2016 at 15:49	history	edited	amWhy	CC BY-SA 3.0	added 21 characters in body
Oct 21, 2016 at 15:46	review	Suggested edits
S Oct 21, 2016 at 15:49
May 9, 2015 at 14:23	answer	added	Kevin Zakka		timeline score: 8
Dec 11, 2013 at 22:59	vote	accept	Scott Odle
Dec 11, 2013 at 22:59	vote	accept	Scott Odle
Dec 11, 2013 at 22:59
Dec 11, 2013 at 22:59	vote	accept	Scott Odle
Dec 11, 2013 at 22:59
Dec 11, 2013 at 22:22	answer	added	André Nicolas		timeline score: 4
Dec 11, 2013 at 21:58	answer	added	Ross Millikan		timeline score: 2
Dec 11, 2013 at 21:48	comment	added	Thomas Kalinowski		In your approach, $count(Y\cap Z)$ looks wrong. For a direct approach you could count (1) the n-bit strings with two consecutive 1's that start with 0, (2) the n-bit strings with two consecutive 1's that start with 10, and (3) the n-bit strings with two consecutive 1's that start with 11.
Dec 11, 2013 at 21:46	review	First posts
Dec 11, 2013 at 22:02
Dec 11, 2013 at 21:45	comment	added	André Nicolas		The standard way is to find a recurrence for the number $b_n$ of bit strings that have no two consecutive $1$'s, then $a_n=2^n-b_n$.
Dec 11, 2013 at 21:35	history	edited	Scott Odle	CC BY-SA 3.0	added 47 characters in body
Dec 11, 2013 at 21:27	history	asked	Scott Odle	CC BY-SA 3.0