A247833 Least number k such that Pi - v(k) < 1/5^n, where v is defined using the Borchardt-Pfaff algorithm; see Comments.

1, 1, 1, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 17, 18, 19, 20, 21, 22, 23, 25, 26, 27, 28, 29, 30, 32, 33, 34, 35, 36, 37, 39, 40, 41, 42, 43, 44, 46, 47, 48, 49, 50, 51, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 68, 69, 70, 71, 72, 73, 75, 76

Offset

0,4

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(3) = 3 because v(3) < 1/5^3 < v(2).

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. A247832.

Sequence in context: A138928 A116587 A057903 * A047310 A184530 A304804

Adjacent sequences: A247830 A247831 A247832 * A247834 A247835 A247836

Keyword

nonn,easy

Author

Clark Kimberling, Sep 26 2014

Status

approved