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A248106 Least k such that ((k+1)/(k-1))^k - e^2 < 1/n^2. 3

%I #4 Oct 02 2014 22:36:14

%S 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 56,58,60,63,65,67,69,72,74,76,78,80,83,85,87,89,92,94,96,98,100,103,

%U 105,107,109,111,114,116,118,120,123,125,127,129,131,134,136

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

%C 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 Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 14.

%H Clark Kimberling, <a href="/A248106/b248106.txt">Table of n, a(n) for n = 1..1000</a>

%e Approximations are shown here:

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

%e 2 ... 1.610943901 ............. 0.25

%e 3 ... 0.610943901 ............. 0.11111

%e 4 ... 0.326993283 ............. 0.0625

%e 5 ... 0.204693901 ............. 0.04

%e 6 ... 0.140479901 ............. 0.02777

%e a(2) = 5 because p(5) - e^2 < 1/4 < p(4) - e^2.

%t z = 1200; p[k_] := p[k] = ((k + 1)/(k - 1))^k; (* Finch p. 15 *);

%t N[Table[p[n] - E^2, {n, 2, z/20}]]

%t f[n_] := f[n] = Select[Range[z], # > 1 && p[#] - E^2 < 1/n^2 &, 1]

%t u = Flatten[Table[f[n], {n, 1, z/4}]] (* A248106 *)

%Y Cf. A248103, A248111.

%K nonn,easy

%O 1,1

%A _Clark Kimberling_, Oct 02 2014

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