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
- References:
- Re: 3D Pascal's Triangle (Cone?)
- From: "Roger L. Bagula" <rlbtftn@netscape.net>
- Re: 3D Pascal's Triangle (Cone?)