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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Sep 28 2014
STATUS
approved
