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Source. Theorem 3.1, Section 3, pp. 6--7 of the author's version named on the source card; proof pp. 7--8. Read on the PDF page images.

Statement

Setting as on the Theorem 2.2 page.

Theorem 3.1 (pp. 6--7). For fixed odd $(q,r)$ and each $n\ge1$,

$$f_{n,q,r}(x)=\sum_{s^{2^n}=1}\left(\frac{s^2}{(x-s)^2}B_{n,q,r}(s) +\frac{s}{x-s}\bigl(A_{n,q,r}(s)+B_{n,q,r}(s)\bigr)\right),$$

(display (4)), where for each $2^n$-th root of unity $s$

$$A_{n,q,r}(s)=-\frac{1}{2^n}\sum_{j=1}^{2^n}T_{q,r}^{(n)}(j)s^j +\frac{1}{4^n}\sum_{j=1}^{2^n}q^{O_{q,r}^{(n)}(j)}js^j , \qquad B_{n,q,r}(s)=\frac{1}{4^n}\sum_{j=1}^{2^n}q^{O_{q,r}^{(n)}(j)}s^j .$$

The paper calls $A_{n,q,r}(s)$ the residue term and $B_{n,q,r}(s)$ the double pole contribution at $s$ (p. 7); Section 3's title calls them interpolating polynomials.

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

Proof pointer

Expand display (1) of Theorem 2.2 to second order about each root of unity $s$ to read off the principal part there. The difference between $f_{n,q,r}$ and the sum of principal parts is then a polynomial, which vanishes because $f_{n,q,r}(0)=0$ and $f_{n,q,r}(x)\to0$ as $|x|\to\infty$, the latter from (1) since $T_{q,r}^{(n)}(2^n)=1$ and $O_{q,r}^{(n)}(2^n)=0$ (pp. 7--8).

Dependencies

Theorem 2.2.

Bears on

Problem 1135: with $(q,r)=(3,1)$ it expands the generating function of the problem's map. Its coefficients are the quantities whose behaviour in $n$ is recorded on the Theorem 4.1 page. It says nothing about whether orbits reach $1$.