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A248106 Least k such that ((k+1)/(k-1))^k - e^2 < 1/n^2.

%I A248106 #4 Oct 02 2014 22:36:14
%S A248106 3,5,7,9,12,14,16,18,20,23,25,27,29,32,34,36,38,40,43,45,47,49,52,54,
%T A248106 56,58,60,63,65,67,69,72,74,76,78,80,83,85,87,89,92,94,96,98,100,103,
%U A248106 105,107,109,111,114,116,118,120,123,125,127,129,131,134,136
%N A248106 Least k such that ((k+1)/(k-1))^k - e^2 < 1/n^2.
%C A248106 In general, for fixed positive m, the limit of ((m*x+1)/(m*x-1))^x is e^(2/m), as illustrated by A248103, A248106, A248111.
%D A248106 Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 14.
%H A248106 Clark Kimberling, <a href="/A248106/b248106.txt">Table of n, a(n) for n = 1..1000</a>
%e A248106 Approximations are shown here:
%e A248106 n ... ((n+1)/(n-1))^n - e^2 ... 1/n^2
%e A248106 2 ... 1.610943901 ............. 0.25
%e A248106 3 ... 0.610943901 ............. 0.11111
%e A248106 4 ... 0.326993283 ............. 0.0625
%e A248106 5 ... 0.204693901 ............. 0.04
%e A248106 6 ... 0.140479901 ............. 0.02777
%e A248106 a(2) = 5 because p(5) - e^2 < 1/4 < p(4) - e^2.
%t A248106 z = 1200; p[k_] := p[k] = ((k + 1)/(k - 1))^k; (* Finch p. 15 *);
%t A248106 N[Table[p[n] - E^2, {n, 2, z/20}]]
%t A248106 f[n_] := f[n] = Select[Range[z], # > 1 && p[#] - E^2 < 1/n^2 &, 1]
%t A248106 u = Flatten[Table[f[n], {n, 1, z/4}]]  (* A248106 *)
%Y A248106 Cf. A248103, A248111.
%K A248106 nonn,easy
%O A248106 1,1
%A A248106 _Clark Kimberling_, Oct 02 2014