This is from Gosper; it was somehow trapped by the mail filtering software. I've edited it slightly. --rich ------------ Date: Sat, 19 Apr 2008 12:15:02 -0700 (PDT) Subject: Re: [math-fun] Help! From: Bill Gosper <rwmgosper@yahoo.com> To: math-fun <math-fun@mailman.xmission.com>, Bill Gosper <rwmgosper@yahoo.com>, rkg@cpsc.ucalgary.ca All: I sent this (retrieved from my outbox) yesterday, and Yahoo somehow poofed it:
Hi Richard, at least for k=3, z needs a factor of (k^n-1)^(2*k^n-1). I haven't checked k>3. I think this differs slightly from Gareth's solution, so we'll need a third opinion. --Bill
Macsyma has subsequently confirmed this result for general k and n, and they can't both be right... --rwg Fred: [Come to think of it, I could have done without them inventing fireworks as well ...] But then we wouldn't have space travel: During the first and second stage flights of the vehicle, if a serious irretrievable fault should occur and the deviation of the flight attitude of the vehicle exceeds a predetermined value, the attitude self-destruction system will make the vehicle self-destroyed. --Long March 3 User's Manual Ministry of Astronautics, People's Republic of China (1985) --- On Fri, 4/18/08, Richard Guy <rkg@cpsc.ucalgary.ca> wrote: <I clipped out a copy of RKGs reply to Gareth McCaughan --rich>
Hi Richard, at least for k=3, z needs a factor of (k^n-1)^(2*k^n-1). I haven't checked k>3. I think this differs slightly from Gareth's solution, so we'll need a third opinion.
Same as mine. 2(k^n-1)+1 = 2k^n-1. I wrote it the former way because that's how it came out of the calculations. (k^n-1 is one of the key quantities.) -- g
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Gareth McCaughan -
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