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