Dick Hess, Puzzles from around the world, 1997, Problem 116:
Extension: Torsten Sillke, FRA, Nov. 1997
Using two positive real c and d (except for (c,d)=(1,1)) you can approximate
each 0= x > 1/2**((n+1)/q) with q = 2**k.
Then (k, n) approximates from below.
Therefore we choose n = round( - 2**k log2(x) ).
If k increases the approximation become arbitrary close.
For c > 1 we would get
1/2**(n/q) <= x/p < 1/2**((n-1)/q) with q = 2**k and p = log(c)^(1/q).
As p = 1 + 1/q log(log(c)) + O(q^2) we can forget c for large k.
Case: c = exp(-1), d=1
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VVVVVVV 1 - VVVVVVVV 1/e
k sqrts n sqrts
can approximate x in (0,1) with arbitrary accuracy.
So if we fix k, n will be set as
1/2**(n/q) >= x > 1/2**((n+1)/q) with q = 2**k.
For c < 1 we would get
1/2**(n/q) <= x/p < 1/2**((n-1)/q) with q = 2**k and p = log(1/c)^(1/q).
As p = 1 + 1/q log(log(1/c)) + O(q^2) we can forget c for large k.
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Torsten Sillke, 17 Nov 1997, FRA