A247911 Least number k such that (2*k+1)/u(2*k+1) - e < 1/n^n, where u is defined as in Comments.

1, 2, 2, 3, 3, 4, 5, 5, 6, 6, 7, 8, 8, 9, 10, 10, 11, 11, 12, 13, 13, 14, 14, 15, 16, 16, 17, 17, 18, 19, 19, 20, 20, 21, 22, 22, 23, 23, 24, 25, 25, 26, 26, 27, 28, 28, 29, 29, 30, 31, 31, 32, 32, 33, 34, 34, 35, 35, 36, 37, 37, 38, 38, 39, 39, 40, 41, 41

Offset

1,2

Comments

The sequence u is define recursively by u(n) = u(n-1) + u(n-2)/(n-2), with u(1) = 0 and u(2) = 1. Let d(n) = a(n+1) - a(n). It appears that d(n) is in {2,3} for n >= 1, that d(n+1) - d(n) is in {-1,0,1}, and that similar bounds hold for higher differences.

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 (2*n+1)/u(2*n+1) - e and 1/n^n are shown here:
n ... (2*n+1)/u(2*n+1)-e .. 1/n^n
1 ... 0.28171817 .......... 1
2 ... 0.0089908988 ........ 0.25
3 ... 0.0001647734 ........ 0.037037
4 ... 0.0000018654 ........ 0.00390625
5 ... 0.0000000143 ........ 0.00032000
a(2) = 2 because 5/u(5) - e < 1/3^3 < 3/u(3).

Mathematica

$RecursionLimit = 1000; $MaxExtraPrecision = 1000;
z = 300; u[1] = 0; u[2] = 1; u[n_] := u[n] = u[n - 1] + u[n - 2]/(n - 2);
f[n_] := f[n] = Select[Range[z], (2 # + 1)/u[2 # + 1] - E < n^-n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A247911 *)
w = Differences[u]
Flatten[Position[w, 0]] (* A247912 *)
Flatten[Position[w, 1]] (* A247913 *)

Crossrefs

Cf. A247908, A247912, A247913, A247914.

Sequence in context: A195119 A077467 A005378 * A103355 A347707 A029092

Adjacent sequences: A247908 A247909 A247910 * A247912 A247913 A247914

Keyword

nonn,easy

Author

Clark Kimberling, Sep 27 2014

Status

approved