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Re: Re: All Factors of a number

By "double trunk" I meant to have to consecutive rows where the 2^p-1 
has no factors. Like n=6 and 7, or n=7 and 8. To reformat it as a 
better mathematical question: Is there two consecutive primes Prime[n] 
and Prime[n+1] such that 2^Prime[n]-1 and 2^Prime[n+1]-1 are also 
primes for n > 8, that is they have no factors other than themselves 
and 1.

If there would be such numbers, they would give a nice trunk to the 
"christmas tree" to stand upright.

Thanks ahead,
János

On Dec 22, 2004, at 12:52 PM, DrBob wrote:



> 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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