Review: A069554

[David Wilson]

----- Original Message -----
From: "Don Reble" <djr at nk.ca>
To: "Number Sequence Mailing List" <seqfan at ext.jussieu.fr>
Sent: Friday, May 31, 2002 6:41 AM
Subject: Re: Review: A069554


> > The questionable elements are:
> > a(37) = 0.  Empirical evidence indicates that
> > 37 | gcf(k, rev(k)) ==> 111 | gcf(k, rev(k)),
> > in which case gcf(k, rev(k)) = 37 has no solution.
>
>     Questionable indeed! For k=1009009009009, gcd(k,rev(k))=37.
>     So we just have to find the smallest one.
> --
> Don Reble       djr at nk.ca

Actually, a more direct method occurred to me this morning.

Create an FA that accepts multiples of 37.  Create an FA that accepts the
reverse of multiples
of 37.  Intersect these to obtain an FA for numbers such that 37 divides
both the number and
its reverse.  Intersect this FA with an FA for the non-multiplies of 3.
This will be an FA for the
for which 37 | gcd(k, rev(k)) while not 111 | gcd(k, rev(k)).  Hopefully,
one of the smaller
numbers accepted by this FA will have gcd(k, rev(k)) = 37.