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%I #9 Oct 15 2014 20:58:14
%S 1,2,3,5,6,7,9,10,11,13,14,16,17,19,20,21,23,24,26,27,29,30,32,33,35,
%T 36,38,39,41,42,43,45,46,48,49,51,52,54,55,57,58,60,61,63,64,66,67,69,
%U 70,72,73,75,76,78,79,81,82,84,85,86,88,89,91,92,94,95
%N Least k such that log(4/3) - sum{1/(h*4^h), h = 1..k} < 1/8^n.
%C This sequence provides insight into the manner of convergence of sum{1/(h*4^h), h = 1..k} to log(4/3). Since a(n+1) - a(n) is in {1,2} for n >= 1, the sequences A248566 and A248567 partition the positive integers.
%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 15.
%H Clark Kimberling, <a href="/A248565/b248565.txt">Table of n, a(n) for n = 1..1000</a>
%e Let s(n) = log(4/3) - sum{1/(h*4^h), h = 1..n}. Approximations follow:
%e n ... s(n) ........ 1/8^n
%e 1 ... 0.037682 ... 0.125
%e 2 ... 0.006432 ... 0.015625
%e 3 ... 0.001223 ... 0.001953
%e 4 ... 0.000247 ... 0.000244
%e 5 ... 0.000051 ... 0.000030
%e a(4) = 5 because s(5) < 1/8^4 < s(4).
%t z = 2500; p[k_] := p[k] = Sum[1/(h*4^h), {h, 1, k}];
%t N[Table[p[k], {k, 1, z/5}], 12];
%t N[Table[Log[4/3] - p[n], {n, 1, z/5}]];
%t f[n_] := f[n] = Select[Range[z], Log[4/3] - p[#] < 1/8^n &, 1];
%t u = Flatten[Table[f[n], {n, 1, z}]] ; (* A248565 *)
%t Flatten[Position[Differences[u], 1]]; (* A248566 *)
%t Flatten[Position[Differences[u], 2]]; (* A248567 *)
%Y Cf. A083679 (log(4/3)), A248566, A248567, A248559, A248565.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Oct 09 2014
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