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Least k such that r - sum{1/Binomial[2h, h], h = 0..k} < 1/3^n, where r = 1/3 + 2*Pi/Sqrt(243).
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%I #5 Oct 02 2014 22:36:42

%S 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 23,23,24,25,26,27,27,28,29,30,31,31,32,33,34,35,35,36,37,38,39,39,40,

%U 41,42,43,43,44,45,46,47,47,48,49,50,51,51,52,53,54,55

%N 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 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 Clark Kimberling, <a href="/A248148/b248148.txt">Table of n, a(n) for n = 1..2000</a>

%e Let s(n) = sum{1/Binomial[2h, h], h = 0..n}. Approximations are shown here:

%e n ... r - s(n) ..... 1/3^n

%e 1 ... 0.2364 ....... 0.33333

%e 2 ... 0.0697332 .... 0.11111

%e 3 ... 0.0197332 .... 0.037037

%e 4 ... 0.00544748 ... 0.012345

%e 5 ... 0.00147922 ... 0.004115

%e a(3) = 3 because r - s(3) < 1/27 < r - s(2).

%t z = 400; p[k_] := p[k] = Sum[1/Binomial[2 h, h], {h, 1, k}]; r = 1/3 + 2 Pi/Sqrt[243];

%t N[Table[r - p[n], {n, 1, z/50}]]

%t f[n_] := f[n] = Select[Range[z], r - p[#] < 1/3^n &, 1]

%t u = Flatten[Table[f[n], {n, 1, z}]] (* A248148 *)

%t v = Flatten[Position[Differences[u], 0]] (* A248149 *)

%Y Cf. A248149, A248111.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Oct 02 2014