A247971 Least k such that 4*k/v(2*k)^2 - Pi < 1/n, where the sequence v is defined in Comments.

1, 2, 3, 4, 5, 5, 6, 7, 8, 8, 9, 10, 11, 12, 12, 13, 14, 15, 16, 16, 17, 18, 19, 19, 20, 21, 22, 23, 23, 24, 25, 26, 27, 27, 28, 29, 30, 30, 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

1,2

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. 19.

Links

Clark Kimberling, Table of n, a(n) for n = 1..1000

Example

Approximations for the first few terms w(n) = 4*n/v(2*n)^2 - Pi and 1/n:
n ... 4*n/v(2*n)^2-Pi ... 1/n
1 ... 0.858407 .......... 1
2 ... 0.413963 .........  0.5
3 ... 0.271741 .......... 0.333333
4 ... 0.202081 .......... 0.25
5 ... 0.160801 .......... 0.2
6 ... 0.133508 .......... 0.166666
a(2) = 2 because w(2) < 1/2 < 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[4 n/(v[2 n]^2) - Pi], N[1/n]}, {n, 1, 10}]]
f[n_] := f[n] = Select[Range[z], 4 #/(v[2 #]^2) - Pi < 1/n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]] (* A247971 *)
d = Differences[u]
v = Flatten[Position[d, 0]] (* A247972 *)

Crossrefs

Cf. A247972, A247973, A247974.

Sequence in context: A254828 A091863 A163296 * A317334 A331266 A172103

Adjacent sequences: A247968 A247969 A247970 * A247972 A247973 A247974

Keyword

nonn,easy

Author

Clark Kimberling, Sep 28 2014

Status

approved