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A248148 Least k such that r - sum{1/Binomial[2h, h], h = 0..k} < 1/3^n, where r = 1/3 + 2*Pi/Sqrt(243).

%I A248148 #5 Oct 02 2014 22:36:42
%S A248148 1,2,3,4,5,6,6,7,8,9,10,11,11,12,13,14,15,15,16,17,18,19,19,20,21,22,
%T A248148 23,23,24,25,26,27,27,28,29,30,31,31,32,33,34,35,35,36,37,38,39,39,40,
%U A248148 41,42,43,43,44,45,46,47,47,48,49,50,51,51,52,53,54,55
%N A248148 Least k such that r - sum{1/Binomial[2h, h], h = 0..k} < 1/3^n, where r = 1/3 + 2*Pi/Sqrt(243).
%C A248148 It is well known that sum{1/Binomial[2h, h], h = 0..infinity} = r (approximately 0.7363998); this sequence gives a measure of the convergence rate.  It appears that a(n+1) - a(n) is in {0,1} for n >= 1.
%H A248148 Clark Kimberling, <a href="/A248148/b248148.txt">Table of n, a(n) for n = 1..2000</a>
%e A248148 Let s(n) = sum{1/Binomial[2h, h], h = 0..n}.  Approximations are shown here:
%e A248148 n ... r - s(n) ..... 1/3^n
%e A248148 1 ... 0.2364 ....... 0.33333
%e A248148 2 ... 0.0697332 .... 0.11111
%e A248148 3 ... 0.0197332 .... 0.037037
%e A248148 4 ... 0.00544748 ... 0.012345
%e A248148 5 ... 0.00147922 ... 0.004115
%e A248148 a(3) = 3 because r - s(3) < 1/27 < r - s(2).
%t A248148 z = 400; p[k_] := p[k] = Sum[1/Binomial[2 h, h], {h, 1, k}]; r = 1/3 + 2 Pi/Sqrt[243];
%t A248148 N[Table[r - p[n], {n, 1, z/50}]]
%t A248148 f[n_] := f[n] = Select[Range[z], r - p[#] < 1/3^n &, 1]
%t A248148 u = Flatten[Table[f[n], {n, 1, z}]]  (* A248148 *)
%t A248148 v = Flatten[Position[Differences[u], 0]] (* A248149 *)
%Y A248148 Cf. A248149, A248111.
%K A248148 nonn,easy
%O A248148 1,2
%A A248148 _Clark Kimberling_, Oct 02 2014