From: rcutler@banach.WPI.EDU (Robert Cutler)
Subject: Pascals triangle (mod infite prime)
Originator: rcutler@banach.WPI.EDU
Organization: Worcester Polytechnic Institute
Date: Tue, 28 Jul 1992 17:42:50 GMT




	Due to Paul Palmers request here is a neat way to conceptualize
pascals triangle.  (This was presented at a math lecture at Bard College
by a mathematician.  When I remember his name I will give him dredit)




	As you know pascals tringle is as follows:



		1
		1  1
		1  2  1
		1  3  3  1
		1  4  6  4  1
		1  5 10 10  5  1 
		1  6 15 20 15  6 1
		.                  .    
		.                     .
		.                        .



So lets extend the triange.  We know that all places to the right and
left of the terminating 1 at each row is zero. (In his case we are 
looking at the degree of each dimensional shape.  Start by taking a 
vertex.  There is one vertex, zero lines, zero faces....   Next take 
another vertex extend it a dimension away and connect to all previous 
vertices.  This gives 2 vertices, 1 edge.  Then 3 vertices, 3 edges, 
1 face.  Then 4 vertices, 6 edges, 4 faces, 1 solid......  The row of 
1's is a -1 dimensional solid that all of these forms have in common. 
Except this for now, it creates a justification for later work.)




 . . . 0  0  0  1  0  0  0  0  0  0  .  .  . 
 . . . 0  0  0  1  1  0  0  0  0  0  .  .  .
 . . . 0  0  0  1  2  1  0  0  0  0  .  .  .
 . . . 0  0  0  1  3  3  1  0  0  0  .  .  .
 . . . 0  0  0  1  4  6  4  1  0  0  .  .  .
 . . . 0  0  0  1  5 10 10  5  1  0  .  .  .
 . . . 0  0  0  1  6 15 20 15  6  1  .  .  .
                .                    .    
                .                       .
                .                          .



	
From here The question is what would the triangle look like above the 
first line?  Extending the 1's column up and the diagonal over we can 
fill in the triangle as follows. 



 
 . . .  -5  1  0  0  0  0  1 -5 15 -35 70 -126 .  .  . 
 . . .  10 -4  1  0  0  0  1 -4 10 -20 35 -56  .  .  . 
 . . . -10  6 -3  1  0  0  1 -3  6 -10 15 -21  .  .  . 
 . . .   5 -4  3 -2  1  0  1 -2  3  -4  5  -6  .  .  . 
 . . .  -1  1 -1  1 -1  1  1 -1  1  -1  1  -1  .  .  . 
 . . .   0  0  0  0  0  0  1  0  0   0  0   0  .  .  . 
 . . .   0  0  0  0  0  0  1  1  0   0  0   0  .  .  .
 . . .   0  0  0  0  0  0  1  2  1   0  0   0  .  .  .
 . . .   0  0  0  0  0  0  1  3  3   1  0   0  .  .  .
 . . .   0  0  0  0  0  0  1  4  6   4  1   0  .  .  .
 . . .   0  0  0  0  0  0  1  5 10  10  5   1  .  .  .
 . . .   0  0  0  0  0  0  1  6 15  20 15   6  .  .  .
                        .                    .    
                        .                       .
                        .                          . 
 
	The top portion comes from continuing on the relation
that every number plus the one to the left gives then number below
it.

	If you noticed there is a fudge right in the center where
1+1=2 but instead 1 is placed in and this was where the interesting
results started appearing.  



Lets side track for a second.  By taking the sixth row of pascals
triangle and modding it with 7 you get the following result.

   (1  6 15  20 15   6  1)  mod 7 = (1 -1  1 -1  1 -1  1)

If you look at this result and compare it to the -2nd row which is 
made up of all 1's and -1's, you can quickly get the feel of the 
argument.  The -2nd can be thought of as the bottom row if Pascals
Triangle mod (infinate prime)  If you will accept that idea, then
the upper right quadrant corresonds to the bottom left corner of
Pascals Triangle Mod (infinite prime), the upper left corresponds
to the bottom right corner of Pascals Triangle mod (infinite prime),
The lower left corresponds 

         1  0  0   0  0   0  .  .  .   0  0  0  0  0  0   
         1  1  0   0  0   0  .  .  .   0  0  0  0  0  0  
         1  2  1   0  0   0  .  .  .   0  0  0  0  0  0   
         1  3  3   1  0   0  .  .  .   0  0  0  0  0  0   
         1  4  6   4  1   0  .  .  .   0  0  0  0  0  0   
         1  5 10  10  5   1  .  .  .   0  0  0  0  0  0   
         1  6 15  20 15   6  1  .  .   0  0  0  0  0  0   
         .                      .      .   
         .                         .   . 
         .                           
         .                             1  0  0  0  0  0
         1 -5 15 -35 70 -126 .  .  .  -5  1  0  0  0  0
         1 -4 10 -20 35 -56  .  .  .  10 -4  1  0  0  0 
         1 -3  6 -10 15 -21  .  .  . -10  6 -3  1  0  0 
         1 -2  3  -4  5  -6  .  .  .   5 -4  3 -2  1  0 
         1 -1  1  -1  1  -1  .  .  .  -1  1 -1  1 -1  1
 



The fudge from above comes from the fact that at that point there
are actually two triangles terminating at the point and the 1+1 = 2
but we actually use 1 discrepancy is because there are actually two
copies of Pascals Triangle starting at line 0, but we are only looking
at one copy.



	If anything was unclear, or you would like me to go into more
depth at any are I would be happy to do so.



				Rob