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Re: Re: All Factors of a number
I'm not sure what you mean by a "double trunk", but there are Mersennes with two factors at n=5, 9, 12, 13, 19, 23... and others. Either I'm misunderstanding you, or numbers are running together on your screen.
Timing on my wife's 2GHz dual-G5 is 818.86 seconds... not a huge improvement over your G4. It's only 42% faster despite a 60% faster clock speed!
Bobby
On Wed, 22 Dec 2004 04:52:49 -0500 (EST), János <janos.lobb at yale.edu> wrote:
> 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
>>
>
>
>
>
--
DrBob at bigfoot.com
www.eclecticdreams.net
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