%N Reduced numerators of the ratios of Pi(2^n+1)/Pi(2^(n)).
%C The ratios Pi(2^n)/Pi(2^(n-1)) ~ 2. This follows directly from the Prime Number Theorem: Pi(x) ~ x/log(x). If we substitute b for 2, we have the general asymptotic Pi(b^n)/Pi(b^(n-1)) ~ b for any base b. For example, using Li(x) ~ Pi(x), Li(2^10000)/Li(2^9999) = 1.9997999711... Similarly, for b=13, Li(13^100000)/Li(13^99999) = 12.9998699994...Of course direct substitution of x=b^n in the PNT will, after some manipulation and taking limits, give us the exact limit b.
%F Pi(n) is the number of primes less than or equal to n.
%e Pi(2^12)/Pi(2^11) = 564/309 = 188/103. So 188 is in the sequence.
%o (PARI) /* Copy and paste the table in A007053 to a text file say, c:\work\test.txt.
%o Edit out the index leaving only a left wall of values. Start a new gp session. Read the file into gp: gp > \r c:/work/test.txt. This fills the %1 to %76 pari variables with successive primes <= 2^n
%Y Cf. A007053
%A _Cino Hilliard_, Feb 26 2009