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Statement

The two maps (p. 1). The Collatz function is $C(x)=3x+1$ for $x\equiv1\pmod2$ and $C(x)=x/2$ for $x\equiv0\pmod2$. The $3x+1$ function is $T(x)=(3x+1)/2$ for $x\equiv1\pmod2$ and $T(x)=x/2$ for $x\equiv0\pmod2$. The survey records the relation $T(x)=C(C(x))$ for odd $x$ and $T(x)=C(x)$ for even $x$, so that iterating $T$ omits some of the steps of iterating $C$; it credits the observation that $T$ is the more convenient map for analysis to Terras (its [88], [89]) and Everett (its [27]), independently.

$3x+1$ Conjecture (p. 1, unnumbered), quoted: "Starting from any positive integer $n$, iterations of the function $C(x)$ will eventually reach the number 1. Thereafter iterations will cycle, taking successive values $1,4,2,1,\ldots$."

Reformulations stated in the survey. Backwards (p. 4): let $S_0$ be the smallest set of integers that contains $1$ and is closed under the maps $x\mapsto2x$ and $3x+2\mapsto2x+1$, the second applied only to inputs $3x+2$ for which $2x+1$ is an integer; the conjecture then says that $S_0$ is the set of all positive integers. Through powers of $2$ (p. 13): the conjecture can be restated as saying that from every positive integer $n$ some iterate $C^{(k)}(n)$ of the Collatz function, or of the $3x+1$ function, is a power of $2$.

Source. J. C. Lagarias, The $3x+1$ problem: an overview, in The Ultimate Challenge: The $3x+1$ Problem (AMS, 2010), 3--29; the arXiv:2111.02635v1 copy, p. 1 (the maps and the conjecture), p. 4 (the set $S_0$) and p. 13 (the power-of-2 form), read on the page images. The edition read is identified on the source card.

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

Proof pointer

None: the statement is a conjecture. The equivalences with the backward and power-of-2 forms are asserted in the survey without proof. The backward form rests on the observation that the inverse images of $y$ under $T$ are $2y$ and, when $y=3x+2$, the odd integer $2x+1$; so the two maps generate from $1$ exactly the positive integers whose $T$-orbit contains $1$ (a remark of this page, not of the survey).

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 $3x+1$ Conjecture stated for $T$ in place of $C$. The survey poses the conjecture for $C$; by the relation $T(x)=C(C(x))$ ($x$ odd), $T(x)=C(x)$ ($x$ even) the $T$-orbit of $m$ is the $C$-orbit with some terms left out, and the two orbits reach $1$ together, as the problem page's map remark explains. The survey states the conjecture and proves nothing about it.