

A247914


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


8



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
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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 {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



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



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



