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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.

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    $\begingroup$ unfortunately my references aren't within reach at the moment, but I believe many modular forms for congruence subgroups have integral coefficients and they all satisfy the Ramanujan-Petersen bound, which should imply polynomial growth in the coefficients of their q-expansions... $\endgroup$
    – Will Chen
    Commented Feb 16, 2015 at 21:21
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    $\begingroup$ Yes, but $1/f$ should also have polynomial growth and this is very stringent (and $1/f$ is generally not a modular from what I remember). $\endgroup$
    – Roland Bacher
    Commented Feb 16, 2015 at 21:31

3 Answers 3

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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.

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    $\begingroup$ You'd better include $0$ in $S$ and $T$: otherwise, $2^{n^2}$ gets no representation in $S+T$. Then $(fg)(x) = 1/(1-x)$. $\endgroup$
    – Vesselin Dimitrov
    Commented Feb 17, 2015 at 2:51
  • $\begingroup$ @Vesselin Thanks. I have included this in the edited version $\endgroup$
    – Venkataramana
    Commented Feb 17, 2015 at 2:56
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    $\begingroup$ In some sense these are also "products of cyclotomic polynomials", you just need to allow for infinite products. See my answer here mathoverflow.net/questions/50798/… $\endgroup$
    – Gjergji Zaimi
    Commented Feb 17, 2015 at 3:43
  • $\begingroup$ Is a "square power of 2" a square number that's a power of 2, or is it the exponent that's a square? If the latter, then $n=4+16$ doesn't qualify, as $4=2^2$. If the former, the other example doesn't qualify, as $2^{25}$ is not a square number. $\endgroup$
    – Gerry Myerson
    Commented Feb 17, 2015 at 4:37
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    $\begingroup$ This example can even be simplified: $S$ given by $0$ and the union of all integers involving only even powers of two, $T$ union of $0$ and integers involving only odd powers of $2$. Since both $S$ and $T$ have arbitrary large gaps, their characteristic functions are not algebraic and multplying one of them by $1-x$, one gets an example (one might even replace even, odd numbers by an arbitrary partition of $\mathbb N$ into two infinite subsets). $\endgroup$
    – Roland Bacher
    Commented Feb 17, 2015 at 11:09
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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 $\alpha_0=1$.

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  • $\begingroup$ And the Hardy space $H^\infty$ should work as well. $\endgroup$
    – Pietro Majer
    Commented Feb 16, 2015 at 23:02
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    $\begingroup$ This can only produce polynomials: The $\alpha_j$ can be viewed as Fourier coefficients of the boundary function, so $\alpha_j\in\ell^2$ if the boundary values are in $H^{\infty}$. $\endgroup$
    – Christian Remling
    Commented Feb 16, 2015 at 23:40
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More examples follow from the paper of Duffin and Schaeffer, 1945.enter link description here

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    $\begingroup$ The above paper seems to give a correct answer to a different question, if I am not mistaken. $\endgroup$
    – Roland Bacher
    Commented Feb 16, 2015 at 18:41

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