John included a great puzzle:

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.Hmmmm.This will be the onlyMath Teachers at Playever 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 thenth Rubik's Cube Number?

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!

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