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A247833
Least number k such that Pi - v(k) < 1/5^n, where v is defined using the Borchardt-Pfaff algorithm; see Comments.
2
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
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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Sep 26 2014
STATUS
approved