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a(1) = 1, a(n) = (max(m) : Sum_{s=r..m} 1/s <= 1)-r+1, r = Sum_{k=1..n-1} a(k). - Pablo Hueso Merino, Feb 16 2020
a(1) = 1 because 1 <= 1, 1 is one term (if you added 1/2 the sum would be greater than 1);.
a(2) = 2 because 1/2 + 1/3 = 0.8333... <= 1, 1/2 and 1/3 are two terms (if you added 1/4 the sum would be greater than one);.
a[1] = 1;
a[n_] := a[n] = Module[{sum = 0}, r = 1 + Sum[a[k], {k, n - 1}];
p = x - 2;
p - r + 1];
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If the harmonic series is divided into the longest possible consecutive groups so that the sum of each group is <= 1, then a(n) is the number of terms in the n-th group. - Pablo Hueso Merino, Feb 16 2020
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a(n) ~ c * exp(n), where c = (exp(1)-1) * A300897 = 0.290142809280953235916025... - Vaclav Kotesovec, Apr 05 2020
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