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
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Sep 27 2014
STATUS
approved