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A248629 Least k such that 6 - sum{(h^2)/2^h, h = 1..k} < 1/3^n.

%I A248629 #5 Oct 15 2014 21:02:25
%S A248629 9,11,13,15,17,19,21,22,24,26,28,29,31,33,35,36,38,40,41,43,45,47,48,
%T A248629 50,52,53,55,57,58,60,62,63,65,67,68,70,72,73,75,76,78,80,81,83,85,86,
%U A248629 88,90,91,93,95,96,98,99,101,103,104,106,108,109,111,112,114
%N A248629 Least k such that 6 - sum{(h^2)/2^h, h = 1..k} < 1/3^n.
%C A248629 This sequence provides insight into the manner of convergence of sum{(h^2)/2^h, h = 1..k} to 6.
%H A248629 Clark Kimberling, <a href="/A248629/b248629.txt">Table of n, a(n) for n = 1..1000</a>
%e A248629 Let s(n) = 6 - sum{(h^2)/2^h, h = 1..n}.  Approximations follow:
%e A248629 n ... s(n) ........ 1/3^n
%e A248629 1 ... 5.50000 ... 0.333333
%e A248629 2 ... 4.50000 ... 0.111111
%e A248629 3 ... 3.37500 ... 0.037037
%e A248629 4 ... 2.37500 ... 0.012345
%e A248629 5 ... 1.59375 ... 0.004115
%e A248629 6 ... 1.03125 ... 0.001371
%e A248629 7 ... 0.64843 ... 0.000457
%e A248629 8 ... 0.39843 ... 0.000152
%e A248629 9 ... 0.24023 ... 0.000050
%e A248629 10 .. 0.14257 ... 0.000018
%e A248629 11 .. 0.08349 ... 0.000006
%e A248629 a(2) = 11 because s(11) < 1/9 < s(10).
%t A248629 z = 300; p[k_] := p[k] = Sum[(h^2/2^h), {h, 1, k}]
%t A248629 d = N[Table[6 - p[k], {k, 1, z/5}], 12]
%t A248629 f[n_] := f[n] = Select[Range[z], 6 - p[#] < 1/3^n &, 1]
%t A248629 u = Flatten[Table[f[n], {n, 1, z}]]  (* A248629 *)
%t A248629 d = Differences[u]
%t A248629 v = Flatten[Position[d, 1]]  (* A248630 *)
%Y A248629 Cf. A248630, A248631.
%K A248629 nonn,easy
%O A248629 1,1
%A A248629 _Clark Kimberling_, Oct 10 2014