## Can you find the missing number sequence?
Don't be misled by the simple linear thinking of other |

## Solution: The 5th Dimension

Yeah, it's all about dimensions!Before jumping to the mere formula, let's have a look at what above matrix represents in a multidimensional world.

The first number on top of our matrix (1) represents

**a dot**, a single point with no sizes in a world without dimensions, the whole universe in a dot, a singularity, for some the beginning of our universe itself, but this is a different story. Clone, duplicate our dot and you'll get two dots, connect the dots together and here you have

**a line**lying on a one dimensional space, and that's the second sequence on our matrix (1 - 2), do the same on the third sequence: clone the line, connect the edges and...

**got it**?

Yeah, the third sequence represents

**a square**in a two dimensional space: 4 dots or vertices connected by 4 lines shaping 1 square (1 - 4 - 4). Logically the fourth sequence (1 - 6 - 12 - 8) represents

**a cube**composed by 8 vertices, 12 lines and 6 faces in a three dimensional space.

And now it becomes interesting: using the same analogy, the next sequence would represents a cube in a four dimensional space (a tesseract) and therefore the missing sequence represents a hypercube in the fifth dimension!
Left, a 3D image of a Tesseract
shadow while rotating in its own 4th dimension. Yes, we can only visualize its shadow!Fascinating, isnt it? |

**The Formula**

Each cell in our matrix represents the property (in terms of edges, lines, faces, etc.) of a shape in a given spatial

*n*dimension. And we've seen that a shape in a

*n*th dimensional space is created by cloning/duplicating the corresponding shape in the previous (

*n*- 1) dimension and by connecting all vertices of the two shapes. Take the number 6 in the the 3rd dimension, it represents the six faces of a cube and can be constructed by cloning the corresponding square in the 2nd dimension (1 + 1) and by connecting all edges with 4 lines. At the same way, the 32 lines of our tesseract in the 4th dimension are given by cloning the 12 lines from the shape of the previous dimension (12 + 12) and by adding the corresponding connections (8). So

**6**= 2*

**1**+

**4**and

**32**= 2*

**12**+

**8**.

**S(p,n) = 2 * S(p, n-1) + S(p-1 ,n-1)**

**S**= Shape in a cell

**p**= Property coordinate (column index in our matrix - from right to left)

**p**= Dimension coordinate (row index in our matrix - from top to bottom)

Or simply: each number in a cell (ie 32) is given by doubling the number in the cell of the previous row (2 * 12) and adding the one from this latter right adjacent cell (8). The missing sequence is then:

**1 10 40 80 80 32**