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 >> > > >