Integer sequence generated by using transcendental numbers Given the integer sequence in which the $n^{\rm th}$ term is defined by: $$ f(n)=\lfloor nt \rfloor, $$ where $t$ is a transcendental number and $n$ is a positive integer, are infinitely many terms of the sequence comprime? More specifically, which terms of the sequence are coprime, and how may one go about proving/disproving such a property? It seems like Dirichlet's theorem on arithmetic progressions is the way to go, but the presence of the floor function and transcendetal numbers is a big hurdle. In general, what can be said about the co-primality of the terms in such sequences? Idem for the functions: * $f(n)=\lfloor n^t\rfloor$, * $f(n)=\lfloor t^n\rfloor$.

$\
ewcommand{\fl}[1]{\lfloor #1 \rfloor}$ Let's reason about the negative. Suppose that for some $t$ only finitely many terms of $\fl{tn}$ are coprime with some other term. Then there must be some constant integer $s$ and constant integer common divisor $d > 1$ such that $\fl{t(n+s)} = dm$ for all $n>0$.

Now write $i = \fl t$ and $x = t - i$. Then we have:

$$\fl{(i + x)(n+s)} = dm$$ $$\fl{x(n+s)} = dm - i(n+s)$$

But also $$\fl{x(n+s+1)} = dm' - i(n+s+1)$$

Subtracting these two we find:

$$\fl{x(n+s+1)} - \fl{x(n+s)} = d(m'-m) - i$$

If we look at this equation $\bmod d$ we find:

$$\fl{x(n+s+1)} - \fl{x(n+s)} = - i$$

Since $0 \leq x < 1$, the left hand side can only take on values $1$ and $0$. And for any $x > 0$, the left hand side varies between $0$ and $1$ infinitely often for various $n$. But the right hand side is constant, therefore $x = 0$ and thus $t$ must be integer.