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Re: All Factors of a number
Bobby,
Nice !!
You must have a minimum 17 inch monitor as I see :)
With the best,
János
P.S: G4 1.25G Total Time: 1161.98 Second. Not bad for a Mac ! I
expected 1277.44 Second. Here is a naive question for you. Will you
ever have a "trunk" of length 2, so this tree can stand on it ? If I am
counting from the bottom up, the first single trunk is at n=31, there
is no double trunk, the only triple is at n={6,7,8} and well the top
with length 4 is the top. I skipped lectures when number theory was
taught, - which might says, there will be never a double "trunk", may
be not even a single one with n > 54 -, so I am not a good candidate to
answer it :)
On Dec 21, 2004, at 5:19 AM, DrBob wrote:
> FactorInteger does this already.
>
> Just for grins, here's code to list factorizations of Mersenne numbers
> in Xmas tree format.
>
> toPowers = {{a_Integer, 1} -> HoldForm[a], {a_Integer, b_Integer} ->
> HoldForm[a]^HoldForm[b],
> List -> Times};
> toStars = StringReplace[ToString[#1 /. toPowers], " " -> " * "] & ;
> n = 1;
> {totalTime, results} =
> Timing[First[Last[Reap[While[n <= 54, p = Prime[n];
> Sow[Timing[{n, p, FactorInteger[2^p - 1]}]]; n++]]]]] /.
> {(s_)*Second, {n_, p_, f_}} :> {n, p, s, toStars[f]};
> TableForm[results, TableAlignments -> {Center, Center, Right},
> TableHeadings -> {None, {"n", "p = Prime[n]"*1, Seconds, "Factors of
> 2^p-1"}}]
> Print["Total Time: ", totalTime]
>
> That took 499 seconds on my AMD 3200+ machine, so reduce 54 to a
> smaller number if you're using a slower machine or don't have the
> patience. Of those 499 seconds, 125 were used for the 54th Mersenne,
> 201 for the 47th Mersenne, and 63 for the 50th. The rest go pretty
> fast, so reducing 54 to 46 will work well.
>
> Bobby
>
> On Mon, 20 Dec 2004 06:34:49 -0500 (EST), Ulrich Sondermann
> <usondermann at earthlink.net> wrote:
>
>
>> 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}
>>
>>
>>
>>
>>
> -- DrBob at bigfoot.com
> www.eclecticdreams.net
>
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