|
|
|
|
|
|
|
|
|
#30 by Max Alekseyev at Sat Apr 25 21:13:25 EDT 2015
|
|
|
|
|
|
|
|
#29 by Max Alekseyev at Sat Apr 25 21:13:23 EDT 2015
|
|
| COMMENTS
|
The number of solutions to the puzzle is 2^a(n). If a(n)=0 then the puzzle has ana unique solution.
|
|
| STATUS
|
approved
editing
|
|
|
|
|
|
|
#28 by M. F. Hasler at Tue Apr 21 22:35:15 EDT 2015
|
|
|
|
|
|
|
|
#27 by M. F. Hasler at Tue Apr 21 22:19:24 EDT 2015
|
|
|
|
|
|
|
|
#26 by M. F. Hasler at Tue Apr 21 22:18:02 EDT 2015
|
|
| LINKS
|
Max Alekseyev and Thomas Buchholz, <a href="/A165738/b165738.txt">Table of n, a(n) for n = 1..1000</a>. [Recomputed>. (Terms n=91 thesethrough a(n), =1000 from Thomas Buchholz, May 20 2014])
|
|
|
|
|
|
|
#25 by M. F. Hasler at Tue Apr 21 22:15:12 EDT 2015
|
|
| COMMENTS
|
a(n) <= 2n.
|
|
| LINKS
|
Max Alekseyev and Thomas Buchholz, <a href="/A165738/b165738.txt">Table of n, a(n) for n = 1..1000</a> [recomputed>. [Recomputed these a(n), May 20 2014]
Andries E. Brouwer, <a href="http://www.win.tue.nl/~aeb/ca/madness/madrect.html">Lights Out and Button Madness Games</a> [>. [Gives theory and a(n) for n = 1..1000, Jun 19 2008]
|
|
| FORMULA
|
a(n) <= 2n.
a(n) is a multiple of 4 and satisfies a(2n) = 2a(n). a(n+1) = 2 * A159257(n) + 4 if n = 2 (mod 3) and a(n+1) = 2 * A159257(n) otherwise. - Thomas Buchholz, May 22 2014
|
|
| STATUS
|
proposed
editing
|
|
|
|
|
|
|
#24 by Michel Marcus at Tue Apr 21 09:08:41 EDT 2015
|
|
|
|
|
|
Discussion
|
| Tue Apr 21
| 14:53
| N. J. A. Sloane: The deleted comment looks OK to me!
|
|
| 15:33
| Andries E. Brouwer: Yes, nothing wrong with that comment. But everything on this page attributed to Thomas Buchholz was taken from my webpage. No information is lost by deleting it. The webpage contains much more, so it is just a random factoid from there.
|
|
| 22:11
| M. F. Hasler: Andries, here on OEIS signatures do not mean that the person has found the fact, but that he added it to the OEIS. Credits to the discoverer are given via references and maybe additional comments etc.
|
|
|
|
|
|
|
#23 by Michel Marcus at Tue Apr 21 09:07:50 EDT 2015
|
|
| LINKS
|
AndriesM. Anderson and ET. BrouwerFeil, <a href="http://www.win.tuejstor.nl/~aeb/caorg/madnessstable/madrect.html2690705">Turning Lights Out and Buttonwith MadnessLinear GamesAlgebra</a> [Gives theory and>, Mathematics a(n) forMagazine, 71 (1998), 300-303.
Andries E. Brouwer, <a href="http://www.win.tue.nl/~aeb/ca/madness/madrect.html">Lights Out and Button Madness Games</a> [Gives theory and a(n) for n = 1..1000, Jun 19 2008]
M. Anderson and T. Feil, <a href="http://www.jstor.org/stable/2690705">Turning Lights Out with Linear Algebra</a>, Mathematics Magazine, 71 (1998), 300-303
|
|
| FORMULA
|
a(n+1) = 2 * A159257(n) + 4 if n = 2 (mod 3) and a(n+1) = 2 * A159257(n) otherwise. - . - _Thomas Buchholz, _, May 22 2014
|
|
| STATUS
|
proposed
editing
|
|
|
|
|
Discussion
|
| Tue Apr 21
| 09:08
| Michel Marcus: The comment was incorrect ?
|
|
|
|
|
|
|
#22 by Andries E. Brouwer at Tue Apr 21 08:22:34 EDT 2015
|
|
|
|
|
|
|
|
#21 by Andries E. Brouwer at Tue Apr 21 08:18:30 EDT 2015
|
|
| COMMENTS
|
a(n) is a multiple of 4 and satisfies a(2n) = 2a(n). - Thomas Buchholz, May 22 2014
|
|
| LINKS
|
Max Alekseyev and Thomas Buchholz, <a href="/A165738/b165738.txt">Table of n, a(n) for n = 1..1000</a> [terms 91 through 1000 were computed byrecomputed Thomasthese Buchholz, a(n), May 20 2014]
Andries E. Brouwer, <a href="http://www.win.tue.nl/~aeb/ca/madness/madrect.html">Lights Out and Button Madness Games</a> [Gives theory and a(n) for
n = 1..1000, Jun 19 2008]
Andries E. Brouwer, <a href="http://www.win.tue.nl/~aeb/ca/madness/madrect.html">Lights Out and Button Madness Games</a>
|
|
| STATUS
|
approved
editing
|
|
|
|
|
|
