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

%I #5 Oct 02 2014 22:35:47

%S 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 19,20,21,21,22,23,23,24,25,25,26,27,27,28,29,29,30,31,31,32,33,33,34,

%U 35,36,36,37,38,38,39,40,40,41,42,42,43,44,44,45,46,46

%N Least k such that ((2k+1)/(2k-1))^k < 1/(2n^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="/A248103/b248103.txt">Table of n, a(n) for n = 1..1000</a>

%e Approximations are shown here:

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

%e 1 ... 0.281718 ............ 0.5

%e 2 ... 0.0595959 ........... 0.125

%e 3 ... 0.0257182 ........... 0.05555

%e 4 ... 0.0143296 ........... 0.3125

%e a(4) = 3 because p(4) - e < 1/32 < p(3) - e.

%t z = 1200; p[k_] := p[k] = ((2 k + 1)/(2 k - 1))^k;

%t N[Table[p[n] - E, {n, 1, z/8}]]

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

%t u = Flatten[Table[f[n], {n, 1, z/10}]] (* A248103 *)

%Y Cf. A248106, A248111.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Oct 02 2014