Which power series have bounded integral coefficients and have an inverse given by a series having bounded integral coefficients Let $A=1+\sum_{n=1}^\infty \alpha_nx^n\in\mathbb Z[[x]]$ and $B=\frac{1}{A}=1+\sum_{n=1}^\infty\beta_n x^n$ two mutually inverse power series
having bounded integral coefficients (ie. $\vert \alpha_n\vert,\vert \beta_n\vert<C$ for some constant $C$ and for all $n$). 
Examples are given by $A=\frac{P}{Q}$ where $P$ and $Q$ are both finite products of cyclotomic polynomials having only simple roots.
Are there other, exotic, examples?
More generally, consider again $A=1+\sum_{n=1}^\infty \alpha_nx^n\in\mathbb Z[[x]]$ with inverse $B=\frac{1}{A}=1+\sum_{n=1}^\infty\beta_n x^n$ and require that the integral coefficients $\alpha_n,\beta_n$ have at most polynomial growth (ie. there exists a constant $C$ such that $\vert\alpha_n\vert,\vert\beta_n\vert<Cn^C+C$ for all $n$). 
Examples are now arbitrary rational fractions $A=\frac{P}{Q}$ involving only cyclotomic polynomials. Again, I know of no other examples. Do they exist?
Added: This question is closely linked to The sum of integers being a bijection, see Zaimi's comment after Venkataramana's anwser.
 A: Not a complete answer, just to suggest that there should be many examples: assume $f(z)$ is a holomorphic function on the open unit disc $D$, with continuous non-vanishing extension up to $\overline D$ and $f(0)=1$. Then $g(z):=1/f(z)$ is also continuous and non-vanishing up to $\overline D$,
and by the Cauchy path integral formula, both have bounded coefficients $\alpha_n$, respectively $\beta_n$. If moreover for some reason the $\alpha_n$ are integers, then so are the $\beta_n$, since $\alpha_0=1$ and $\sum_{j=0}^n\alpha_j\beta_{n-j}=0$ for $n\ge1$.
In other words, a source of examples (hopefully not too empty) are the coefficients $(\alpha_n)$ of the power series expansions of the invertible elements  of the algebra $H(D)\cap C^0(\overline D)$, whenever they are integers with $\alpha_0=1$.
A: Consider the set $S$ of nonnegative integers whose $2$-adic expansion involves only square powers of two (e.g. $n=1+16$, and $n=2^{25}+2^{49}+2^{64}$ belong to $S$). Let $T$ be the set of integers whose $2$-adic expansion never involves any square power of two. Then, $(S\cup\{0\})+(T\cup\{0\})$ represents  each positive integer only once. Write $f(x)=\sum _{k\in S\cup\{0\}}x^k$ and $g(x)=\sum _{k\in T\cup\{0\}}x^k$. Clearly,
$f(x)g(x)=\frac{1}{1-x}$.  This shows that $f(x)$ and $g(x)(1-x)$ have bounded integer coefficients, but are not ratios of products of cyclotomic polynomials. 
A: More examples follow from the paper of Duffin and Schaeffer, 1945.enter link description here
