For real irrational $C > 1 $ and natural $n,b$, define $a(C,n,b)=\lfloor C^n \rfloor \mod b$

Q1 For which $C,b$ is $a(C,n,b)$ computable in time polynomial in $\log{n}$?

Searching in OEIS suggests that for $C \in \{1+\sqrt{2},1+\sqrt{3},(1+\sqrt{5})/2\}$, $a(C,n,b)$ satisfy linear recurrence with constant coefficients and so it is efficiently computable over the integers and all bases $b$.

In OEIS: a(n) = floor(phi^n)

a(n) = floor((1+sqrt(2))^n)

a(n) = floor((1+sqrt(3))^n)

For natural $k$, $a(b^{1/k},n,b)$ is related to the base $b$ representation of $b^{1/k}$ so it is probably hopeless.

Q2 Is $a(1+\sqrt{6},n,b)$ efficiently computable in some base $b$?

(We couldn't find linear recurrence for it)

Q3 Except linear recurrences, are there other islands of tractability for algebraic $C$?

For real irrational $C > 1 $ and natural $n,b$, define $a(C,n,b)=\lfloor C^n \rfloor \mod b$

Q1 For which $C,b$ is $a(C,n,b)$ computable in time polynomial in $\log{n}$?

Searching in OEIS suggests that for $C \in \{1+\sqrt{2},1+\sqrt{3},(1+\sqrt{5})/2\}$, $a(C,n,b)$ satisfy linear recurrence with constant coefficients and so it is efficiently computable over the integers and all bases $b$.

In OEIS: a(n) = floor(phi^n)

a(n) = floor((1+sqrt(2))^n)

a(n) = floor((1+sqrt(3))^n)

For natural $k$, $a(b^{1/k},n,b)$ is related to the base $b$ representation of $b^{1/k}$ so it is probably hopeless.

Q2 Is $a(1+\sqrt{6},n,b)$ efficiently computable in some base $b$?

(We couldn't find linear recurrence for it)

Q3 Except linear recurrences, are there other islands of tractability for algebraic $C$?

In comments @user44191 asked about specific constant near $1.75$.

We couldn't find linear recurrence, but got degree 2 relation factoring into linear factors, which might be hint:

0 == (2*a(n + 2) - 3*a(n + 1) - 3*a(n - 1) - a(n - 3) + a(n) - 2) *
         (a(n + 1) - a(n - 1) - a(n - 3) - a(n) - 1)

Computational bugs are possible.