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A248559 Least k such that log(2) - sum{1/(h*2^h), h = 1..k} < 1/3^n.

%I A248559 #9 Oct 15 2014 20:56:30
%S A248559 1,2,3,4,6,7,8,10,11,13,14,15,17,18,20,21,23,24,26,27,29,30,32,33,35,
%T A248559 36,38,40,41,43,44,46,47,49,50,52,53,55,56,58,60,61,63,64,66,67,69,70,
%U A248559 72,74,75,77,78,80,81,83,84,86,88,89,91,92,94,95,97,98
%N A248559 Least k such that log(2) - sum{1/(h*2^h), h = 1..k} < 1/3^n.
%C A248559 This sequence provides insight into the manner of convergence of  sum{1/(h*2^h), h = 1..k} to log 2.  Since a(n+1) - a(n) is in {1,2} for n >= 1, the sequences A248560 and A248561 partition the positive integers.
%D A248559 Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 15.
%H A248559 Clark Kimberling, <a href="/A248559/b248559.txt">Table of n, a(n) for n = 1..1000</a>
%e A248559 Let s(n) = log(2) - sum{1/(h*2^h), h = 1..n}.  Approximations follow:
%e A248559 n ... s(n) ........ 1/3^n
%e A248559 1 ... 0.193147 .... 0.33333
%e A248559 2 ... 0.0681472 ... 0.11111
%e A248559 3 ... 0.0264805 ... 0.037037
%e A248559 4 ... 0.0108555 ... 0.0123457
%e A248559 5 ... 0.0046066 ... 0.004115
%e A248559 6 ... 0.0020013 ... 0.00137174
%e A248559 a(5) = 6 because s(6) < 1/3^5 < s(5).
%t A248559 z = 200; p[k_] := p[k] = Sum[1/(h*2^h), {h, 1, k}]
%t A248559 N[Table[Log[2] - p[n], {n, 1, z/5}]]
%t A248559 f[n_] := f[n] = Select[Range[z], Log[2] - p[#] < 1/3^n &, 1]
%t A248559 u = Flatten[Table[f[n], {n, 1, z}]]    (* A248559 *)
%t A248559 Flatten[Position[Differences[u], 1]]   (* A248560 *)
%t A248559 Flatten[Position[Differences[u], 2]]   (* A248561 *)
%Y A248559 Cf. A002162 (log(2)), A248560, A248561.
%K A248559 nonn,easy
%O A248559 1,2
%A A248559 _Clark Kimberling_, Oct 09 2014