# [seqfan] Re: Proof For A269254

Mon Oct 23 07:49:02 CEST 2017

Hi Neil, Edson and Seqfans,

Edson, thank you for putting in the comment. Hopefully future readers will
benefit from the hyperlinks, especially if we can't find a better textbook
reference.

Neil, many people, including me, will have no idea where your preferred
"gfun" comes from or what it means. It is nice to have different methods of
proof, but computational artifice does seem to over-complicate the
situation and introduce readers to unnecessary confusion. That is to say,
the case 110 is not significantly different from the regular cases, so why
no just slightly modify the previous proof ? ?

Thanks to Don's nice observation, I also calculate the following:

u_k = 110 * u_{k - 1} - u_{k - 2},
v_k = 12099 * v_{k - 1} - 12099 * v_{k - 2} + v_{k - 3}

As the mirror symmetry is broken, these definitions exchange previous
special symbols L&R for generic u&v. We then propose a recursion for the
product

a_{k} = 1330670 * a_{k-1} - a_{k-2},

which directly implies the zero-sum

u_{k} * v_{k} - 1330670 * u_{k-1} * v_{k-1} + u_{k-2} * v_{k-2} = 0.

Applying the recursion relations we obtain

{220, -1330890, 110, -12099, 12100, -1} \cdot UV = 0,

where UV is a vector of bilinear variables, decreasing first along v index,
then decreasing along u index,

UV ={
u_{k-1} * v_{k-1}, u_{k-1} * v_{k-2}, u_{k-1} * v_{k-3},
u_{k-2} * v_{k-1}, u_{k-2} * v_{k-2}, u_{k-2} * v_{k-3}
} ;

This zero-sum will act as a first induction hypothesis. We increment k ->
k+1, to obtain the next higher condition. Applying the recursion relations
returns to the basis provided by UV. The coefficients are not the same:

{ 24199, -146397900, 12101, -1330890, 2661670, -220 } \cdot UV = 0,

The saying is that: "One mistaken hypothesis is no reason to end a research
career". Amend the hypothesis to include both linearly-independent
zero-sums:

{220, -1330890, 110, -12099, 12100, -1} \cdot UV = v_1 \cdot UV = 0 ,
{ 24199, -146397900, 12101, -1330890, 2661670, -220 } \cdot UV = v_2 \cdot
UV = 0 .

Repeat the procedure of incrementing both by k -> k+1, and reducing to
variables UV by the recursion relations. Again the first goes to the
second, and the second goes to a third

{2661670, -16102438110 <%28610%29%20243-8110>, 1331000, -146385801,
292771600, -24199} \cdot UV = v_3 \cdot UV = 0 .

It is not immediately apparent, but the hypothesis does admit valid
recursion. The coefficient vectors are linearly dependent by another zero
sum:

110 * v_2 - v_1 - v_3 = 0 .

When the hypothesis holds at k, it also holds at k+1, thus telescopes to
every value of k, off toward infinity. The base case and the strictly
increasing property must be verified for all three of the trisections. The
complete proof should be no less true than it was previously.

Regards,

On Sun, Oct 22, 2017 at 11:12 PM, Neil Sloane <njasloane at gmail.com> wrote:

> PS
>
> The missing steps in the proof can be filled in like this.
>
> For example, we know by construction that b(n) satisfies a second-order
> recurrence,
> and c(n) a third-order recurrence.  We want to prove that the componentwise
> product b(n)*c(n) satisfies a second order recurrence. Well, a basic
> theorem about linear recurrences tells us that the product satisfies a
> linear recurrence of order at most 2*3 =  6. And NOW we can use Gfun in
> Maple to find it,
> and it turns out to satisfy a second-order recurrence.  All that gfun does
> in a case like this is to make a call to Maple's convert[ratpoly], which
> uses Pade approximations.  I assume this can be made rigorous if anyone has
> doubts.
>
> 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.
> Email: njasloane at gmail.com
>
>
> On Sun, Oct 22, 2017 at 10:38 PM, Neil Sloane <njasloane at gmail.com> wrote:
>
> > 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.
> > Email: njasloane at gmail.com
> >
> >
> > On Sun, Oct 22, 2017 at 5:49 PM, L. Edson Jeffery <lejeffery2 at gmail.com>
> > wrote:
> >
> >>
> >> 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
> >>
> >> --
> >> Seqfan Mailing list - http://list.seqfan.eu/
> >>
> >
> >
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>