Terras–Everett parity-vector proof
The argument
Because each of the 2^k parity patterns of length k is realised by exactly one residue class modulo 2^k, the number of odd steps among the first k iterates of the compact Collatz map T is binomially distributed over N, and for the exceptional N with at least k·log 2/log 3 odd steps the Chernoff–Hoeffding tail bound gives a proportion at most η^k for some η < 1. Writing T^k(N) = 3^a N / 2^k + β with β bounded by 3^a shows that any N large relative to k with a < k·log 2/log 3 satisfies T^k(N) < N; letting k grow slowly with N, the exceptional set has natural density zero, so almost all N descend below themselves.
Does it work?
Not yet evaluated.
Premises
- UnassessedNo current assessment. Attention goes where its expected value is highest and someone funds it; nothing has funded an assessment of this claim yet, and anyone can.constitution →The map assigning to each residue class modulo 2^k its first k Collatz parities is a well-defined bijection onto the 2^k parity patterns.
Where it is made
Passages where sources make, criticize or mention this argument. They record where it appears, not how strong it is, and none has precedence over another.
No source is recorded as making it.
Earlier ratings
- Sep 17, 2026
The inference goes through and was checked directly: given the equidistribution of parity patterns, the binomial tail bound and the explicit form of the k-th iterate together force descent below N for all but an exponentially small fraction of residue classes, and letting k grow with N gives density one. The argument rests entirely on the parity-vector bijection modulo 2^k, an elementary induction that nobody disputes; the remaining steps are standard estimates carried in the written form. This is the proof found independently by Terras, Everett, Möller, Heppner and Allouche, and the one gap raised against Terras's original write-up was filled in his 1979 note.