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

• To: mathgroup at smc.vnet.net
• Subject: [mg49597] Re: [mg49575] Re: 3D Pascal's Triangle (Cone?)
• From: DrBob <drbob at bigfoot.com>
• Date: Sat, 24 Jul 2004 03:47:27 -0400 (EDT)
• References: <200407211814.i6LIELN23030@proapp.mathforum.org> <cdnq13\$l3v\$1@smc.vnet.net> <200407231001.GAA20526@smc.vnet.net>
• Reply-to: drbob at bigfoot.com
• Sender: owner-wri-mathgroup at wolfram.com

```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
>>>
>>
>>
>>
>
>

--
DrBob at bigfoot.com
www.eclecticdreams.net

```

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