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Source. Theorem 4.2, Section 4, p. 12 of the author's version named on the source card; proof p. 12. Read on the PDF page image.

Statement

Setting as on the Theorem 3.1 page.

Theorem 4.2 (p. 12). For fixed $(q,r)$ and $m\ge1$,

$$T_{q,r}^{(n)}(m)=\sum_{s^{2^n}=1}\frac{1}{s^m} \bigl(mB_{n,q,r}(s)-A_{n,q,r}(s)\bigr).$$

The statement says only "fixed $(q,r)$"; the coefficients are those of Theorem 3.1, defined there for odd $(q,r)$ and $n\ge1$. Section 5 (pp. 13--14) works the case $q=r=1$, where $T_{1,1}^{(n)}(j)=1$ for $1\le j\le2^n$, as an illustration.

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

Insert the defining sums of $A_{n,q,r}(s)$ and $B_{n,q,r}(s)$, exchange the two sums, and use orthogonality of the $2^n$-th roots of unity, which keeps only the index $k\equiv m\pmod{2^n}$ (p. 12). The printed proof writes that index as $m$, which covers $1\le m\le2^n$; for larger $m$ the surviving index is the residue $j$ of $m$, and Theorem 2.1 with $m=2^nk'+j$ completes the computation, a step the print does not write out.

Dependencies

Theorem 3.1.

Bears on

Problem 1135: with $(q,r)=(3,1)$ it writes each iterate of the problem's map through the polar coefficients of Theorem 4.1. The paper uses the two together only as a heuristic (pp. 12--13) that every orbit is bounded. Nothing is proved about whether orbits reach $1$.