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A248610 Least k such that (Pi^2)/18 - sum{1/(h^2*C(2h,h)), h = 1..k} < 1/3^n.

%I A248610 #6 Dec 25 2023 17:15:00
%S A248610 1,1,2,2,3,3,4,5,5,6,7,7,8,9,9,10,11,12,12,13,14,15,15,16,17,18,18,19,
%T A248610 20,21,21,22,23,24,24,25,26,27,27,28,29,30,30,31,32,33,34,34,35,36,37,
%U A248610 37,38,39,40,40,41,42,43,44,44,45,46,47,47,48,49,50
%N A248610 Least k such that (Pi^2)/18 - sum{1/(h^2*C(2h,h)), h = 1..k} < 1/3^n.
%C A248610 This sequence provides insight into the manner of convergence of sum{1/(h^2*C(2h,h)), h = 1..k} to (Pi^2)/18.  Since a(n+1) - a(n) is in {0,1} for n >= 1, the sequences A248611 and A248612 partition the positive integers.
%D A248610 Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 20.
%H A248610 Clark Kimberling, <a href="/A248610/b248610.txt">Table of n, a(n) for n = 1..2000</a>
%e A248610 Let s(n) = Pi/2 - sum{2^h/((2h+1)*C(2h,h)), h = 1..n}.  Approximations follow:
%e A248610 n ... s(n) ........ 1/3^n
%e A248610 1 ... 0.0483114 ... 0.333333
%e A248610 2 ... 0.0066446 ... 0.111111
%e A248610 3 ... 0.0010891 ... 0.037037
%e A248610 4 ... 0.0001962 ... 0.012345
%e A248610 5 ... 0.00003754 .. 0.004115
%e A248610 a(5) = 3 because s(3) < 1/3^5 < s(2).
%t A248610 z = 300; p[k_] := p[k] = Sum[1/((h^2)*Binomial[2 h, h]), {h, 1, k}]
%t A248610 d = N[Table[Pi^2/18 - p[k], {k, 1, z/5}], 12]
%t A248610 f[n_] := f[n] = Select[Range[z], Pi^2/18 - p[#] < 1/3^n &, 1]
%t A248610 u = Flatten[Table[f[n], {n, 1, z}]]  (* A248610 *)
%t A248610 d = Differences[u]
%t A248610 v = Flatten[Position[d, 0]]  (* A248611 *)
%t A248610 w = Flatten[Position[d, 1]]  (* A248612 *)
%Y A248610 Cf. A248611, A248612, A248607.
%K A248610 nonn,easy
%O A248610 1,3
%A A248610 _Clark Kimberling_, Oct 10 2014