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MathGroup Archive 2004

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

  • To: mathgroup at smc.vnet.net
  • Subject: [mg49625] Re: 3D Pascal's Triangle (Cone?)
  • From: Erich Neuwirth <erich.neuwirth at univie.ac.at>
  • Date: Sun, 25 Jul 2004 02:55:27 -0400 (EDT)
  • References: <200407211814.i6LIELN23030@proapp.mathforum.org> <cdnq13$l3v$1@smc.vnet.net> <200407231001.GAA20526@smc.vnet.net> <cdt5gl$6du$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Perhaps it should be
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]]


DrBob wrote:

> That was one typo down, and three to go (missing "[", missing "]", and "." where you should have ",").
> 
> It should be:
> 
> 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]]
> 
> But the resulting plot is just a rectangular brick wall, not a triangle of any kind.
> 
> Bobby
> 
> On Fri, 23 Jul 2004 06:01:47 -0400 (EDT), Roger L. Bagula <rlbtftn at netscape.net> wrote:
> 
> 
>>typo: extra "}" in the cuboid:
>>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]]
>>
>>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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