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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 (list; graph; refs; listen; history; text; internal format)
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 {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

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

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

Cf. A247908, A247911, A247915, A247916.

Sequence in context: A039235 A247782 A039179 * A184518 A304802 A288228

Adjacent sequences:  A247911 A247912 A247913 * A247915 A247916 A247917

KEYWORD

nonn,easy

AUTHOR

Clark Kimberling, Sep 27 2014

STATUS

approved

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Last modified June 15 22:06 EDT 2021. Contains 345053 sequences. (Running on oeis4.)