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Statement

Setting: the notation of Lemma 5 (SlS_l, l(s)l(s), n(s)n(s), φ\varphi and the rotation class σ(s)\sigma(s)), and the right shift ρl:(s1,…,sl)↦(sl,s1,…,sl−1)\rho_l:(s_1,\ldots,s_l)\mapsto(s_l,s_1,\ldots,s_{l-1}) on SlS_l (p. 12; the print ends the image tuple with s2s_2).

Lemma 9 (p. 12). Let s∈Sls\in S_l with s≠(0,…,0)s\neq(0,\ldots,0), let sˉ1∈σ(s)\bar s1\in\sigma(s), and suppose 2l(s)−3n(s)>02^{l(s)}-3^{n(s)}>0. Then

gcd⁡(φ(sˉ1),φ(ρl(sˉ1)))=gcd⁡(φ(sˉ1),2l(s)−3n(s))\gcd(\varphi(\bar s1),\varphi(\rho_l(\bar s1)))=\gcd(\varphi(\bar s1),2^{l(s)}-3^{n(s)})

and in particular

gcd⁡(φ(t):t∈σ(s))=gcd⁡(φ(s),2l(s)−3n(s)).\gcd(\varphi(t):t\in\sigma(s))=\gcd(\varphi(s),2^{l(s)}-3^{n(s)}).

The displays are (14) and (15).

Reformulation (p. 13). The paper concludes from Lemma 9 that the negation of (A), the statement that (1,2)(1,2) is the only Collatz cycle in N\mathbb N (p. 1), is equivalent to

(A′') there exists a non-periodic s∈Sls\in S_l, l>3l>3, with gcd⁡{φ(t):t∈σ(s)}=2l(s)−3n(s)\gcd\{\varphi(t):t\in\sigma(s)\}=2^{l(s)}-3^{n(s)}.

The paper gives this equivalence in one sentence, without a separate proof. It offers it (p. 12) as the kind of number-theoretic argument it thinks is needed for (A), since the minimum of rational cycles grows at least linearly in their length (Remark 1, p. 8) and bounds of the kind in Theorem 4 therefore cannot prove (A).

Source. Lorenz Halbeisen and Norbert Hungerbühler, Optimal bounds for the length of rational Collatz cycles, Acta Arith. 78 (1997), 227--239; Lemma 9 on p. 12, its proof on pp. 12--13 and (A′') on p. 13 of the authors' preprint named on the source card, numbered 1--13 rather than by the journal's pagination.

Read depth. Claims checked: the statement, the formulas (13) and the reformulation (A′') were read on the print, and the proof on pp. 12--13 was read. Nothing here is independently reviewed.

Proof pointer

Pages 12--13. From (4), φ(ρl(s0))=2φ(s0)\varphi(\rho_l(s0))=2\varphi(s0) and φ(ρl(s1))=13(2φ(s1)+3n(s1)−2l(s1))\varphi(\rho_l(s1))=\frac13(2\varphi(s1)+3^{n(s1)}-2^{l(s1)}) (13). For x=φ(sˉ1)x=\varphi(\bar s1) the integrality of 13(2x+3n−2l)\frac13(2x+3^n-2^l) shows that 3∤x3\nmid x, so the gcd of xx with that number equals gcd⁡(x,3n−2l)\gcd(x,3^n-2^l), which is (14). A rotation ending in 00 only doubles φ\varphi under ρl\rho_l, and following the rotation class around gives (15).

Dependencies

The decomposition formula (4) (p. 3) and the definition (2) of φ\varphi (p. 2).

Bears on

  • #1135: background only. A cycle of the problem's ff in the positive integers other than {1,2}\{1,2\} would answer the problem in the negative; (A′') restates the existence of such a cycle as a divisibility condition on φ\varphi and decides nothing about it.