%I #41 Jul 15 2023 05:33:51
%S 3,5,8,21,76,389,2679,23969,269777,3717613,61326301,1188642478,
%T 26651213526,682263659097,19720607003199,637490095320530,
%U 22857266906194526,902495758030572213,38993221443197045348,1833273720522384358862
%N a(n) = smallest integer x such that Sum_{k = 2..x} 1/(k*log(log(k))) > n.
%C Because lim_{x->oo} (Sum_{k=2..x} 1 / (k*log(log(k)))) - li(log(x)) = 2.7977647035208... (see A363078) then a(n) = round(w) where w is the solution of the equation li(log(w)) + 2.7977647035208... = n.
%H Pascal Sebah, <a href="/A361089/b361089.txt">Table of n, a(n) for n = 2..35</a>
%F For n >= 3, a(n) = round(w) where w is the solution of the equation li(log(w)) + 2.7977647035208... = n.
%e a(2) = 3 because Sum_{k=2..3} 1/(k*log(log(k))) = 2.18008755... > 2 and Sum_{k=2..2} 1/(k*log(log(k))) = -1.364208386450... < 2.
%e a(7) = 389 because Sum_{k=2..389} 1/(k*log(log(k))) = 7.000345... > 7 and Sum_{k=2..388} 1/(k*log(log(k))) = 6.99890560988... < 7.
%t (*slow procedure*)
%t lim = 2; sum = 0; aa = {}; Do[sum = sum + N[1/(k Log[Log[k]]), 100];
%t If[sum >= lim, AppendTo[aa, k]; Print[{lim, sum, k}];
%t lim = lim + 1], {k, 2, 269777}];aa
%t (*quick procedure *)
%t aa = {3}; cons = 2.79776470352080492766050456553352884330850083202326989577856315;
%t Do[ww = w /. NSolve[LogIntegral[Log[w]] + cons == n, w];
%t AppendTo[aa, Round[ww][[1]]], {n, 3, 21}]; aa
%Y Cf. A077761, A086242, A137245, A138312, A221711, A303493, A319231, A319232, A354887, A354917, A354952, A361972, A362533, A363368, A363078.
%K nonn
%O 2,1
%A _Artur Jasinski_, Jun 11 2023
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