The largest admissible digit sets modulo primes up to 31

Let r_k(ℤ_mⁿ) denote the largest size of a subset of ℤ_mⁿ containing no arithmetic progression of length k. Elsholtz, Klahn and Lipnik construct large progression-free sets from admissible digit sets D ⊆ ℤ_p and tabulate, for primes p ≤ 31 and 3 ≤ k ≤ 8, the largest admissible D they could find. Seventeen entries of their table are recorded only as lower bounds. We determine all of them. Two entries increase: the largest admissible digit sets modulo 31 have size 23 for k = 5 and 24 for k = 6, rather than at least 22 and 23, so that r₅(ℤ₃₁ⁿ) ≫ 23ⁿ/n¹¹ and r₆(ℤ₃₁ⁿ) ≫ 24ⁿ/n^(23/2). The other fifteen tabulated values are optimal. For three-term progressions we prove that the largest admissible digit set modulo any odd prime p has exactly (p + 1)/2 elements, so the digit-set construction cannot improve the exponential base (p + 1)/2 for k = 3. The tool is an observation about the decision problem. Admissibility is the non-existence of a non-zero non-negative integral vector in a rational subspace, and has accordingly been treated as an integer program. The feasible set is a rational cone, so it contains a non-zero integral point precisely when it contains a non-zero real one, and admissibility is a linear program. Gordan's theorem then supplies a short certificate of admissibility, verifiable in exact integer arithmetic, and non-admissibility is certified by an explicit integral solution. Every lower bound in the table is certified in this way, and every upper bound is either a consequence of the three-term theorem or certified by a list of non-admissible sets with a symmetry-free covering check.