Congruent Products Under XOR; Fibbinary Numbers

[franktaw at netscape.net]

Yes, this is true.  Take n = 2^k for any k.  More generally, n can be any integer whose 1 bits are all separated by at least the number of bits in p and q.  Or even more generally, the set of distances between 1 bits in n and the set of distances between 1 bits in (p OR q) should be disjoint.  This doesn't quite cover all the cases, but it gets most of them.  Another class of solutions is any time p=q (p=0 or q=0 would also work, if we allowed zero).  One solution that doesn't fit either class is p=19, q=67, n=3; look at it in binary and you'll see that it's really a combination of the two classes.
 
Franklin T. Adams-Watters
16 W. Michigan Ave.
Palatine, IL 60067
847-776-7645
 
 
-----Original Message-----
From: Paul D. Hanna <pauldhanna at juno.com>
To: seqfan at ext.jussieu.fr
Sent: Mon, 23 Jan 2006 20:19:43 -0500
Subject: Re: Congruent Products Under XOR; Fibbinary Numbers


... 
A better conjecture may be: 
there exists an infinite set of integers n that satisfy 
p*n XOR q*n = (p XOR q)*n for all positive integers p,q, r. 
I find no exceptions to this guess. 
...
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