Problem

[Max Alekseyev]

On Nov 16, 2007 11:37 AM, Max Alekseyev <maxale at gmail.com> wrote:

> On Nov 16, 2007 11:06 AM, Artur <grafix at csl.pl> wrote:
>
> > I'm apologize but but I was do mistake in formula
> > Find number m such that 5n^4(4+n^2) is square
> > Mathemtica sample Code:
> > a={};Do[If[IntegerQ[Sqrt[5n^4 (4+n^2)]],AppendTo[a,n]],{n,1,1364}];a
> >
> > Out: {1,4,11,29,76,199,521,1364}

[...]

> With respect to n, the solution consists of the following 3 sequences,
> satisfying the same recurrent relation n(j+1)=18*n(j)-n(j-1) with
> different initial conditions:
>
> n(0) = 1, n(1) = 29: { 1, 29, 521, 9349, 167761, 3010349, ... }
>
> n(0) = 4, n(1) = 76: { 4, 76, 1364, 24476, 439204, 7881196, ... }
>
> n(0) = 11, n(1) = 199:  { 11, 199, 3571, 64079, 1149851, 20633239, ... }
>
> Your sequence is the union of these three.

You can describe their union as the following 6th order recurrent sequence:
n(j) = 18*n(j-3) - n(j-6)
with
n(0)=1, n(1)=4, n(2)=11, n(3)=29, n(4)=76, n(5)=199.

Max