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