Consecutive prime gaps, reduced mod K=3, can NEVER repeat class 1 or class 2. Three doors, but if you came through door 1 it locks behind you. If you came through door 2 it locks too. Only door 0 stays open. K=3 forces alternation.
Proof: Suppose g_n = g_{n+1} = r (mod 3) with r != 0. Then p, p + g_n, p + g_n + g_{n+1} are three consecutive primes. Their residues mod 3 form: p mod 3, (p + r) mod 3, (p + 2r) mod 3. Since r != 0 (mod 3), these three residues are a permutation of {0, 1, 2}. One of the three primes is divisible by 3. But all three are > 3, so that one is composite. Contradiction.
The proof uses ONLY that K=3 is prime and that three terms of an arithmetic progression mod a prime hit all residues. This is K=3 doing what K does: closing. The minimum prime that forces a modular constraint.
| Aspect | Standard view | Through the axiom |
|---|---|---|
| Gap classes | Gaps seem random, unpredictable | K=3 forces T[1][1]=T[2][2]=0. Exact. Proved |
| The bias | Lemke Oliver 2016: empirical, unexplained | K=3 closure: 3 AP terms mod prime must hit all residues |
| Information | No quantification | 0.17 bits mutual information from K alone (94% of total) |
| Why K? | No structural reason given | K=3 = minimum closure. D=2 only splits parity. K is the first that gates |
Enter a prime limit. The widget computes the 3x3 transition matrix for consecutive gap classes (gap mod 3) across all primes up to that limit. Watch T[1][1] and T[2][2] stay at ZERO.
Primes up to:
Try: 100 (25 primes), 1000 (168 primes), 10000 (1229 primes), 50000 (5133 primes).
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