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Least number k such that (2*k+1)/u(2*k+1) - e < 1/n^n, where u is defined as in Comments.
8

%I #4 Sep 27 2014 19:02:50

%S 1,2,2,3,3,4,5,5,6,6,7,8,8,9,10,10,11,11,12,13,13,14,14,15,16,16,17,

%T 17,18,19,19,20,20,21,22,22,23,23,24,25,25,26,26,27,28,28,29,29,30,31,

%U 31,32,32,33,34,34,35,35,36,37,37,38,38,39,39,40,41,41

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

%C 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 {2,3} for n >= 1, that d(n+1) - d(n) is in {-1,0,1}, and that similar bounds hold for higher differences.

%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 19.

%H Clark Kimberling, <a href="/A247911/b247911.txt">Table of n, a(n) for n = 1..1000</a>

%e Approximations for the first few terms (2*n+1)/u(2*n+1) - e and 1/n^n are shown here:

%e n ... (2*n+1)/u(2*n+1)-e .. 1/n^n

%e 1 ... 0.28171817 .......... 1

%e 2 ... 0.0089908988 ........ 0.25

%e 3 ... 0.0001647734 ........ 0.037037

%e 4 ... 0.0000018654 ........ 0.00390625

%e 5 ... 0.0000000143 ........ 0.00032000

%e a(2) = 2 because 5/u(5) - e < 1/3^3 < 3/u(3).

%t $RecursionLimit = 1000; $MaxExtraPrecision = 1000;

%t z = 300; u[1] = 0; u[2] = 1; u[n_] := u[n] = u[n - 1] + u[n - 2]/(n - 2);

%t f[n_] := f[n] = Select[Range[z], (2 # + 1)/u[2 # + 1] - E < n^-n &, 1];

%t u = Flatten[Table[f[n], {n, 1, z}]] (* A247911 *)

%t w = Differences[u]

%t Flatten[Position[w, 0]] (* A247912 *)

%t Flatten[Position[w, 1]] (* A247913 *)

%Y Cf. A247908, A247912, A247913, A247914.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Sep 27 2014