A247832 Least number k such that u(k) - Pi < 1/5^n, where u is defined using the Borchardt-Pfaff algorithm; see Comments.

1, 1, 2, 3, 4, 5, 7, 8, 9, 10, 11, 12, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 25, 26, 27, 29, 30, 31, 32, 33, 34, 36, 37, 38, 39, 40, 41, 43, 44, 45, 46, 47, 48, 50, 51, 52, 53, 54, 55, 56, 58, 59, 60, 61, 62, 63, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76

Offset

0,3

Comments

The sequences u and v are defined by joint recurrences: u(n) = 2*u(n-1)*v(n-1)/(u(n-1) + v(n-1)) and v(n) = sqrt(u(n)*v(n-1)), with initial values u(0) = 2*sqrt(3) and v(0) = 3. This Borchardt-Pfaff algorithm is used to approximate Pi; see Finch.

Is A247832(n) - A247833(n) in {0,1} for n >= 0?

References

Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 19.

Links

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

Example

Approximations for the first few terms of u - Pi, Pi - v, and 1/5^n are shown here:
n ... u(n) - Pi ... Pi - v(n) ... 1/5^n
0 ... 0.322509 .... 0.1415930 ... 1
1 ... 0.0737977 ... 0.035764 .... 0.2
2 ... 0.0180673 ... 0.008964 .... 0.04
3 ... 0.00449356 .. 0.002242 .... 0.008
4 ... 0.00112195 .. 0.000560 .... 0.0016
5 ... 0.00028039 .. 0.000140 .... 0.00032
6 ... 0.00007009 .. 0.000035 .... 0.000064
7 ... 0.00001752 .. 0.000008 .... 0.0000128
a(6) = 7 because u(7) < 1/5^6 < u(6).

Mathematica

$RecursionLimit = 1000; z = 200; u[0] = N[2*Sqrt[3], 100]; v[0] = 3;
u[n_] := u[n] = 2*u[n - 1]*v[n - 1]/(u[n - 1] + v[n - 1]); v[n_] := v[n] =
Sqrt[u[n]*v[n - 1]]; f[n_] := f[n] = Select[Range[z], u[#] - Pi < 5^(-n) &, 1];
Flatten[Table[f[n], {n, 0, z}]]  (* A247832 *)
g[n_] := g[n] = Select[Range[z], Pi - v[#] < 5^(-n) &, 1]
Flatten[Table[g[n], {n, 0, z}]]  (* A247833 *)

Crossrefs

Cf. A247833.

Sequence in context: A359908 A230872 A346151 * A047368 A020657 A037470

Adjacent sequences: A247829 A247830 A247831 * A247833 A247834 A247835

Keyword

nonn,easy

Author

Clark Kimberling, Sep 26 2014

Status

approved