#789: h(28) = h(29) = 5 — the block-of-eight pattern breaks at 28, the lemma that carried the table is false, and h ≤ 6 holds through n = 39
The problem (erdosproblems.com/789): h(n) is the largest k such that every n-element set of integers contains a k-element subset in which equal subset sums force equal subset sizes ("admissible"). Our entry of 11 October pinned h(n) = 5 for 20 ≤ n ≤ 27, noticed that the whole table through 27 reads ⌊(n+20)/8⌋, and left h(28) ∈ {5, 6} with a named lemma running for the lower side.
The upper side: a 29-point obstruction. h is non-decreasing in n, so one n-set with no admissible 6-subset proves h(n′) ≤ 5 for every n′ ≤ n. Finding the largest such set inside a window [−N, N] is pure SAT: one Boolean per integer, one clause per admissible 6-subset of the window ("not all of these"), and a cardinality constraint, with cadical raising the cardinality until UNSAT. Admissible 6-subsets are rare among small integers — none at all in [−11,11], 16 of 475,020 in [−14,14], 83,788 of 22,957,480 in [−26,26] — which is why the search is cheap. Windows [−17,17], [−20,20], [−23,23] and [−26,26] all return the same maximum, 29, and all are UNSAT at 30. The 29-set is symmetric: {0} ∪ ±{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16}; every one of its 475,020 six-subsets was re-checked against the literal definition and none is admissible. With the standing lower bound h(n) ≥ 5 for n ≥ 20, h(28) = h(29) = 5. The ⌊(n+20)/8⌋ pattern, which predicted 6 at 28, is dead at the first step past the table it was read off — as the previous entry said it had to be eventually, only sooner. (The search also reproduces the known pieces: the whole interval [−11,11] has no admissible 6-subset, and [−14,14] gives back the 27-point obstruction.)
The lower side: the lemma is false. The route that produced h(20 … 27) was a positive-reals lemma — "every m distinct positive reals contain an admissible k-subset", decided by SMT — plus pigeonhole (an n-set has at least ⌈(n−1)/2⌉ nonzero elements of one sign), giving h(2m) ≥ k. The next rung, (m, k) = (14, 6), had run 36 hours on z3 without a verdict. The same SAT search restricted to positive integers settles it in 3 seconds: {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 14, 15, 16} ⊂ [1,17] has no admissible 6-subset (all 3,003 six-subsets re-checked), so the lemma fails at 14 — and at 15: {1, …, 10, 12, 14, 18, 19, 23} ⊂ [1,24]. No 16-set exists in [1,24], [1,28] or [1,32]. So pigeonhole cannot certify h(30) ≥ 6 or h(31) ≥ 6 by any solver run; the first value it can still reach is h(32) ≥ 6, via (16, 6) over the reals, which we ran on an order-pruned encoding (dropping the 153 of 231 bad equations that the ordering alone makes impossible; it finds the (12, 6) witness in 116 s where the old encoding took 11,251 s — but even pruned it did not find the (14, 6) witness within an hour). Updated 24 October: z3 stopped on (16, 6) after 25.6 hours (92,230 s) and returned unknown. No time limit was set, so it most likely ran out of its 20 GB memory cap (the reason was not logged). So this route does not certify h(32) ≥ 6, and we have not relaunched it. Whether h(30) and h(31) are 5 or 6 is open; every window we searched says no 30-set avoids an admissible 6-subset, but a window is not a proof.
The next ceiling. The same search with 7-subsets gives sets with no admissible 7-subset of size 29 in [−14,14] (the whole interval), 33 in [−17,17], 37 in [−20,20], 39 in [−23,23], 41 in [−26,26] and 43 in [−29,29] — {0} ∪ ±{1, …, 12, 14, 15, 16, 18, 20, 21, 25, 27, 28}, all 32,224,114 seven-subsets re-checked — so h(n) ≤ 6 for all n ≤ 43. The windows [−32,32] (696 million seven-subsets classified) and [−35,35] (1.33 billion) also stop at 43 and are UNSAT at 44, the same plateau shape the 6-subset search showed at 29 — evidence, not proof, that h(44) = 7.
Controls and what is not claimed. The integer search recovers the known (12, 6), (13, 6) and (9, 5) witnesses and returns UNSAT for the proved (10, 5) lemma; the signed search reproduces H([−11,11]) = 5 and the 27-point obstruction; the pruned encoding was checked to drop no equation any known witness uses. All upper bounds are exhaustive checks of explicit sets; the only real-arithmetic verdict used is the already-published (10, 5). Nothing here touches the exponent of h. Artefacts: verification/789/h28-29.