%I #6 Sep 27 2014 19:02:25
%S 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,
%T 18,19,19,20,20,21,22,22,23,23,24,25,25,26,26,27,28,28,29,29,30,30,31,
%U 32,32,33,33,34,35,35,36,36,37,38,38,39,39,40,41,41,42
%N Least number k such that e  2*k/u(2*k) < 1/n^n, where u is defined as in Comments.
%C 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.
%D Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 19.
%H Clark Kimberling, <a href="/A247908/b247908.txt">Table of n, a(n) for n = 1..1000</a>
%e Approximations for the first few terms of e  2*n/u(2*n) and 1/n^n are shown here:
%e n ... e2*n/u(2*n) .... 1/n^n
%e 1 ... 0.71828 ........ 1
%e 2 ... 0.0516152 ....... 0.25
%e 3 ... 0.0013007 ....... 0.037037
%e 4 ... 0.0000184967 .... 0.00390625
%e a(2) = 2 because e  4/u(4) < 1/2^2 < e  2/u(2).
%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], E  2 #/u[2 #] < 1/n^n &, 1];
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A247908 *)
%t w = Differences[u]
%t Flatten[Position[w, 0]] (* A247909 *)
%t Flatten[Position[w, 1]] (* A247910 *)
%Y Cf. A247909, A247910, A247911, A247914.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Sep 27 2014
