arXiv2026
Let $T(n)=n/2$ if $n$ is even and $T(n)=(3n+1)/2$ if $n$ is odd. We prove that for each $m\ge1$, exactly $F(m+1)$ odd integers $n$ in $\{1,\ldots,2^m\}$ have the property that none of the iterates $T(n),T^2(n),\ldots,T^{m-1}(n)$ lies in the residue class $4\pmod6$, where $F(m+1)$ is the $(m+1)$-th Fibonacci number; the proportion decays at rate $(φ/2)^m$, $φ=(1+\sqrt{5})/2$. Equivalently, these are the odd $n\le2^m$ for which no two consecutive terms of $n,T(n),\ldots,T^{m-1}(n)$ are even. The proof uses the directed graph $G$ of Collatz transitions modulo $6$ and its unique absorbing strongly connected component $G'=G[\{1,2,4,5\}]$. Removing vertex $4$ from $G'$ yields a subgraph of spectral radius $φ$, against $ρ(G')=2$; the Fibonacci count follows from this spectral gap. We construct an explicit bijection $Ψ_m:\{1,\ldots,6\cdot2^m\}\to\mathcal{P}_m(G)$ onto the directed paths of length $m$ in $G$. We further show that no vertex of $G'$ is dispensable: removing any single vertex reduces the spectral radius strictly below $2$, with hierarchy $1<\sqrt{2}<φ<2$. In particular, every positive cycle of $T$ must visit residue class $2\pmod6$, and a flow conservation identity forces this class to account for more than $18\%$ of the steps in any such cycle.