Théorème

T1 — Forbidden auto-transitions mod 3

The sieve forbids two consecutive identical residues mod 3.

Statement

Let $T_3$ be the transfer matrix of the sieve of 6-rough integers (integers divisible by neither 2 nor 3) on residue classes modulo 3. Then:

\boxed{T_3[1, 1] = T_3[2, 2] = 0.}

In other words: three consecutive survivors $n_1 < n_2 < n_3$ cannot all belong to the same class $\{1, 2\} \pmod 3$ .

Théorème

Plain reading. List integers divisible by neither 2 nor 3: 1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, … Look at their remainder mod 3: 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, … It always alternates. Three in a row with the same remainder is arithmetically impossible. That’s T1.

The sieve and the forbidden rule

Primes emerge from the sieve. The gaps between sieve survivors do not distribute arbitrarily across residue classes mod 3 — that is rule T1.

prime multiple eliminated current prime gap ≡ 1 mod 3 gap ≡ 2 mod 3

Why it matters

T1 is the first exact constraint on the dynamics of the sieve. It is what forces $s = 1/2$ (zero-diagonal theorem) and opens the way to T3 (the matrix $T_3$ is purely antidiagonal), T6 (the holonomy sin²θ_p emerges from restricted matrices after T1), and T2 (spectral conservation).

Without T1, the transfer matrix would have a positive diagonal term and the symmetry $s = 1/2$ would not be forced — it would have to be imposed as a postulate.

T1 holds at the sieve level, not at the prime level: twin primes like $(61, 67)$ may share the same class mod 3, but the determining dynamics of PT is sieve-level dynamics, not prime-level (cf. Cor. T1-primes, ch. 1).

Proof — outline

Characterise 6-rough integers as an arithmetic sequence mod 6.
List survivors mod 6: $\{1, 5\} \pmod 6$ .
Reduce mod 3: $1 \to 1$ , $5 \to 2$ .
Observe that the sequence $\{1, 5, 7, 11, 13, 17, \ldots\} \pmod 3$ strictly alternates between 1 and 2.
Conclude: the transfer matrix on $\{1, 2\}$ has only off-diagonal non-zero entries.

Detailed proof

Step 1 — Characterisation of 6-rough integers

An integer $n$ is 6-rough if $\gcd(n, 6) = 1$ . This is equivalent to:

n \equiv 1 \pmod 6 \quad\text{or}\quad n \equiv 5 \pmod 6.

The 6-rough integers form two arithmetic progressions mod 6: $\{1, 7, 13, 19, 25, \ldots\}$ and $\{5, 11, 17, 23, 29, \ldots\}$ .

Step 2 — Reduction mod 3

Reduce both progressions modulo 3:

$1, 7, 13, 19, 25, \ldots \equiv 1, 1, 1, 1, 1, \ldots \pmod 3$ ,
$5, 11, 17, 23, 29, \ldots \equiv 2, 2, 2, 2, 2, \ldots \pmod 3$ .

Step 3 — Forced alternation

But sorting their union in increasing order yields:

1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, \ldots

with mod-3 classes:

1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, \ldots

The alternation is rigorous: between two consecutive 6-rough integers, the mod-3 class must change.

Formal proof: if $n$ and $n + g$ are two consecutive 6-rough (with $g \in \{2, 4\}$ — the possible gaps between adjacent 6-rough), then:

$g = 2$ sends $1 \pmod 3 \to 0 \pmod 3$ or $2 \pmod 3 \to 1 \pmod 3$ : the first option contradicts “6-rough” (multiple of 3), so $g = 2$ gives $2 \to 1$ .
$g = 4$ gives $1 \to 2$ or $2 \to 0$ : the second option contradicts “6-rough”, so $g = 4$ gives $1 \to 2$ .

In both cases, the class changes. QED.

Step 4 — Transfer matrix

The transfer matrix $T_3$ on states $\{1, 2\}$ (the non-zero classes mod 3) captures the probability of transitioning from one class to the other in one sieve step:

T_3 = \begin{pmatrix} P(1 \to 1) & P(1 \to 2) \\ P(2 \to 1) & P(2 \to 2) \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}.

This is precisely the statement of T1 (the diagonal coefficients are zero). T3 follows: $T_3$ is antidiagonal.

Step 5 — Consequence: $s = 1/2$

The antidiagonal matrix $T_3 = \mathrm{antidiag}(1, 1)$ has eigenvalues $\pm 1$ . The stationary distribution is uniform on $\{1, 2\}$ with weight $1/2$ each. This stationarity, combined with the Dirichlet exchange symmetry, forces the fundamental value $s = 1/2$ .

Thus the “fundamental sieve symmetry” is not a postulate but a consequence of T1.

Sieve / primes distinction

At the sieve level (6-rough integers), T1 is exact: no mod-3 transition repeats. At the prime level (subsequence of 6-rough that are prime), T1 is only approximate: pairs like $(61, 67)$ or $(151, 157)$ share mod-3 class (both $\equiv 1$ ). But these violations are of order $O(1/\ln N)$ — they vanish in the asymptotic limit.

All later PT derivations use the sieve-level matrix, not the empirical prime-level matrix (cf. Remark T0_scope, ch. 1).

For the complete derivation and the sieve/primes discussion, see chapter 1 of the monograph.