Re: All Factors of a number
- To: mathgroup at smc.vnet.net
- Subject: [mg53049] Re: [mg53025] All Factors of a number
- From: Bob Hanlon <hanlonr at cox.net>
- Date: Tue, 21 Dec 2004 05:19:39 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
A shorter method than my first response:
Needs["DiscreteMath`Combinatorica`"];
bt=Table[KSubsets[{1,2,3,5},a],{a,3}];
Times@@@#&/@bt
{{1, 2, 3, 5}, {2, 3, 5, 6, 10, 15}, {6, 10, 15, 30}}
Bob Hanlon
>
> From: Bob Hanlon <hanlonr at cox.net>
To: mathgroup at smc.vnet.net
> Date: 2004/12/20 Mon PM 07:52:19 EST
> To: "Ulrich Sondermann" <usondermann at earthlink.net>,
<mathgroup at smc.vnet.net>
> Subject: [mg53049] Re: [mg53025] All Factors of a number
>
> Needs["DiscreteMath`Combinatorica`"];
>
> bt=Table[KSubsets[{1,2,3,5},a],{a,3}];
>
> ((Times@@#)&/@#)&/@bt
>
> {{1, 2, 3, 5}, {2, 3, 5, 6, 10, 15}, {6, 10, 15, 30}}
>
>
> Bob Hanlon
>
> >
> > From: "Ulrich Sondermann" <usondermann at earthlink.net>
To: mathgroup at smc.vnet.net
> > To: mathgroup at smc.vnet.net
> > Subject: [mg53049] [mg53025] All Factors of a number
> >
> > Following is a test code that I am trying to get down to a set of
> > instructions that will allways put all of the factors of a number in a table
> > or list.
> > bt contains all of the factors if I multiply each list entry, however I
> > cannot accomplish that with a single line, as an example I have broken
into
> > the three lists needed for this example. The results of each "Times@@" is
> > what I am after all placed into one table. All of my attempts have proved
> > disasterous, I am new to Mathematica and could do this with nested
loops
> in
> > any programming language, but this has me stumped.
> > Thanx!
> >
> > <<DiscreteMath`Combinatorica`
> > bt=Table[KSubsets[{1,2,3,5},a],{a,3}]
> > {{{1},{2},{3},{5}},{{1,2},{1,3},{1,5},{2,3},{2,5},{3,5}},{{1,2,3},{1,2,5},{1
> > ,3,5},{2,3,5}}}
> > bt1=bt[[1,All]]
> > {{1},{2},{3},{5}}
> > Table[Times@@bt1[[a]],{a,4}]
> > {1,2,3,5}
> > bt2=bt[[2,All]]
> > {{1,2},{1,3},{1,5},{2,3},{2,5},{3,5}}
> > Table[Times@@bt2[[a]],{a,6}]
> > {2,3,5,6,10,15}
> > bt3=bt[[3,All]]
> > {{1,2,3},{1,2,5},{1,3,5},{2,3,5}}
> > Table[Times@@bt3[[a]],{a,4}]
> > {6,10,15,30}
> >
> >
>
Bob Hanlon
Chantilly, VA