==> pickover/pickover.16.s <==
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In article <1992Oct30.175102.142177@watson.ibm.com> you write:
: 1. Are there any undulating square numbers?
11^2 = 121
: 2. Are there any undulating cube numbers?
7^3 = 343
(yes, I know they're short, but they qualify!)
--
Michael Neylon aka Masem the Great and Almighty Thermodynamics GOD!
// | Senior, Chemical Engineering, Univ. of Toledo
\\ // Only the | Summer Intern, NASA Lewis Research Center
\ \X/ AMIGA! | mneylon@jupiter.cse.utoledo.edu /
--------+ How do YOU spell 'potato'? How 'bout 'lousy'? +----------
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In article <1992Oct30.204134.97881@watson.ibm.com> you write:
>Hi, I was interested in non-trivial cases. Those with greater
>than 3 digits. Award goes to the person who finds the largest
>undulating square or cube number. Thanks, Cliff
343 and 676 aren't trivial (unlike 121 and 484 it doesn't come from
obvious algebraic identities). The chance that a "random"
number around x should be a perfect square is about 1/sqrt(x);
more generally, x^(-1+1/d) for a perfect d-th power. Since
there are for each k only 90 k-digit undulants you expect
to find only finitely many of these that are perfect powers,
and none that are very large. But provably listing all cases
is probably only barely, if at all, possible by present-day
methods for treating exponential Diophantine equations, unless
(as was shown in a rec.puzzles posting re your puzzles on
arith. prog. of squares with common difference 10^k) there is
some ad-hoc trick available. At any rate the largest undulating
power is probably 69696=264^2, though 211^3=9393931 comes
remarkably close.
--Noam D. Elkies
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In article <1992Oct30.175102.142177@watson.ibm.com>, you write...
>1. Are there any undulating square numbers?
>
Other than the obvious 11**2, 22**2, and 26**2, there is 264**2
which equals 69696.
>2. Are there any undulating cube numbers?
>
Just 7**3 as far as I can tell, though I'm limited to IEEE computational
reals.
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