[seqfan] Re: Proof For A269254

[Neil Sloane]

The case 110 is certainly always composite. I didn't work out all the
details, but, first, a(3n+1) is zero mod 3 (just run the recurrence mod 3)

Second, it seems that a(3n) = b(n)*c(n),
where Don Reble found b(n) = 110*b(n-1)-b(n-2), and using gfun I get
c(n) = 12099*c(n-1)-12099*c(n-2)+c(n-3)

Third, similarly, it seems that a(3n+2) = d(n)*e(n), where d(n) and e(n)
satisfy the same recurrences as b(n) and c(n), except with different
initial conditions.

This may not be a legal proof, but one can now use gfun to verify that
a(3n) and b(n)*c(n) satisfy the recurrence
A(n) = 1330670*A(n-1) - A(n-2),
and so must be equal.
Likewise,  a(3n+2), d(n)*e(n) satisfy the same recurrence
(with different initial conditions)
and so are also equal.



Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com


On Sun, Oct 22, 2017 at 5:49 PM, L. Edson Jeffery <lejeffery2 at gmail.com>
wrote:

> Brad,
>
> Nice proof. I just submitted the array used for computing A269254. Here is
> the draft if anyone wants to add to it:
>
> https://oeis.org/draft/A294099
>
> I also stated the theorem in A269254 and gave a link to your proof on
> seqfan.
>
> Meanwhile, a(110) is still unknown. My Mathematica program ran all night,
> and is still running, to no avail.
>
> Ed Jeffery
>
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