A069659 => A061433 ?
David Wilson
davidwwilson at comcast.net
Sat Aug 12 18:54:00 CEST 2006
My initial reaction to this problem was that since almost all perfect powers
are squares, we should expect that for large numbers n, the largest perfect
power <= n should be a square.
Certainly, for any individual prime p > 2, the pth powers are of density 0
over the perfect powers, being overwhelmed by the squares. However, I
realized that I'm not whether all the powers with prime exponent p > 2
together might overwhelm the squares. If they do, then I would expect Zak's
two sequences to differ at some point, otherwise, I would expect them to
differ at only a finite number of elements (probably 0 given the terms
provided). This is of course all gut feeling.
----- Original Message -----
From: "zak seidov" <zakseidov at yahoo.com>
To: <seqfan at ext.jussieu.fr>
Sent: Saturday, August 05, 2006 12:41 AM
Subject: A069659 => A061433 ?
> Definitions are different:
>
> A061433 a(n) = largest n-digit square
> A069659 a(1) = 1; then the largest n-digit perfect
> power
>
> while all entries are the same(?):
>
> 9, 81, 961, 9801, 99856, 998001, 9998244, 99980001,
> 999950884, 9999800001, 99999515529, 999998000001,
> 9999995824729, 99999980000001, 999999961946176,
> 9999999800000001, 99999999989350756,
> 999999998000000001.
>
> Shouldn't the valuable comment in A069659:
> "Are there terms which are not perfect squares?"
> be included (with rewording) into A061433,
> and A069659 deleted?
>
> Simple adding 1 (but why?} in A069659
> may "save" A069659 as well.
> Thanks, Zak
>
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