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