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A248180 Least k such that r - sum{1/C(2h+1,h), h = 0..k} < 1/2^n, where r = (2/27)*(9 + 2*sqrt(3)*Pi).

%I A248180 #4 Oct 06 2014 23:00:06
%S A248180 1,1,1,2,2,3,3,4,4,5,5,6,6,7,7,8,8,9,10,10,11,11,12,12,13,13,14,14,15,
%T A248180 15,16,16,17,17,18,18,19,19,20,20,21,21,22,22,23,23,24,24,25,25,26,26,
%U A248180 27,27,28,28,29,29,30,30,31,31,32,32,33,33,34,34,35
%N A248180 Least k such that r - sum{1/C(2h+1,h), h = 0..k} < 1/2^n, where r = (2/27)*(9 + 2*sqrt(3)*Pi).
%C A248180 This sequence gives a measure of the convergence rate of sum{1/C(2h+1,h), h = 0..k}.  Since a(n+1) - a(n) is in {0,1} for n >= 0, the sequences A248195 and A248196 partition the positive integers.
%H A248180 Clark Kimberling, <a href="/A248180/b248180.txt">Table of n, a(n) for n = 0..3000</a>
%e A248180 Let s(n) = sum{1/C(2h+1,h), h = 0..n}.  Approximations are shown here:
%e A248180 n ... r - s(n) ..... 1/2^n
%e A248180 0 ... 0.47289 ...... 1
%e A248180 1 ... 0.139466 ..... 0.5
%e A248180 2 ... 0.0394664 .... 0.25
%e A248180 3 ... 0.010895 ..... 0.125
%e A248180 4 ... 0.00295845 ... 0.0625
%e A248180 a(3) = 2 because r - s(2) < 1/8 < r - s(1).
%t A248180 $MaxExtraPrecision = Infinity;
%t A248180 z = 300; p[k_] := p[k] = Sum[1/Binomial[2 h + 1, h], {h, 0, k}] ;
%t A248180 r = Sum[1/Binomial[2 h + 1, h], {h, 0, Infinity}]  (* A248179 *)
%t A248180 r = 2/27 (9 + 2 Sqrt[3] \[Pi]); N[r, 20]
%t A248180 N[Table[r - p[n], {n, 0, z/10}]]
%t A248180 f[n_] := f[n] = Select[Range[z], r - p[#] < 1/2^n &, 1]
%t A248180 u = Flatten[Table[f[n], {n, 0, z}]]  (* A248180 *)
%t A248180 Flatten[Position[Differences[u], 0]] (* A248195 *)
%t A248180 Flatten[Position[Differences[u], 1]] (* A248196 *)
%Y A248180 Cf. A248179, A248195, A248196.
%K A248180 nonn,easy
%O A248180 0,4
%A A248180 _Clark Kimberling_, Oct 03 2014