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A247908
Least number k such that e - 2*k/u(2*k) < 1/n^n, where u is defined as in Comments.
10
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
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