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

Author

Clark Kimberling, Sep 27 2014

Keywords

• nonn
• easy

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.

Examples

			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).
		

References

• Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 19.

Links

• Clark Kimberling, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A247909, A247910, A247911, A247914.

Programs

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 *)