A247973 Least k such that Pi - (4*k+2)/v(2*k+2)^2 < 1/n, where the sequence v is defined in Comments.

1, 1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 9, 10, 11, 12, 12, 13, 14, 15, 16, 16, 17, 18, 19, 20, 20, 21, 22, 23, 23, 24, 25, 26, 27, 27, 28, 29, 30, 31, 31, 32, 33, 34, 34, 35, 36, 37, 38, 38, 39, 40, 41, 41, 42, 43, 44, 45, 45, 46, 47, 48, 49, 49, 50, 51, 52, 52, 53

Offset

1,3

Comments

The sequence v is defined as follows: v(1) = 0, v(2) = 1, v(n) = v(n-1)/(n-2) + v(n-2). It appears that a(n+1) - a(n) is in {0,1} for n >= 2.

References

Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 21.

Links

Table of n, a(n) for n=1..68.

Example

Approximations for the first few terms w(n) = Pi - (4*n+2)/v(2*n+2)^2 and 1/n:
n ... Pi-(4*n+2)/v(2*n+2)^2 ... 1/n
1 ... 0.474926 ................ 1
2 ... 0.297148 ...............  0.5
3 ... 0.215878 ................ 0.333333
4 ... 0.169438 ................ 0.25
5 ... 0.139417 ................ 0.2
6 ... 0.118422 ................ 0.166666
a(3) = 2 because w(2) < 1/3 < w(1).

Mathematica

$RecursionLimit = Infinity; z = 400; v[1] = 0; v[2] = 1;
v[n_] := v[n] = v[n - 1]/(n - 2) + v[n - 2];
TableForm[Table[{n, N[Pi - (4 n + 2)/(v[2 (n + 1)]^2)], N[1/n]}, {n, 1, 10}]]
g[n_] := g[n] = Select[Range[z], Pi - (4 # + 2)/(v[2 (# + 1)]^2) < 1/n &, 1];
u = Flatten[Table[g[n], {n, 1, z}]]  (* A247973 *)
d = Differences[u]
Flatten[Position[d, 0]] (* A247974 *)

Crossrefs

Cf. A247971, A247972, A247974.

Sequence in context: A331266 A172103 A123731 * A352241 A195181 A003005

Adjacent sequences: A247970 A247971 A247972 * A247974 A247975 A247976

Keyword

nonn,easy

Author

Clark Kimberling, Sep 28 2014

Status

approved