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A247983 Least number k such that log(2) - sum{1/(h*2^h), h=1..k} < 1/2^n.

%I A247983 #11 Sep 30 2014 15:56:42
%S A247983 1,1,2,3,3,4,5,6,7,7,8,9,10,11,12,13,14,15,15,16,17,18,19,20,21,22,23,
%T A247983 24,25,26,27,28,29,30,31,31,32,33,34,35,36,37,38,39,40,41,42,43,44,45,
%U A247983 46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63
%N A247983 Least number k such that log(2) - sum{1/(h*2^h), h=1..k} < 1/2^n.
%C A247983 It appears that a(n+1) - a(n) if and only if n is in A083706(n), for n >= 1.
%D A247983 Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 15.
%H A247983 Clark Kimberling, <a href="/A247983/b247983.txt">Table of n, a(n) for n = 1..1000</a>
%e A247983 Let w(n) = log(2) - sum{1/(h*2^h), h=1..n}.  Approximations are shown here:
%e A247983 n .... w(n)  ...... 1/2^n
%e A247983 1 ... 0.193147 .... 0.5
%e A247983 2 ... 0.0681472 ... 0.25
%e A247983 3 ... 0.0264805 ... 0.125
%e A247983 4 ... 0.0108555 ... 0.0625
%e A247983 5 ... 0.0046055 ... 0. 03125
%e A247983 a(4) = 3 because w(3) < 1/2^4 < w(2).
%t A247983 z = 200; s[k_] := s[k] = Sum[1/(h*2^h), {h, 1, k}];
%t A247983 N[Table[Log[2] - s[n], {n, 1, z/8}]]
%t A247983 f[n_] := f[n] = Select[Range[z], Log[2] - s[#] < 1/2^n &, 1];
%t A247983 u = Flatten[Table[f[n], {n, 1, z}]]   (* A247983 *)
%Y A247983 Cf. A083706, A247968.
%K A247983 nonn,easy
%O A247983 1,3
%A A247983 _Clark Kimberling_, Sep 28 2014