[seqfan] Re: amicable numbers - Conjecture

[Max Alekseyev]

It's known - see:
http://en.wikipedia.org/wiki/Amicable_number#Rules_for_generating

Max

On Sat, Aug 8, 2009 at 9:16 AM,
vincenzo.librandi at tin.it<vincenzo.librandi at tin.it> wrote:
>  Let p = prime number
>
> if 2^n=[2*(p+1)/3]
>
> b=2*p+1 (with b=prime)
>
> c=2*p^2+4*p+1 (with c=prime)
>
> then
>
> p*b*2^n   and   c*2^n
>
> are amicable numbers
>
> Best Regards
>
> Vincenzo Librandi
>
>
>
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