Prime factors for your numbers :
2
5
2122943609450249057621
405620061201508643988375124643817406843652803
34444340671702467488860342883734479061576134487870291451728421045657
And
17222170335851233744430171441867239530788067265164581820366701099039
17222170335851233744430577061928441039432055599953297058218812657049
Jean-Claude Arbaut
On Wed, Sep 14, 2011 at 2:21 AM, Alasdair McAndrew <amca01 at gmail.com> wrote:
> The number is simply too big to be factored by Maxima's algorithm. You
> might like to try other software with better support for computational
> number theory, such as Pari/GP. I suspect that the numbers in question
> (from here <http://www.immortaltheory.com/NumberTheory/RuthAaron.htm>;)
> were produced by a polynomial, so all you need to do is to solve the
> appropriate polynomial f(x)=n, where n is the smaller of the two numbers.
>
> -Alasdair
>
> On Wed, Sep 14, 2011 at 7:25 AM, Matematicas <metodosu at gmail.com> wrote:
>
>> Numbers are called Ruth-Aaron pair, such as {714, 715}. Are consecutive
>> numbers {n, n +1} such that the sum of the prime factors of each are
>> equal.
>>
>> Let's see
>>
>> 714 = 2x3x7x17
>> 715 = 5x11x13
>>
>> Sum of divisors of 714 = 29 = Sum of divisors of 715
>>
>>
>>
>> FACTORS FOR EACH OF THE FOLLOWING TWO NUMBERS. COULD NOT DERIVE
>>
>> A number of 135 digits es1 Aoron
>>
>> *
>> 296603151077074197308684124715622560525294501766069587161483751777320391554230009758029564174782423278870115094641644418265512090475910
>> *
>>
>>
>> The following of the former is
>>
>> *
>> 296603151077074197308684124715622560525294501766069587161483751777320391554230009758029564174782423278870115094641644418265512090475911
>> *
>>
>> The problem is that the command ifactor(n) don't work, generate an error,
>> can help me??
>>
>> 1 <#132654fce756ae77_13264ae44588bf30_sdfootnote1anc>
>> http://www.immortaltheory.com/NumberTheory/RuthAaron.htm
>>
>>
>>
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>>
>>
>
>
> --
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>
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