A248182 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}.

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, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69

Offset: 0

Author

Clark Kimberling, Oct 04 2014

Keywords

• nonn
• easy

Comments

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.

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 ... 2.2092 ..... 1
1 ... 1.2092 ..... 0.5
2 ... 0.7092 ..... 0.25
3 ... 0.375866 ... 0.125
4 ... 0.2092 ..... 0.0625
5 ... 0.1092 ..... 0.0635
6 ... 0.05919 .... 0.0156
7 ... 0.03063 .... 0.007812
8 ... 0.01634 .... 0.003906
9 ... 0.00840 .... 0.001953
a(6) = 9 because r - s(9) < 1/64 < r - s(8).
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A248181, A248148.

Programs

Mathematica

z = 300; p[k_] := p[k] = Sum[1/Binomial[h, Floor[h/2]], {h, 0, k}];
r = N[Sum[1/Binomial[h, Floor[h/2]], {h, 0, 1000}], 20]  (* A248181 *)
N[Table[r - p[n], {n, 0, z/10}]]
f[n_] := f[n] = Select[Range[z], r - p[#] < 1/2^n &, 1]
Flatten[Table[f[n], {n, 0, z}]]  (* A248182 *)