Roland Bacher Roland.Bacher at ujf-grenoble.fr
Tue Sep 7 15:18:19 CEST 2004

```Dear Creighton,

your sequences jes,les,tes and ves seem to be recursively defined by

jes(n+1)=-4*jes(n)-jes(n-1)
les(n+1)=les(n-1)+jes(n)
ves(n+1)=les(n-1)-jes(n-1)+tes(n-1)
tes(n+1)=les(n-1)+3*jes(n)

Plus initial conditions for n=0,1.

They satisfy thus all linear recursion relations (by elimination)
the associated characteristic polynomials.

Best whishes,  Roland Bacher

On Mon, Sep 06, 2004 at 08:01:38PM +0100, creigh at o2online.de wrote:
> Please excuse all the postings, below is a IX 'th  property
> (without proof - although,
> Limit as n-> infinity of A001353(n)/A001353(n-1) = 2 + sqrt(3).
> -Gregory V. Richardson (omomom(AT)hotmail.com),
> Oct 06 2002 has been proven- I was informed by Neil that
> "jes(n)" =   [1, -4, 15, -56, ...] is (-1)^(n+1)*A001353(n+1)).
> I also fixed a mistake at the end.
>
> *************************
>
> Just found an additional property (VIII) which perplexes
> me. I also fixed a mistake at the end.
>
> *********************
>
> Apparently, none of the four sequences
>
> jes seq: 1(0), -4(1), 15(2), -56(3), 209(4), -780(5), 2911(6), -10864(7),
> 40545(8), -151316(9), 564719(10)
>
> les seq: -2(0), 1(1), -6(2), 16(3), -62(4), 225(5), -842(6), 3136(7), -11706
> (8), 43681(9), -163022(10)
>
> tes seq: -1(0), 4(1), -13(2), 49(3), -181(4), 676(5), -2521(6), 9409(7), -35113
> (8), 131044(9), -489061(10)
>
> ves seq: -2(0), 1(1), -4(2), 9(3), -34(4), 121(5), -452(6), 1681(7), -6274(8),
> 23409(9), -87364(10)
>
> are listed at OEIS. However, notice the symmetries (for all n):
>
> I: jes(n) + les(n) + tes(n) = ves(n)
>
> II: les(2n+1); tes(2n+1); ves(2n+1);
> ves(2n+1) - jes(2n+1) - 1 = les(2n+1) + tes(2n+1) - 1;
> 3*les(2n+1) + 1 = 3*jes(n)^2 + 1 (see IV and VII) are perfect squares
>
> III: les(2n+1) divides ves(2n+1) - jes(2n+1) - 1 = les(2n+1) + tes(2n+1) -
>
> 1
>
> IV: (jes(n))^2 = les(2n+1)
>
> V: tes(2n) = A001570(n), sqrt( tes(2n+1) ) = A001075(n) [both are "nice"
> sequences]
>
> VI: ves(2n), ves(2n)/2 do not exist in OEIS, however: sqrt( ves(2n+1) )
> = A001835(n) [ "nice" ]
>
> VII: sqrt( les(2n+1) ) = A001353(n) [ Comments: 3*a(n)^2 + 1 is a perfect square.
> ]
>
> VIII: les(n) + tes(n) = ves(2+n) + jes(n)
>
> IX: lim n |jes(n+1)/jes(n)| =
>      lim n |les(n+1)/les(n)| =
>      lim n |tes(n+1)/tes(n)| =
>      lim n |ves(n+1)/ves(n)| = 2 + sqrt(3)
>
> It follows immediately from properties I + II that...
> 1^2 (les(1) + 2^2 (tes(1)) = 2^2 (ves(1) - jes(1) - 1) + 1
> 4^2 (les(3))+ 7^2 (tes(3)) = 8^2 (ves(3) - jes(3) - 1)  + 1
> 15^2 (les(5)) + 26^2 (tes(5)) = 30^2 (ves(5) - jes(5) - 1)  + 1
>
> Sincerely,
> Creighton

```