[seqfan] Re: amicable numbers - Conjecture

[Richard Guy]

I'm copying this to the number theory
list, to see if this is well-known to
those who well know it.

It appears to work for

2^4 = 4, p = 5, b = 11, c = 71
2^n = 16, p = 41, b = 83, c = 3527

Are there other examples?   R.

On Sat, 8 Aug 2009, 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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