Divisibility Obstruction Conjecture

The Conjecture (Refined)

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

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

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

Equivalently: the only Collatz cycle is $1 \to 4 \to 2 \to 1$.

Why This Is the Right Formulation

The original conjecture stated: "no valid parity word has $g \mid 2^E \cdot C$." Testing all $(S, E)$ pairs with $K \leq 30$ revealed apparent counterexamples — but every one turned out to be the trivial cycle in disguise.

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 $\{1, 2, 4\}$.

This pattern can only tile a cycle of length $K$ when $3 \mid K$. For convergents of $\log_2 3$:

$(S, E)$	$K$	$3 \mid K$?	Trivial fits?	Non-trivial zeros?
$(1, 2)$	3	Yes	Yes → $n = 1, 2, 4$	None
$(5, 8)$	13	No	Can't tile	None (verified)
$(41, 65)$	106	No	Can't tile	Conjectured none
$(665, 1054)$	1719	Yes	Could tile — but ascending	Eliminated by sign

The ordering is what blocks non-trivial zeros

Key finding. Without the ordering constraint $q_0 < q_1 < \cdots < q_{S-1}$, the sum $T$ equals zero with probability exactly $1/g$ — perfectly random. But with the ordering constraint, $T = 0$ occurs only for the trivial pattern.

Verified for gap $= 13$: assigning the 5 coefficients $[3^4, 3^3, 3^2, 3^1, 3^0]$ to 5 distinct exponents in any order gives $T \equiv 0$ for 7.7% of assignments ($\approx 1/13$). But restricting to the monotone assignment $q_0 < q_1 < q_2 < q_3 < q_4$ eliminates all zeros.

Evidence

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

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

• When $3 \nmid K$: zero parity words give $g \mid T$. Every case is blocked.
• When $3 \mid K$: the only words giving $g \mid T$ produce $n \in \{1, 2, 4\}$ — the trivial cycle.

No non-trivial cycle candidates found in any case.

Gap = 13 (detailed)

For $(S=5, E=8, K=13)$: 91 valid parity words, distribution of $T \bmod 13$:

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 structurally excluded.

The Algebraic Structure

The sum decomposes as:

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

Working mod $g$ where $2^E \equiv 3^S$:

• The coefficients $3^{S-1-j}$ cycle with period $\text{ord}(3 \bmod g)$
• The bases $2^{q_j}$ cycle with period $\text{ord}(2 \bmod g)$
• The monotonicity of exponents creates a positional lock: larger indices get smaller 3-coefficients and larger 2-exponents

For the trivial pattern $q_j = 2j$: the sum telescopes because each term $3^{S-1-j} \cdot 2^{2j} = 3^{S-1-j} \cdot 4^j$, and $\sum 3^{S-1-j} \cdot 4^j = (4^S - 3^S)/(4 - 3) = 4^S - 3^S = g$. So $T \equiv 0$ automatically.

For any other monotone sequence: the sum no longer telescopes, and the empirical evidence shows it never hits zero.

The Challenge Ahead

Proving the ordering obstruction reduces to:

Show that $\sum_{j=0}^{S-1} 3^{S-1-j} \cdot 2^{q_j} \equiv 0 \pmod{2^E - 3^S}$ with $0 \leq q_0 < q_1 < \cdots < q_{S-1} \leq E-1$ implies $q_j = 2j$ for all $j$.

This is a statement about weighted power sums modulo a specific composite number. The monotonicity of exponents, combined with the geometric decay of 3-coefficients, prevents the kind of cancellation that would make the sum vanish.

Approaches:

1. Induction on $S$: show that adding one more term to a non-trivial monotone sum can't restore cancellation
2. $p$-adic analysis: the sum has specific $p$-adic properties for primes $p \mid g$
3. Connection to digital representations: the sum resembles a mixed-radix representation in bases 2 and 3, where the ordering prevents "carrying"

Related

• Convergent Elimination — the computational results this conjecture refines
• abc Conjecture — the number theory connection
• Path to Proof — how this fits in the roadmap