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
>>
>
>
>