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%I #13 Oct 15 2014 20:57:25
%S 1,2,3,5,6,8,9,11,12,14,15,17,18,20,22,23,25,26,28,29,31,33,34,36,37,
%T 39,41,42,44,45,47,49,50,52,53,55,57,58,60,61,63,65,66,68,69,71,73,74,
%U 76,77,79,81,82,84,86,87,89,90,92,94,95,97,98,100,102,103
%N Least k such that log(3/2) - sum{1/(h*3^h), h = 1..k} < 1/6^n.
%C This sequence provides insight into the manner of convergence of sum{1/(h*3^h), h = 1..k} to log(3/2). Since a(n+1) - a(n) is in {1,2} for n >= 1, the sequences A248563 and A248564 partition the positive integers.
%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 15.
%H Clark Kimberling, <a href="/A248562/b248562.txt">Table of n, a(n) for n = 1..1000</a>
%e Let s(n) = log(3/2) - sum{1/(h*3^h), h = 1..n}. Approximations follow:
%e n ... s(n) ........ 1/6^n
%e 1 ... 0.0721318 ... 0.166667
%e 2 ... 0.0165762 ... 0.0277777
%e 3 ... 0.0042305 ... 0.0046296
%e 4 ... 0.0011441 ... 0.0007716
%e 5 ... 0.0003210 ... 0.0001286
%e a(4) = 5 because s(5) < 1/6^4 < s(4).
%t z = 300; p[k_] := p[k] = Sum[1/(h*3^h), {h, 1, k}];
%t N[Table[Log[3/2] - p[n], {n, 1, z/5}]]
%t f[n_] := f[n] = Select[Range[z], Log[3/2] - p[#] < 1/6^n &, 1];
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248562 *)
%t Flatten[Position[Differences[u], 1]] (* A248563 *)
%t Flatten[Position[Differences[u], 2]] (* A248564 *)
%Y Cf. A016578 (log(3/2)), A248563, A248564, A248559, A248565.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Oct 09 2014
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