A247908 Least number k such that e - 2*k/u(2*k) < 1/n^n, where u is defined as in Comments.

1, 2, 3, 3, 4, 4, 5, 6, 6, 7, 8, 8, 9, 9, 10, 11, 11, 12, 12, 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, 30, 31, 32, 32, 33, 33, 34, 35, 35, 36, 36, 37, 38, 38, 39, 39, 40, 41, 41, 42

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 {0,1} for n >= 1, that d(n+1) - d(n) is in {2,3}, 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 of e - 2*n/u(2*n) and 1/n^n are shown here:
n ... e-2*n/u(2*n) .... 1/n^n
1 ... 0.71828 ........  1
2 ... 0.0516152 ....... 0.25
3 ... 0.0013007 ....... 0.037037
4 ... 0.0000184967 .... 0.00390625
a(2) = 2 because e - 4/u(4) < 1/2^2 < e - 2/u(2).

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], E - 2 #/u[2 #] < 1/n^n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A247908 *)
w = Differences[u]
Flatten[Position[w, 0]] (* A247909 *)
Flatten[Position[w, 1]] (* A247910 *)

Crossrefs

Cf. A247909, A247910, A247911, A247914.

Sequence in context: A079952 A055930 A090638 * A057363 A073869 A060143

Adjacent sequences: A247905 A247906 A247907 * A247909 A247910 A247911

Keyword

nonn,easy

Author

Clark Kimberling, Sep 27 2014

Status

approved