I don’t know if you’re still interested in another answer after nearly five years, but since I found this question through a Google search and noticed you haven’t got the definitive answer yet (or rather, Omphaloskeptic gave it but it may not have been concrete enough with actual numbers): the explanation comes from continued fractions.
Basically, the continued fraction representation of the number is the following: write the number as [an integer] + [fractional part], then write the [fractional part] as the reciprocal of a number, and repeat. So for instance,
3.14159 = 3 + 0.14159 = 3 + 1/7.0625
= 3 + 1/(7 + 0.0625) = 3 + 1/(7 + 1/15.9966)
= 3 + 1/(7 + 1/(15 + 0.9966)) = 3 + 1/(7 + 1/(15 + 1/1.0034))
= 3 + 1/(7 + 1/(15 + 1/(1 + 0.0034))) = 3 + 1/(7 + 1/(15 + 1/(1 + 1/292.6346)))
= 3 + 1/(7 + 1/(15 + 1/(1 + (1/(292 + ...))))
(I’ve cheated a bit, using the actual value of π at each step instead of 3.14159.)
This representation is called the continued fraction. Since this is unwieldy to write, another notation used is:
π = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14, 2…]
Now all the fractions you listed come from the continued fraction, in the following way. Truncate the continued fraction, and replace the final term successively by all numbers from half the final term to the final term itself. Thus, the approximations you get are precisely:
3/1 = [3]
13/4 = [3; 4]
16/5 = [3; 5]
19/6 = [3; 6]
22/7 = [3; 7]
179/57 = [3; 7, 8]
201/64 = [3; 7, 9]
223/71 = [3; 7, 10]
245/78 = [3; 7, 11]
267/85 = [3; 7, 12]
289/92 = [3; 7, 13]
311/99 = [3; 7, 14]
333/106 = [3; 7, 15]
355/113 = [3; 7, 15, 1]
52163/16604 = [3; 7, 15, 1, 146]
52518/16717 = [3; 7, 15, 1, 147]
*… (terms from [3; 7, 15, 1, 148] to [3; 7, 15, 1, 291])...*
103993/33102 = [3; 7, 15, 1, 292]
104348/33215 = [3; 7, 15, 1, 292, 1]
...
etc. So the “island” of 146 fractions you see from 52163/16604 to 103993/33102 is because of the large term 292 in the continued fraction of π, and similarly the “island” of 7 fractions from 179/57 to 333/106 is because of the large term 15, etc.
Using continued fractions is also the best way of finding all these best fractional approximations, immensely faster than looping through all denominators.
Hope this helps (and sorry for bumping up an old thread),