

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
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OFFSET

1,2


COMMENTS

The sequence u is define recursively by u(n) = u(n1) + u(n2)/(n2), 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 ... e2*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



STATUS

approved



