Cycles as a Divisibility Question

This page states the no-cycles question as a divisibility problem and shows some computations. The statement below is equivalent to "there are no nontrivial Collatz cycles": it is just the Böhm–Sontacchi cycle equation written out term by term. So it is exactly as hard as the cycle problem, which is open.

The statement

Conjecture (Divisibility Obstruction). For any $(S, E)$ with gap $g = 2^E - 3^S > 0$, the sum

$$T = \sum_{j=0}^{S-1} 3^{S-1-j} \cdot 2^{q_j}$$

with strictly increasing exponents $0 = q_0 < q_1 < \cdots < q_{S-1} \leq E - 1$ is divisible by $g$ only when $q_j = 2j$ for all $j$ — the trivial cycle pattern $(1,0,0)$ repeated.

Why this is the cycle problem: an odd number $n$ starts a cycle with $S$ odd steps and $E$ halvings exactly when $n = T / g$ for the exponents $q_j$ recording how many halvings have happened before the $j$-th odd step (Böhm and Sontacchi, 1978). So "$g \mid T$ only for the trivial pattern" says "the only cycle is $1 \to 4 \to 2 \to 1$".

The trivial cycle has period 3

The cycle $1 \to 4 \to 2 \to 1$ has parity word $(1, 0, 0)$: one odd step, two even steps. When repeated $m$ times, it gives $(S, E) = (m, 2m)$ with $K = 3m$ and gap $g = 4^m - 3^m$. The resulting $n$ values are always $\{1, 2, 4\}$.

For that pattern the sum telescopes: $\sum_{j} 3^{S-1-j} \cdot 4^j = 4^S - 3^S = g$, so $T \equiv 0$ automatically. For any other increasing sequence it no longer telescopes, and in every case I have tested it never hits zero.

Evidence

Exhaustive check ($K \leq 30$, gap $< 10{,}000$)

All $(S, E)$ pairs with $K = S + E \leq 30$ and $0 < g < 10{,}000$ were tested:

• When $3 \nmid K$: no parity word gives $g \mid T$.
• When $3 \mid K$: the only words giving $g \mid T$ produce $n \in \{1, 2, 4\}$ — the trivial cycle.

This is consistent with, and much weaker than, the known results on cycles (Steiner 1977, Eliahou 1993, Simons and de Weger 2005, Hercher 2023), which rule out every nontrivial cycle of this size and far beyond.

Gap = 13

For $(S=5, E=8, K=13)$, taking all 91 circular parity words (every starting point, odd or even), the residue of the cycle equation mod 13 is distributed as:

Remainder	0	1	2	3	4	5	6	7	8	9	10	11	12
Count	0	7	6	9	7	6	10	7	6	10	6	8	9

Roughly uniform over $\{1, \ldots, 12\}$, with zero missing.

Ordered versus unordered exponents

For gap 13, assigning the five coefficients $[3^4, 3^3, 3^2, 3^1, 3^0]$ to five distinct exponents from $0$ to $7$ in any order gives $T \equiv 0$ for 7.7% of the 6,720 assignments, close to $1/13$. None of the 56 increasing assignments $q_0 < q_1 < \cdots < q_4$ gives zero. That is a small sample: the 56 come from just 7 distinct cyclic words, and seven tries at a 1-in-13 chance all miss more than half the time. The increasing order — larger 2-exponents paired with smaller 3-coefficients — is what the cycle equation actually requires, but the order alone is not an obstruction: for the negative gap $2^{11} - 3^7 = -139$ there are increasing assignments with $g \mid T$, and they are the cycle through $-17$. An argument that ignores the sign of the gap cannot work.

Ideas worth trying

1. Induction on $S$: understand how adding one more term to an increasing sum changes its residue mod $g$.
2. $p$-adic analysis: look at the sum modulo the individual primes dividing $g$.
3. Mixed-radix digits: the sum resembles a representation in bases 2 and 3 at once, where the ordering might prevent the "carrying" that cancellation needs.

None of these has produced a proof.

Related

• The Cycle Equation: Small Cases — the cycle equation and the first few cases
• No Loops? — the tour chapter on cycles
• abc Conjecture — the number theory connection