The Proof (Prime Alternation Theorem)
Claim: T[1][1] = T[2][2] = 0 for primes > 3.
Proof: Suppose gn ≡ gn+1 ≡ r (mod 3), with r ≠ 0.
Then p, p+r, p+2r form an arithmetic progression mod 3.
Three consecutive AP terms mod 3 must hit all residues {0,1,2}.
So one of the three "primes" is divisible by 3.
But all three are > 3, so the one divisible by 3 is composite.
Contradiction. █
Translation: Three doors labeled 0, 1, 2.
If you came through door 1, that door LOCKS behind you.
If you came through door 2, THAT door locks too.
Only door 0 can stay open. K=3 forces alternation.
The wall with three doors.