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

[Max Alekseyev]

CORRECTION:
In the latter case, there are two zeros in the period modulo p, ie. if a(m)
== 0 (mod p) and q(p)/4 is even, then m == q(p)/4 or q(p)/2 (mod q(p)) and
thus q(p)/4 divides m.

On Sat, Apr 29, 2023 at 8:58 AM Max Alekseyev <maxale at gmail.com> wrote:

> It all follows from the explicit formula:
> a(n) = (1 + sqrt(2))^n + (1 - sqrt(2))^n.
>
> For an odd prime p, we have
> (1 +- sqrt(2))^(p-1) == 1 (mod p) whenever 2 is a quadratic residue modulo
> p, ie. p == 1,7 (mod 8);
> and
> (1 +- sqrt(2))^(p+1) == 1 (mod p) whenever 2 is a non-quadratic residue
> modulo p, ie. p == 3,5 (mod 8).
> It follows that the period length q(p) of a() modulo p divides p-1 or p+1,
> respectively.
>
> Then, a(m) == 0 (mod p) is equivalent to
> (1 + sqrt(2))/(1-sqrt(2))^m = (1 + sqrt(2))^(2m) * (-1)^m == -1 (mod p).
> If m is the smallest, ie. m = r(p) is the rank of apparition, then r(p) =
> q(p)/2 when it's odd; or r(p) = q(p)/4 when it's even.
> In the former case, there is only one zero in the period modulo p, ie. if
> a(m) == 0 (mod p) and q(p)/2 is odd, then m == q(p)/2 (mod q(p)) and thus
> q(p)/2 divides m.
> In the latter case, there are two zeros in the period modulo p, ie. if
> a(m) == 0 (mod p) and q(p)/4 is even, then m == q(p)/4 (mod q(p)/2) and
> thus q(p)/4 divides m.
>
> Let me know if anything is still unclear.
>
> Regards,
> Max
>
> On Sat, Apr 29, 2023 at 2:51 AM Robert Dougherty-Bliss <
> robert.w.bliss at gmail.com> wrote:
>
>> 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
>> > >
>> > > --
>> > > Seqfan Mailing list - http://list.seqfan.eu/
>> > >
>> >
>> > --
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>>
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