A248610 Least k such that (Pi^2)/18 - sum{1/(h^2*C(2h,h)), h = 1..k} < 1/3^n.

1, 1, 2, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 9, 9, 10, 11, 12, 12, 13, 14, 15, 15, 16, 17, 18, 18, 19, 20, 21, 21, 22, 23, 24, 24, 25, 26, 27, 27, 28, 29, 30, 30, 31, 32, 33, 34, 34, 35, 36, 37, 37, 38, 39, 40, 40, 41, 42, 43, 44, 44, 45, 46, 47, 47, 48, 49, 50

Offset

1,3

Comments

This sequence provides insight into the manner of convergence of sum{1/(h^2*C(2h,h)), h = 1..k} to (Pi^2)/18. Since a(n+1) - a(n) is in {0,1} for n >= 1, the sequences A248611 and A248612 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..2000

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 ... 0.0483114 ... 0.333333
2 ... 0.0066446 ... 0.111111
3 ... 0.0010891 ... 0.037037
4 ... 0.0001962 ... 0.012345
5 ... 0.00003754 .. 0.004115
a(5) = 3 because s(3) < 1/3^5 < s(2).

Mathematica

z = 300; p[k_] := p[k] = Sum[1/((h^2)*Binomial[2 h, h]), {h, 1, k}]
d = N[Table[Pi^2/18 - p[k], {k, 1, z/5}], 12]
f[n_] := f[n] = Select[Range[z], Pi^2/18 - p[#] < 1/3^n &, 1]
u = Flatten[Table[f[n], {n, 1, z}]]  (* A248610 *)
d = Differences[u]
v = Flatten[Position[d, 0]]  (* A248611 *)
w = Flatten[Position[d, 1]]  (* A248612 *)

Crossrefs

Cf. A248611, A248612, A248607.

Sequence in context: A319433 A253671 A209081 * A168581 A024699 A083479

Adjacent sequences: A248607 A248608 A248609 * A248611 A248612 A248613

Keyword

nonn,easy

Author

Clark Kimberling, Oct 10 2014

Status

approved