John included a great puzzle:
Hmmmm. This will be the only Math Teachers at Play ever directly after a square and before a cube. (Proof.) Of course, in a 3x3x3 cube, only 26 cubes are visible, so you're really taking the 27th on faith. I guess that makes 26 the third Rubik's Cube number? If 8 is the previous and 56 is the next, what is the fifth Rubik's Cube number? What is the closed form for the nth Rubik's Cube Number?I liked Princess Kitten's post on playing an online game called Cops and Robbers, and from a comment there, I found this cool site - ThinkQuest: "Over 7,000 websites created by students around the world who have participated in a ThinkQuest Competition". I really like seeing math content from kids.
I loved finding out in this post that someone has used a bit of math (or engineering) know-how to help with the difficult situation the people of Haiti are facing.
Lots more goodies over there - Check it out!
Posts
There are 8 corners, each of which has 1 cube.
ReplyDeleteThere are 12 edges, each of which has n-2 cubes.
There are 6 sides, each of which has (n-2)^2 cubes not on edges or corners.
So the nth Rubics Cube number is 6(n-2)^2 + 12(n-2) + 8
Check: A 2x2x2 would have 6*0^2+12*0+8 = 8, which is correct. A 3x3x3 would have 6*1^2+12*1+8=26. A 4x4x4 would have 6*4+12*2+8 = 24+24+8 = 56. A 5x5x5 would have 6*3^2 + 12*3 + 8 = 54+36+8 = 98.
Sounds reasonable to me.
Of course, multiplying it all out yields 6n^2 - 12n + 8, I think. Three data points will check it: 2x2x2: 6*4-12*2+8=8, 3x3x3: 6*9-12*3+8=26, 4x4x4: 6*16-12*4+8=96-48+8=56. It's right.
I held Blaise's comment for a few days, so his answer wouldn't spoil people's fun. He and I discussed by email, and he also wrote (and suggested I post):
ReplyDeleteSloane's Encyclopedia of Integer Sequences describes a slightly
different version of the sequence, where it counts the number of
points on the surface of a cube where each face is marked with an n+1
square lattice of points extending to the edge of the cube (or in my
words, the number of distinct visible vertices on an n-Rubik's cube,
where a vertex is defined as the (possibly shared) corner of a
sub-cube). By this definition, a 0-Cube would have 1 point, a 1-Cube
would have 8, etc. Essentially, it shifts the function down by one.
This has the pleasing result of changing your n^3-(n-2)^3 to the more
symmetric (n+1)^3 - (n-1)^3 and my quadratic formula from 6n^2-12n+8
to a more wieldy and easy to remember 6n^2+2 -- both results are
listed in the entry
http://www.research.att.com/~njas/sequences/A005897