A248607 Least k such that Pi/2 - sum{2^h/((2h+1)*C(2h,h)), h = 1..k} < 1/3^n.

1, 2, 4, 5, 7, 8, 10, 11, 13, 14, 16, 17, 19, 20, 22, 23, 25, 26, 28, 30, 31, 33, 34, 36, 37, 39, 40, 42, 44, 45, 47, 48, 50, 51, 53, 54, 56, 58, 59, 61, 62, 64, 65, 67, 69, 70, 72, 73, 75, 76, 78, 80, 81, 83, 84, 86, 87, 89, 91, 92, 94, 95, 97, 98, 100, 102

Offset

1,2

Comments

This sequence provides insight into the manner of convergence of sum{2^h/((2h+1)*C(2h,h)), h = 1..k} to Pi/2. Since a(n+1) - a(n) is in {1,2} for n >= 1, the sequences A248608 and A248609 partition the positive integers.

References

Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 20.

Links

Clark Kimberling, Table of n, a(n) for n = 1..1000

Example

Let s(n) = Pi/2 - sum{2^h/((2h+1)*C(2h,h)), h = 1..n}.  Approximations follow:
n ... s(n) ...... 1/3^n
1 ... 0.23746 ... 0.333333
2 ... 0.10413 ... 0.111111
3 ... 0.04698 ... 0.037037
4 ... 0.02159 ... 0.012345
5 ... 0.01004 ... 0.004115
6 ... 0.00471 ... 0.001371
7 ... 0.00223 ... 0.000472
a(5) = 7 because s(7) < 1/3^5 < s(6).

Mathematica

z = 300; p[k_] := p[k] = Sum[2^h/((2 h + 1) Binomial[2 h, h]), {h, 0, k}]
d = N[Table[Pi/2 - p[k], {k, 1, z/5}], 12]
f[n_] := f[n] = Select[Range[z], Pi/2 - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248607 *)
d = Differences[u]
v = Flatten[Position[d, 1]] (* A248608 *)
w = Flatten[Position[d, 2]] (* A248609 *)

Crossrefs

Cf. A248608, A248609, A248610.

Sequence in context: A224999 A274384 A195175 * A191266 A003253 A119905

Adjacent sequences: A248604 A248605 A248606 * A248608 A248609 A248610

Keyword

nonn,easy

Author

Clark Kimberling, Oct 10 2014

Status

approved