What solutions are there to the functional equation $R(1/x) = x^kR(x)$, where $k$ is a non-negative integer?
This is a generalization of the earlier question Functional equation: $R(1/x)/x^2 = R(x) $ which had $k=2$.
A more general question is this:
What solutions are there to the functional equation $p(x)R(1/x) = q(x)R(x)$, where $p(x)$ and $q(x)$ are pre-specified polynomials?
I have only looked at the first equation, though my technique below might be a starting point for the second question.
Here is what I have done so far.
I will look for solutions to $R(1/x) = x^kR(x)$ of the form $R(x) = A(x)/B(x)$, where $A$ and $B$ are relatively prime polynomials of respective degrees $n$ and $m$,
Some definitions:
If $C(x)$ is a polynomial of degree $n$ (so $C(x) = \sum_{i=0}^n c_i x^i$), $[x^i]C(x)$ is the coefficient of $x^i$ in $C(x)$ (i.e., $[x^i]C(x) = c_i$), $deg(C(x)$ is the largest $i$ such that $[x^i]C(x)$ is non-zero, $codeg(C(x))$ is the smallest $i$ such that $[x^i]C(x)$ is non-zero, $\rho(C(x))$ is the reciprocal polynomial of $C(x)$ so that $\rho(C(x)) = x^n C(1/x) = \sum_{i=0}^n [x^i]C(x) x^{n-i} = \sum_{i=0}^n c_i x^{n-i}$.
A polynomial $C(x)$ is $symmetric$ if $C(x) = \rho(C(x))$.
Suppose $R(x) = A(x)/B(x)$ where $A(x)$ and $B(x)$ are relatively prime polynomials of degrees $n$ and $m$, respectively. Let $\rho(A(x)) = a(x)$ and $\rho(B(x)) = b(x)$, so that $a(x) = x^nA(1/x)$ and $b(x) = x^mB(1/x)$. Let $u = deg(a(x))$ and $v = deg(b(x))$.
We have $R(1/x)= A(1/x)/B(1/x)= (a(x)x^{-n})/(b(x)x^{-m})= x^{m-n}a(x)/b(x)$, so that $x^{m-n}a(x)/b(x) = x^k A(x)/B(x)$ or $x^ma(x)B(x) = x^{n+k}A(x)b(x)$.
The degree of the left side is $m+u+m = 2m+u$ and the degree of the right side is $n+k+n+v = 2n+k+v$, so $2m+u=2n+k+v$.
I now look at various possibilities for $A(x)$ and $B(x)$.
If the degrees of $a(x)$ and $b(x)$ are the same as $A$ and $B$, respectively, (i.e., the constant terms of $A$ and $B$ are non-zero), then $u=n$ and $v=m$, so that $2m+n=2n+k+m$ or $m = n+k$. This means that $a(x)B(x) = A(x)b(x)$. Since $A(x)$ and $B(x)$ are relatively prime, so are $a(x)$ and $b(x)$, so that $a(x) = A(x)$ and $b(x) = B(x)$ which means that $A(x)$ and $B(x)$ are symmetric.
Conversely, if $A(x)$ and $B(x)$ are symmetric and $deg(B(x)) = k+deg(A(x))$, it is easy to verify that $R(x)=A(x)/B(x)$ satisfies $R(1/x) = x^kR(x)$.
As a special case, if $B(x)$ is symmetric with degree $k$, then $1/B(x)$ satisfies the equation (i.e., $A(x) = 1$).
Suppose $A(x) = x^n$, so that $a(x) = 1$ and $u = 0$. The condition $2m+u=2n+k+v$ becomes $2m=2n+k+v$ (or $v = 2m-2n-k$) and $x^ma(x)B(x) = x^{n+k}A(x)b(x)$ becomes $x^mB(x) = x^{2n+k}b(x)$. Since $v \le m$, $2m=2n+k+v \le 2n+k+m$ or $m \le 2n+k$. so $B(x) = x^{2n+k-m}b(x)$. Therefore the part of $B(x)$ with the low-order zero coefficients removed is symmetric. The converse also holds.
If $B(x) = x^m$, so $b(x) = 1$ and $v=0$, $2m+u=2n+k$ and $x^ma(x)B(x) = x^{n+k}A(x)b(x)$ becomes $x^{2m}a(x) = x^{2n+k}A(x)$ or $x^{2m-2n-k}a(x) = A(x)$. Therefore, as with $B(x)$, above, the part of $A(x)$ with the low-order zero coefficients removed is symmetric. The converse also holds.
