%I #4 Sep 27 2014 19:03:08
%S 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,
%T 32,33,34,36,37,38,39,40,42,43,44,45,46,48,49,50,51,52,54,55,56,57,58,
%U 59,61,62,63,64,65,67,68,69,70,71,73,74,75,76,77,78,80
%N Least number k such that (k+1)/u(k+1)  e < 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 {1,2} 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="/A247914/b247914.txt">Table of n, a(n) for n = 1..1000</a>
%e Approximations for the first few terms (n+1)/u(n+1)  e and 1/n^n are shown here:
%e n ... (n+1)/u(n+1)e .. 1/n^n
%e 1 ... 0.7182818285 ...... 1
%e 2 ... 0.28171817 ........ 0.25
%e 3 ... 0.051615161 ....... 0.037037
%e 4 ... 0.0089908988 ...... 0.00390625
%e 5 ... 0.0013006963 ...... 0.00032000
%e a(2) = 3 because 4/u(4)  e < 1/2^2 < 3/u(3)  e.
%t $RecursionLimit = Infinity; $MaxExtraPrecision = Infinity;
%t z = 500; 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], Abs[(# + 1)/u[# + 1]  E] < n^n &, 1];
%t u = Flatten[Table[f[n], {n, 1, z}]] (* A247914 *)
%t w = Differences[u]
%t f1 = Flatten[Position[w, 1]] (* A247915 *)
%t f2 = Flatten[Position[w, 2]] (* A247916 *)
%Y Cf. A247908, A247911, A247915, A247916.
%K nonn,easy
%O 1,2
%A _Clark Kimberling_, Sep 27 2014
