A247914 Least number k such that |(k+1)/u(k+1) - e| < 1/n^n, where u is defined as in Comments.

1, 3, 4, 5, 6, 7, 9, 10, 11, 12, 14, 15, 16, 17, 19, 20, 21, 22, 23, 25, 26, 27, 28, 30, 31, 32, 33, 34, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 48, 49, 50, 51, 52, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 67, 68, 69, 70, 71, 73, 74, 75, 76, 77, 78, 80

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 {1,2} 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 |(n+1)/u(n+1) - e| and 1/n^n are shown here:
n ... |(n+1)/u(n+1)-e| .. 1/n^n
1 ... 0.7182818285 ...... 1
2 ... 0.28171817 ........ 0.25
3 ... 0.051615161 ....... 0.037037
4 ... 0.0089908988 ...... 0.00390625
5 ... 0.0013006963 ...... 0.00032000
a(2) = 3 because |4/u(4) - e| < 1/2^2 < |3/u(3) - e|.

Mathematica

$RecursionLimit = Infinity; $MaxExtraPrecision = Infinity;
z = 500; 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], Abs[(# + 1)/u[# + 1] - E] < n^-n &, 1];
u = Flatten[Table[f[n], {n, 1, z}]]  (* A247914 *)
w = Differences[u]
f1 = Flatten[Position[w, 1]] (* A247915 *)
f2 = Flatten[Position[w, 2]] (* A247916 *)

Crossrefs

Cf. A247908, A247911, A247915, A247916.

Sequence in context: A039235 A247782 A039179 * A184518 A304802 A288228

Adjacent sequences: A247911 A247912 A247913 * A247915 A247916 A247917

Keyword

nonn,easy

Author

Clark Kimberling, Sep 27 2014

Status

approved