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<DIV>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.</DIV>
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<DIV>Franklin T. Adams-Watters<BR>16 W. Michigan Ave.<BR>Palatine, IL 60067<BR>847-776-7645</DIV>
<DIV> </DIV> <BR>-----Original Message-----<BR>From: Paul D. Hanna <pauldhanna@juno.com><BR>To: seqfan@ext.jussieu.fr<BR>Sent: Mon, 23 Jan 2006 20:19:43 -0500<BR>Subject: Re: Congruent Products Under XOR; Fibbinary Numbers<BR><BR>
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<DIV class=AOLPlainTextBody id=AOLMsgPart_0_2fd5e835-257b-4e75-b571-135ead4a72e2><PRE><TT>...
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.
...</TT></PRE></DIV><!-- end of AOLMsgPart_0_2fd5e835-257b-4e75-b571-135ead4a72e2 --></DIV></DIV>
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