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Statement

Setting (pp. 1--2). $\Phi$ is the permutation of the 2-adic integers defined on the conjecture's page, and $C$ is the map $C(N)=N/2$ for even $N$, $C(N)=3N+1$ for odd $N$ (see Theorem 1).

Theorem 3 (p. 2): "If $N \in \mathbf{Z}^+$ and $N \in \Phi((1/3)\mathbf{Z})$ then $C^k(N) = 1$ for some $k$."

With Theorem 2, a positive integer $N$ has some iterate $C^k(N)=1$ exactly when $N\in\Phi((1/3)\mathbb Z)$, and the paper concludes (p. 2) that its conjecture is equivalent to the $3N+1$ conjecture.

Proof pointer

Put $Q=\Phi^{-1}(N)$, so $3Q\in\mathbb Z$, and let $d$ be the common exponent sequence of the expansions (1) and (2). The paper first rules out $Q\in\mathbb Z$: $Q=0$ gives $N=0$, positive $Q$ gives a finite $d$ and a negative rational $N$, and negative $Q$ gives exponents that eventually step by $1$ and again a negative rational $N$. So $Q$ differs from an integer by $1/3$, and its exponents eventually step by $2$, say from index $m$ on. On exponent sequences, $C$ subtracts $1$ from every exponent when $d_0>0$ and drops the leading exponent when $d_0=0$; after $d_m+m$ steps the sequence is $\langle0,2,4,\ldots\rangle$, the sequence of $\Phi^{-1}(1)=-1/3$, so $C^{d_m+m}(N)=1$ (p. 2).

Read depth

Claims checked: the statement was read on the page images of the print, and the proof was followed. A second reader checked the statement, hypotheses, label and page against the print; the proof was not independently reviewed.

Dependencies

Theorem 1 and the expansions (1) and (2) (p. 1).

Source. Daniel J. Bernstein, A non-iterative 2-adic statement of the $3N+1$ conjecture, Proceedings of the American Mathematical Society 121 (1994), 405--408. Pages are those of the author's typescript named on the source card, numbered 1--4 rather than by the journal's pagination.

Bears on

• Problem 1135: the converse direction of the equivalence between the problem's question and the conjecture $\mathbb Z^+\subseteq\Phi((1/3)\mathbb Z)$; the problem page records that a $C$-orbit reaches $1$ exactly when the orbit of the problem's shortcut map $f$ does. It proves neither statement.