[seqfan] Re: amicable numbers - Conjecture

[Richard Mathar]

On behalf of http://list.seqfan.eu/pipermail/seqfan/2009-August/002094.html

vl>  Let p = prime number
vl> 
vl> if 2^n=[2*(p+1)/3]

So 3*2^n=2*(p+1), therefore 3*2^(n-1)=p+1 or p=3*2^(n-1)-1.

vl> 
vl> b=2*p+1 (with b=prime)

So b = 3*2^n-1.

vl> 
vl> c=2*p^2+4*p+1 (with c=prime)

So c = 9*2^(2n-1)-1.

vl> 
vl> then
vl> 
vl> p*b*2^n   and   c*2^n
vl> 
vl> are amicable numbers

This seems to be closely related to the statement on page 2 in the Andreescu
"Number Theory Trivia: Amicable Numbers" in the first link of A063990,
just n replaced by n-1.

Richard