1
$\begingroup$

Is there something known about the values of the Kronecker symbol $\left(\frac{a_n}{a_{n+1}}\right)$ if $a_n$ is a recursive sequence? I know something about this if $a_n$ is the Fibonacci sequence or the Lucase sequence and for a given sequence, these symbols can be computed. Nevertheless, it would be great to have some results that hold for all recursive sequences of the form $a_{n+1} = \alpha \cdot a_n + \beta \cdot a_{n-1}$ for some $\alpha, \beta$ independent (or even dependent) of $n$.

$\endgroup$
2
$\begingroup$

At the end of his 2012 paper A. Granville states that finding a "usable" formula is an open problem for (certain) Lucas sequences. Thus there does not seem to be much hope to answer your question with $\alpha,\beta$ depending on $n$ in full generality.

$\endgroup$
2
$\begingroup$

In the cited paper I challenged the reader to find a general usable formula for $(x_m/x_n)$. However this special case should be doable. It seems to me you can simply proceed by induction, though you will need to write $x_n = 2^{k_n} y_n$ where one will need some idea of the value of each $k_n$, and each $y_n \mod 8$. But if one has succeeded with the Fibonacci numbers, these issues arise, so I would guess that that method should generalize with some work.

To give an easy general class of example, I work with $x_{n+2}=ax_{n+1}+bx_n$ for all $n\geq 0$, where we suppose $4$ divides $b$, and $a$ and $x_1$ are odd (else the whole sequence will be even). From the recurrence we have $x_{n+2}\equiv ax_{n+1} \pmod b$, and so $x_n\equiv a^{n-1}x_1 \pmod b$ for all $n\geq 1$. This implies that each $x_n$ is odd, and at least one of $x_n, x_{n+1}$ is $\equiv 1 \pmod 4$ and therefore $(x_n/x_{n+1})=(x_{n+1}/x_n)$ for all $n\geq 1$. Moreover $x_{n+1}\equiv bx_{n-1} \pmod {x_n}$ by the recurrence relation and so $(x_{n+1}/x_n)=(b/x_n)(x_{n-1}/x_n)$. Putting these together yields

$(x_n/x_{n+1})= (b/x_n)(x_{n-1}/x_n)$ for all $n\geq 1$.

Proceeding by induction then yields (assuming $x_0\ne 0$ -- if it is we stop one step earlier)

$(x_n/x_{n+1})= (b/x_nx_{n-1}...x_1)(x_0/x_1)$

As $4|b$ the Kronecker symbol $(b/.)$ has period dividing $b$, and so we use that $x_nx_{n-1}...x_1 \equiv a^{n(n-1)/2}x_1^n \pmod b$. Therefore

$(x_n/x_{n+1})= (b/a)^{n(n-1)/2} \ (b/x_1)^n \ (x_0/x_1)$, as desired.

This formula depends only on the value of $n\pmod 4$ and so we can write

$(x_n/x_{n+1})= (x_r/x_{r+1})$, where $r$ is the least residue of $n \pmod 4$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.