---

Statement

The Collatz function (p. 2). $C(x)=3x+1$ for $x\equiv1\pmod2$ and $C(x)=x/2$ for $x\equiv0\pmod2$.

The conjecture (p. 2, unnumbered). For each $m\in\mathbf Z^+$ there is a $k\in\mathbf Z^+$ with $C^{(k)}(m)=1$; in the survey's gloss, every positive integer eventually iterates to $1$.

The compressed map (p. 2). Since an odd $m$ goes to $3m+1$ and then to $(3m+1)/2$, the survey works with $T(x)=(3x+1)/2$ for $x\equiv1\pmod2$ and $T(x)=x/2$ for $x\equiv0\pmod2$, and notes that $T$ is the map the literature usually favors.

The odd-to-odd map (p. 2). $F:\mathbf Z^+{\mathrm{odd}}\to\mathbf Z^+{\mathrm{odd}}$, $F(x)=(3x+1)/2^{m(3x+1)}$, where $m(3x+1)$ is the number of factors of $2$ in $3x+1$; the survey remarks that the variability of the exponent seems to stand in the way of substantial analysis with $F$.

The Erdős remark (p. 2), quoted: "Paul Erdös was correct when he stated, “Mathematics is not ready for such problems.”" The survey gives no reference for the remark.

Source. M. Chamberland, An Update on the $3x+1$ Problem, author's English version of the survey in Butll. Soc. Catalana Mat. 18 (2003), 19--45; p. 2 of the 32-page English version, read on the page image. The edition read is identified on the source card.

Read depth. Claims checked: the definitions, the conjecture and the quotation were read clause by clause on the page image. The conjecture is open; nothing here is a proof, and nothing here is independently reviewed.

Proof pointer

None: the statement is a conjecture.

Dependencies

None.

Bears on

• Problem 1135: the problem's map $f$ is the survey's $T$, and its question, whether every $m\ge1$ has some $k\ge1$ with $f^{(k)}(m)=1$, is the survey's conjecture stated for $T$ in place of $C$; the two orbits reach $1$ together, as the problem page's map remark explains. The survey states the conjecture and proves nothing about it. Its form of the Erdős remark reads "is not ready" where the site, quoting Guy, has "may not be ready".