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MathGroup Archive 2004

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Re: 3D Pascal's Triangle (Cone?)

  • To: mathgroup at smc.vnet.net
  • Subject: [mg49545] Re: 3D Pascal's Triangle (Cone?)
  • From: snoofly <snoofly at snoofly.net>
  • Date: Fri, 23 Jul 2004 05:59:34 -0400 (EDT)
  • References: <200407211814.i6LIELN23030@proapp.mathforum.org> <cdnq13$l3v$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

corrected:

g = Flatten[Table[If [Mod[Multinomial[x, y, z], 2] == 1, Cuboid[1.2*{x, 
y, -z}], {}], {x, 0, 15}, {y, 0, 15}, {z, 0, 15}]]
Show[Graphics3D[g]]

Roger L. Bagula wrote:
> There is and old Visualization in Mathematica that
> gives a modulo 2 version of a Pascal's triangle.
> It is a right angle version of a tetrahedral 3d Sierpiski triangle.
> Here it is: ( copyright Mathematica):
> 
> g=Flatten[Table[If 
> Mod[Multinomial[x,y,x],2]==1,Cuboid[1.2*{x,y,-z}}],{}],{x,0,15},{y.0,15},{z,0,15}]
> Show[Graphics3D[g]]
> 
> phil wrote:
> 
>>Is there a three dimensional version of Pascal's
>>triangle? If so, I suppose it would be a cone (?).
>>Applications?
>>
>>phil
>>
> 
> 
> 


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