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%I #6 Jan 03 2024 07:44:08
%S 2,3,4,5,6,7,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,
%T 28,29,30,31,32,33,34,35,36,37,39,40,41,42,43,44,45,46,47,48,49,50,51,
%U 52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68,69
%N Least k such that r - sum{1/C(h,[h/2]), h = 0..k} < 1/2^n, where r = sum{1/C(h,[h/2]), h = 0..infinity}.
%C This sequence gives a measure of the convergence rate of sum{1/C(h,[h/2]), h = 0..k}. It appears that a(n+1) - a(n) is in {2,3} for n >= 0.
%H Clark Kimberling, <a href="/A248182/b248182.txt">Table of n, a(n) for n = 0..1000</a>
%e Let s(n) = sum{1/C(2h+1,h), h = 0..n}. Approximations are shown here:
%e n ... r - s(n) ... 1/2^n
%e 0 ... 2.2092 ..... 1
%e 1 ... 1.2092 ..... 0.5
%e 2 ... 0.7092 ..... 0.25
%e 3 ... 0.375866 ... 0.125
%e 4 ... 0.2092 ..... 0.0625
%e 5 ... 0.1092 ..... 0.0635
%e 6 ... 0.05919 .... 0.0156
%e 7 ... 0.03063 .... 0.007812
%e 8 ... 0.01634 .... 0.003906
%e 9 ... 0.00840 .... 0.001953
%e a(6) = 9 because r - s(9) < 1/64 < r - s(8).
%t z = 300; p[k_] := p[k] = Sum[1/Binomial[h, Floor[h/2]], {h, 0, k}];
%t r = N[Sum[1/Binomial[h, Floor[h/2]], {h, 0, 1000}], 20] (* A248181 *)
%t N[Table[r - p[n], {n, 0, z/10}]]
%t f[n_] := f[n] = Select[Range[z], r - p[#] < 1/2^n &, 1]
%t Flatten[Table[f[n], {n, 0, z}]] (* A248182 *)
%Y Cf. A248181, A248148.
%K nonn,easy
%O 0,1
%A _Clark Kimberling_, Oct 04 2014
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