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

[Max Alekseyev]

Hi Robert,

If n is "divisible by just 2", ie. n==2 (mod 4), then in fact we cannot
have a(n) == 2 (mod n) unless n=2.
Here is a brief idea.
Suppose that a(n) == 2 (mod n) for some n=2*m with odd m > 1.
Since a(2m) = a(m)^2 + 2, we have
a(m)^2 == 0 (mod m).
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.
That is, r(p) <= (p+1)/2 < p and r(p) divides m, which together with
minimality of p implies r(p)=1. Then p divides a(1)=2, a contradiction to
existence of m > 1.

PS. We can be a bit more specific and notice that odd primes dividing a(m)
must be == 1,7 (mod 8), and thus m itself must be the product of such
primes, and the period modulo such prime p divides p-1. Although these
details are inessential for the above proof.

Regards,
Max


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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