“Whether halve-once can ever run away fast or settle into a loop is open.”
That’s where I left it in my notebook. The rule is simple: start 1, 2, then every term is the sum of the previous two — except if the sum is even, you halve it. Once. Not until it’s odd, just once. Halve once because repeated halving collapses 64 straight to 1; one halve keeps it interesting.
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. It jumps around. It doesn’t look like it settles. But “doesn’t look like it” isn’t a proof, so I wanted to know for real: can it ever loop?
We did it my way:
“We are just experimenting as if it is true… and then if it isn’t true, it’ll fail over and over.”
Assume the loop. Now look at the smallest value in it — call it m. That’s the whole trick: the minimum, not the maximum. At the maximum the argument stalls; at the minimum the odd case dies instantly. m came from two predecessors, u and v, both at least m. If u + v was odd, then m = u + v ≥ 2m > m. Dead on arrival. So u + v was even, m = (u + v) / 2, and the only way two numbers ≥ m average to m is:
“M.”
That was my entire answer at that step: u = v = m.
Then you walk it backwards:
“Dominoes start falling backwards.”
The term before was m with predecessors m and w, and the same two cases force w = m. Induction all the way around the finite loop: every term is m. So any cycle has to be constant — the same number repeating.
And here’s the thing about this rule: the only constant cycle it allows is doubles. (m, m) → 2m → halved → m. A genuine fixed point. So the whole question reduces to: can doubles ever first appear? Say (m, m) shows up for the first time, with x before it, x ≠ m. The second m came from (x, m). Odd branch: x + m = m, so x = 0 — impossible, terms are ≥ 1. Even branch: (x + m) / 2 = m, so x = m — contradicting x ≠ m.
“They cant.”
Doubles can’t first appear. The seeds 1, 2 aren’t doubles. So doubles never happen, constant cycles never happen, and by the minimum argument no other cycle can either. It never loops. And a sequence that never ends and never loops —
“They go on infinitely and state why.”
My corollary: below any bound B there are only B² possible consecutive pairs; infinitely many terms force a repeat, and the rule is deterministic, so a repeated pair is a loop. No loop, no bound. Unbounded.
One thing that does not say: unbounded is not running away fast. The sequence still falls, constantly — 6,688 dips in the first 20,000 terms. It just can’t be trapped under any ceiling, and it can’t ever come back around.
∎
Before any of this, I ran it: 20,000 terms, no consecutive doubles anywhere, biggest term 2,138 digits by term 19,999. That’s checking. It told me the conjecture was worth attacking and it told me nothing else. The computation is the scout; the minimum argument is the army. From my own notebook: Pólya’s conjecture held for every number below 906,150,257 and was still false. Collatz is verified past 2⁶⁸ and still unproven. Checking is how you decide what to try to prove. It is not the proof.
]]>About the name: ZVJ — Zach, Vera, Joseph. A working name, chosen by the author.
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.
The halving sequence above never becomes periodic. It never loops.
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.
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):
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.
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.
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