More digital silliness

[David W. Wilson]

OK, but the rotations of 1111 are all equal to 1111, which I'm sure you
observed. I should have made it clear that I was not counting the number of
rotations that lead to a square, but the number of distinct squares produced
by rotation.

Still, it is interesting that you can do better than 1111 in this sense. In
base 3, 11111 is a square.

> -----Original Message-----
> From: David W. Cantrell [mailto:DWCantrell at sigmaxi.net]
> Sent: Wednesday, January 09, 2008 3:42 PM
> To: Sequence Fans
> Subject: Re: More digital silliness
> 
> Bob Hearn wrote:
> > There's probably some simple reason that the answer is no, but
> > empirically, there are no such numbers < 10^16.
> 
> I suppose that may be true if you are restricted to the decimal
> system, but DWW said nothing about such a restriction. Using an
> appropriate base, all rotations of 1111 are square, and of course,
> using that same base, so are all rotations of 4444.
> 
> David W. Cantrell
> 
> > On Jan 9, 2008, at 2:33 PM, David W. Wilson wrote:
> >
> >> Let an n-digit number be valid if it does not start or end with 0.
> >>
> >> Let a rotation of n be gotten by rotating its digits. Thus the
> >> rotations of 256 are 256, 562 and 625.
> >>
> >> We note that 256 has two valid square rotations, 256 and 625.
> >>
> >> Is there a number with more than two valid square rotations?
>