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A248631 Least k such that 3/2 - sum{(h^2)/3^h, h = 1..k} < 1/2^n.

%I A248631 #4 Oct 15 2014 21:02:49
%S A248631 3,4,5,6,7,8,8,9,10,11,11,12,13,14,14,15,16,16,17,18,19,19,20,21,21,
%T A248631 22,23,23,24,25,25,26,27,27,28,29,29,30,31,31,32,33,33,34,35,35,36,37,
%U A248631 37,38,39,39,40,41,41,42,43,43,44,45,45,46,47,47,48,49,49
%N A248631 Least k such that 3/2 - sum{(h^2)/3^h, h = 1..k} < 1/2^n.
%C A248631 This sequence provides insight into the manner of convergence of sum{(h^2)/3^h, h = 1..k} to 3/2.
%H A248631 Clark Kimberling, <a href="/A248631/b248631.txt">Table of n, a(n) for n = 1..1000</a>
%e A248631 Let s(n) = 3/2 - sum{(h^2)/3^h, h = 1..n}.  Approximations follow:
%e A248631 n ... s(n) ...... 1/2^n
%e A248631 1 ... 1.16666 ... 0.500000
%e A248631 2 ... 0.72222 ... 0.250000
%e A248631 3 ... 0.38888 ... 0.125000
%e A248631 4 ... 0.03909 ... 0.062500
%e A248631 5 ... 0.08847 ... 0.031250
%e A248631 6 ... 0.03909 ... 0.015625
%e A248631 7 ... 0.01668 ... 0.007812
%e A248631 a(5) = 7 because s(7) < 1/32 < s(6).
%t A248631 z = 200; p[k_] := p[k] = Sum[(h^2/2^h), {h, 1, k}];
%t A248631 d = N[Table[6 - p[k], {k, 1, z/5}], 12];
%t A248631 f[n_] := f[n] = Select[Range[z], 6 - p[#] < 1/3^n &, 1];
%t A248631 u = Flatten[Table[f[n], {n, 1, z}]]; (* A248629 *)
%t A248631 d = Differences[u];
%t A248631 Flatten[Position[d, 1]];  (* A248630 *)
%Y A248631 Cf. A248632, A248630.
%K A248631 nonn,easy
%O A248631 1,1
%A A248631 _Clark Kimberling_, Oct 11 2014