A248148 Least k such that r - sum{1/Binomial[2h, h], h = 0..k} < 1/3^n, where r = 1/3 + 2*Pi/Sqrt(243).

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, 23, 23, 24, 25, 26, 27, 27, 28, 29, 30, 31, 31, 32, 33, 34, 35, 35, 36, 37, 38, 39, 39, 40, 41, 42, 43, 43, 44, 45, 46, 47, 47, 48, 49, 50, 51, 51, 52, 53, 54, 55

Offset: 1

Author

Clark Kimberling, Oct 02 2014

Keywords

• nonn
• easy

Comments

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.

Examples

			Let s(n) = sum{1/Binomial[2h, h], h = 0..n}.  Approximations are shown here:
n ... r - s(n) ..... 1/3^n
1 ... 0.2364 ....... 0.33333
2 ... 0.0697332 .... 0.11111
3 ... 0.0197332 .... 0.037037
4 ... 0.00544748 ... 0.012345
5 ... 0.00147922 ... 0.004115
a(3) = 3 because r - s(3) < 1/27 < r - s(2).
		

Links

• Clark Kimberling, Table of n, a(n) for n = 1..2000

Crossrefs

Cf. A248149, A248111.

Programs

Mathematica

z = 400; p[k_] := p[k] = Sum[1/Binomial[2 h, h], {h, 1, k}]; r = 1/3 + 2 Pi/Sqrt[243];
N[Table[r - p[n], {n, 1, z/50}]]
f[n_] := f[n] = Select[Range[z], r - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248148 *)
v = Flatten[Position[Differences[u], 0]] (* A248149 *)