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<DIV>The formula I gave is deg(D)+deg(N)+1. Now, if deg(D)>=deg(N), this is <= 2*deg(D)+1, so Ralf's claim is rather trivially correct. Actually, you usually want deg(D)>deg(N), which gives the originally quoted bound of 2*deg(D).</DIV>
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<DIV>Sequences a(n) and b(n) have the same e.g.f iff a(n)/n! and b(n)/n! have the same e.g.f., so exactly the same formula applies.</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: Ralf Stephan <ralf@ark.in-berlin.de><BR>To: Seqfan <seqfan@ext.jussieu.fr><BR>Sent: Tue, 17 Jan 2006 18:29:14 +0100<BR>Subject: Re: Basic questions?<BR><BR>
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<DIV class=AOLPlainTextBody id=AOLMsgPart_0_27a79560-a25e-4c42-b68f-0aa964fbaa60><PRE><TT>I'm still of the opinion that, given a true rat.g.f., i.e.,
deg(D)>=deg(N), the degree of the numerator is irrelevant
for the needed number of comparisons.
What about rational egfs?
ralf
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