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).

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, 15, 16, 16, 17, 17, 18, 18, 19, 19, 20, 20, 21, 21, 22, 22, 23, 23, 24, 24, 25, 25, 26, 26, 27, 27, 28, 28, 29, 29, 30, 30, 31, 31, 32, 32, 33, 33, 34, 34, 35

Offset: 0

Author

Clark Kimberling, Oct 03 2014

Keywords

• nonn
• easy

Comments

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.

Examples

			Let s(n) = sum{1/C(2h+1,h), h = 0..n}.  Approximations are shown here:
n ... r - s(n) ..... 1/2^n
0 ... 0.47289 ...... 1
1 ... 0.139466 ..... 0.5
2 ... 0.0394664 .... 0.25
3 ... 0.010895 ..... 0.125
4 ... 0.00295845 ... 0.0625
a(3) = 2 because r - s(2) < 1/8 < r - s(1).
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..3000

Crossrefs

Cf. A248179, A248195, A248196.

Programs

Mathematica

$MaxExtraPrecision = Infinity;
z = 300; p[k_] := p[k] = Sum[1/Binomial[2 h + 1, h], {h, 0, k}] ;
r = Sum[1/Binomial[2 h + 1, h], {h, 0, Infinity}]  (* A248179 *)
r = 2/27 (9 + 2 Sqrt[3] \[Pi]); N[r, 20]
N[Table[r - p[n], {n, 0, z/10}]]
f[n_] := f[n] = Select[Range[z], r - p[#] < 1/2^n &, 1]
u = Flatten[Table[f[n], {n, 0, z}]]  (* A248180 *)
Flatten[Position[Differences[u], 0]] (* A248195 *)
Flatten[Position[Differences[u], 1]] (* A248196 *)