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A248103 Least k such that ((2k+1)/(2k-1))^k < 1/(2n^2).

%I A248103 #5 Oct 02 2014 22:35:47
%S A248103 1,2,3,3,4,5,5,6,7,7,8,9,9,10,11,11,12,13,13,14,15,15,16,17,17,18,19,
%T A248103 19,20,21,21,22,23,23,24,25,25,26,27,27,28,29,29,30,31,31,32,33,33,34,
%U A248103 35,36,36,37,38,38,39,40,40,41,42,42,43,44,44,45,46,46
%N A248103 Least k such that ((2k+1)/(2k-1))^k < 1/(2n^2).
%C A248103 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 A248103 Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, p. 14.
%H A248103 Clark Kimberling, <a href="/A248103/b248103.txt">Table of n, a(n) for n = 1..1000</a>
%e A248103 Approximations are shown here:
%e A248103 n ... ((2n+1)/(2n-1))^n ... 1/(2*n^2)
%e A248103 1 ... 0.281718 ............ 0.5
%e A248103 2 ... 0.0595959 ........... 0.125
%e A248103 3 ... 0.0257182 ........... 0.05555
%e A248103 4 ... 0.0143296 ........... 0.3125
%e A248103 a(4) = 3 because p(4) - e < 1/32 < p(3) - e.
%t A248103 z = 1200; p[k_] := p[k] = ((2 k + 1)/(2 k - 1))^k;
%t A248103 N[Table[p[n] - E, {n, 1, z/8}]]
%t A248103 f[n_] := f[n] = Select[Range[z], p[#] - E < 1/(2 n^2) &, 1]
%t A248103 u = Flatten[Table[f[n], {n, 1, z/10}]]  (* A248103 *)
%Y A248103 Cf. A248106, A248111.
%K A248103 nonn,easy
%O A248103 1,2
%A A248103 _Clark Kimberling_, Oct 02 2014