[seqfan] Re: Sum of Pell numbers and companion Pell numbers

[Robert Dougherty-Bliss]

Dear Max,

Thank you for the proof! I am excited about it, because this is
exactly what I suspected was true. However, I'm having some trouble
following the details of your argument.

> Let p be the smallest prime dividing m and r(p) the rank of apparition of
> a() modulo p, then the period of a() modulo p is either 2*r(p) or 4*r(p),
> and it divides p-1 or p+1.

Why is the period either 2 * r(p) or 4 * r(p)? And why does it divide
p - 1 or p + 1?

> That is, r(p) <= (p+1)/2 < p and r(p) divides m, which together with
> minimality of p implies r(p)=1.

Why does r(p) divide m?

I know some similar facts / arguments about Fibonacci numbers, but I
don't see how the ideas apply here.

Robert

> On Thu, Apr 27, 2023 at 11:05 PM Robert Dougherty-Bliss <
> robert.w.bliss at gmail.com> wrote:
>
> > Dear Sequence Fans,
> >
> > Consider the sequence a(n) which satisfies the recurrence
> >      a(n) = 2 * a(n - 1) + a(n - 2)
> > with initial conditions a(0) = a(1) = 2. (This is A2203, the companion
> > Pell numbers.)
> >
> > The sequence of positive n such that a(n) = 2 (mod n) begins
> > 1, 2, 3, 4, 5, 7, 8, 11, 13, 16, 17, 19, 23, 24, 29.
> > Except for the leading 1 and 2, this looks like A270342, the sequence
> > of n such that n divides the sum of the first n Pell numbers, but
> > there is no comment to this effect. Can anyone prove this?
> >
> > The sum of the first n Pell numbers S(n) turns out to equal (a(n) - 2)
> > / 4, so everything in A270342 is in this sequence. For the other way,
> > I have managed to prove that if a(n) = 2 (mod n) and n is odd or
> > divisible by 4, then n is in A270341, but I cannot figure out the
> > "divisible by just 2" case.
> >
> > Robert
> >
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> >
>
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