OFFSET
1,1
COMMENTS
The sum Sum_{i=2..k} 1/log(i) is asymptotic to li(k) - li(2), where li is the logarithmic integral. Hence a(n) is approximately the inverse logarithmic integral of n + li(2).
Analogous to A004080 (least k such that the harmonic number H(k) >= n), but with 1/log(k) replacing 1/k.
The first differences are nondecreasing and grow approximately as log(n).
FORMULA
a(n) ~ n*log(n) as n -> infinity. More precisely, a(n) is approximately li^(-1)(n + li(2)), where li^(-1) is the inverse of the logarithmic integral.
EXAMPLE
a(1) = 2 because 1/log(2) = 1.4427... >= 1.
a(4) = 6 because 1/log(2) + 1/log(3) + 1/log(4) + 1/log(5) + 1/log(6) = 4.2537... >= 4, while the sum up to k=5 is 3.6956... < 4.
MATHEMATICA
a[n_]:=Module[{k=1}, While[Sum[1/Log[i], {i, 2, k}]<n, k++]; k]; Array[a, 64] (* James C. McMahon, Mar 29 2026 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Firdous Ahmad Mala, Mar 24 2026
STATUS
approved
