[seqfan] Re: amicable numbers - Conjecture
Richard Mathar
mathar at strw.leidenuniv.nl
Fri Aug 14 23:26:44 CEST 2009
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
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