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A096987 Numerator of Sum_{k=1..n} 1/H(k), where H(k) = Sum_{j=1..k} 1/j is the k-th harmonic number.

%I A096987 #101 Jun 12 2023 16:58:12
%S A096987 0,1,5,73,2221,353777,19595573,239046803,198972350083,
%T A096987 1535302297058707,100536661265514127,8974880059175708288297,
%U A096987 818810519369821323965929237,990666575600755815615137883006341,1220749860499992165560973207703210595953
%N A096987 Numerator of Sum_{k=1..n} 1/H(k), where H(k) = Sum_{j=1..k} 1/j is the k-th harmonic number.
%F A096987 From _Thomas Ordowski_, Mar 21 2023: (Start)
%F A096987 Sum_{k=1..n} 1/H(k) ~ Sum_{k=2..n} 1/log(k) ~ Integral_{2..n} dx/log(x) = Li(n).
%F A096987 Sum_{k=1..n} 1/H(k) = Sum_{k=1..n} 1/(log(k + 1/2) + gamma) - C + o(1), where gamma = A001620 = 0.577... is Euler's constant and the constant C = 0.0229825...
%F A096987 Sum_{k=1..n} 1/H(k) = exp(-gamma)*(Ei(log(n) + gamma) - 1) + o(1), where Ei(x) is the exponential integral function of real x, and we have Ei(log(x)) = li(x).
%F A096987 Note that a(n)/A124432(n) ~ pi(n) = A000720(n), see my first formula.
%F A096987 Sum_{k=1..n} 1/H(k) = n/(H(n) - 1 + ...) = n/(log(n) + gamma - 1 + O(1/log(n))).
%F A096987 Theorem: lim_{n->oo} (H(n) - n / Sum_{k=1..n} 1/H(k)) = 1, see my third formula.
%F A096987 Proof: since Integral dx / (log(x) + gamma) = exp(-gamma)*Ei(log(x) + gamma) + c, so we get lim_{n->oo} (log(n) + gamma - n*exp(gamma) / Ei(log(n) + gamma)) = 1, qed. (End)
%e A096987 1/1 + 1/(1 + 1/2) + 1/(1 + 1/2 + 1/3) = 73/33, so a(3) = 73.
%t A096987 f[n_] := Numerator[ Sum[ 1/HarmonicNumber[j], {j, 1, n}]]; Table[ f[n], {n, 0, 14}] (* _Robert G. Wilson v_, Aug 21 2004 *)
%o A096987 (PARI) m=13;for(n=0,m,print1(numerator(sum(k=1,n,1/sum(j=1,k,1/j))),",")) \\ _Klaus Brockhaus_, Aug 21 2004
%Y A096987 Cf. A124432 (denominators), A000720, A001008, A002805, A001620, A080130.
%K A096987 easy,frac,nonn
%O A096987 0,3
%A A096987 _Leroy Quet_, Aug 19 2004
%E A096987 More terms from _Klaus Brockhaus_ and _Robert G. Wilson v_, Aug 21 2004