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Source. Theorem 2.3, Section 2, p. 4 of the author's version named on the source card; proof p. 5. Read on the PDF page images.

Statement

Setting as on the Theorem 2.2 page. The paper writes $T_{+,q,r}=T_{q,r}$ and $T_{-,q,r}=T_{+,q,-r}$, and notes (p. 4) that $T_{+,q,r}^{(n)}(-j)=-T_{-,q,r}^{(n)}(j)$, so the $qx-r$ map on the positive integers is the $qx+r$ map on the negative integers.

Theorem 2.3 (p. 4). For each $n\in\mathbb{Z}^+$, as rational functions,

$$f_{n,q,r}(x)=f_{n,q,-r}\left(\frac{1}{x}\right).$$

Here $q$ and $r$ are odd, as throughout the paper. As power series, the left side converges for $|x|<1$ and the right side for $|x|>1$; the equality is of the rational functions of Theorem 2.2. The paper remarks (p. 5) that Berg and Meinardus state the $3x+1$ case as following from a general theorem, and that as $n\to\infty$ the poles become dense on the unit circle.

Read depth. Claims checked: the statement was read clause by clause on the page image. The proof was read for structure only, and nothing here is independently reviewed.

Proof pointer

A computation from display (1) of Theorem 2.2: rewrite the expression in powers of $1/x$, shift the summation range from $\{1,\dots,2^n\}$ to the residues $\{0,\dots,2^n-1\}$ taken negatively, apply Theorem 2.1 to pass to negative arguments, and use the sign relation above to recognize the series of $T_{-,q,r}^{(n)}$ in $1/x$ (p. 5).

Dependencies

Theorem 2.2 and Theorem 2.1.

Bears on

Problem 1135: with $(q,r)=(3,1)$ it relates the generating function of the problem's map to that of the $3x-1$ map. It says nothing about whether orbits reach $1$.