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A248607 Least k such that Pi/2 - sum{2^h/((2h+1)*C(2h,h)), h = 1..k} < 1/3^n.

%I A248607 #5 Oct 15 2014 21:01:09
%S A248607 1,2,4,5,7,8,10,11,13,14,16,17,19,20,22,23,25,26,28,30,31,33,34,36,37,
%T A248607 39,40,42,44,45,47,48,50,51,53,54,56,58,59,61,62,64,65,67,69,70,72,73,
%U A248607 75,76,78,80,81,83,84,86,87,89,91,92,94,95,97,98,100,102
%N A248607 Least k such that Pi/2 - sum{2^h/((2h+1)*C(2h,h)), h = 1..k} < 1/3^n.
%C A248607 This sequence provides insight into the manner of convergence of sum{2^h/((2h+1)*C(2h,h)), h = 1..k} to Pi/2.  Since a(n+1) - a(n) is in {1,2} for n >= 1, the sequences A248608 and A248609 partition the positive integers.
%D A248607 Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 20.
%H A248607 Clark Kimberling, <a href="/A248607/b248607.txt">Table of n, a(n) for n = 1..1000</a>
%e A248607 Let s(n) = Pi/2 - sum{2^h/((2h+1)*C(2h,h)), h = 1..n}.  Approximations follow:
%e A248607 n ... s(n) ...... 1/3^n
%e A248607 1 ... 0.23746 ... 0.333333
%e A248607 2 ... 0.10413 ... 0.111111
%e A248607 3 ... 0.04698 ... 0.037037
%e A248607 4 ... 0.02159 ... 0.012345
%e A248607 5 ... 0.01004 ... 0.004115
%e A248607 6 ... 0.00471 ... 0.001371
%e A248607 7 ... 0.00223 ... 0.000472
%e A248607 a(5) = 7 because s(7) < 1/3^5 < s(6).
%t A248607 z = 300; p[k_] := p[k] = Sum[2^h/((2 h + 1) Binomial[2 h, h]), {h, 0, k}]
%t A248607 d = N[Table[Pi/2 - p[k], {k, 1, z/5}], 12]
%t A248607 f[n_] := f[n] = Select[Range[z], Pi/2 - p[#] < 1/3^n &, 1]
%t A248607 u = Flatten[Table[f[n], {n, 1, z}]]  (* A248607 *)
%t A248607 d = Differences[u]
%t A248607 v = Flatten[Position[d, 1]] (* A248608 *)
%t A248607 w = Flatten[Position[d, 2]] (* A248609 *)
%Y A248607 Cf. A248608, A248609, A248610.
%K A248607 nonn,easy
%O A248607 1,2
%A A248607 _Clark Kimberling_, Oct 10 2014