Proved: 17 October 2026

About the name: ZVJ — Zach, Vera, Joseph. A working name, chosen by the author.

The sequence

This theorem is about a sequence; it does not claim the sequence itself. The sequence is catalogued in the OEIS as A069202, “A Collatz–Fibonacci mixture,” Benoit Cloitre (2002). The catalogue entry does not contain the theorem below; the proof is the author’s own, dated above.

The rule:

a₁ = 1,  a₂ = 2

For n ≥ 3, let s = aₙ₋₂ + aₙ₋₁

  if s is odd:   aₙ = s        (just add the previous two)
  if s is even:  aₙ = s / 2    (halve the sum — once only, not repeatedly)

The first 23 terms:

1, 2, 3, 5, 4, 9, 13, 11, 12, 23, 35, 29, 32, 61, 93, 77, 85, 81, 83, 82, 165, 247, 206

The open question, in the author’s notebook: whether halve-once can ever settle into a loop.

Theorem (ZVJ)

The halving sequence above never becomes periodic. It never loops.

Proof

Lemma 1 — Any cycle must be constant.

Assume the sequence becomes periodic, and let m be the smallest value in the cycle, with predecessors u and v in the cycle, so u, v ≥ m, and m came from s = u + v.

  • If s was odd, then m = s = u + v ≥ 2m > m — a contradiction.
  • So s was even: m = s / 2, i.e. u + v = 2m, which forces u = v = m.

Step one position back: the previous term is m with predecessors m and w ≥ m, and the same two cases force w = m. Continuing backward around the whole finite cycle, every term is m.

Lemma 2 — Doubles can never first appear.

Suppose (m, m) occurs for the first time, with the term before it x ≠ m (the seeds are 1, 2, so this is not the start). The second m came from (x, m):

  • Odd branch: x + m = m, i.e. x = 0 — impossible, all terms are ≥ 1.
  • Even branch: (x + m) / 2 = m, i.e. x = m — contradicting x ≠ m.

So a first occurrence of doubles is impossible, and since the seeds are not doubles, doubles never occur.

A constant sequence is a genuine fixed point of this rule — (m, m) → 2m → halved → m — so the only way to loop would be to land on doubles. By Lemma 2 that cannot happen, and by Lemma 1 no other cycle is possible. ∎

Corollary — The terms are unbounded.

If every term were ≤ B, there would be only B² possible consecutive pairs. Infinitely many terms would force a repeated pair, and the rule is deterministic, so a repeated pair is a loop — contradicting the theorem. Hence no bound exists.

Checking is not proving

Before the proof, the first 20,000 terms were computed: no consecutive doubles anywhere, and the largest term, at term 19,999, has about 1,000 digits. That computation is consistent with the theorem; it is not the proof.