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There are twice as many primes as even semiprimes:
Select some large number, n. Between 0 and n there are p primes. Of these p primes,
(1) all those greater than n/2 will have no even semiprime counterpart.
(2) all those less than or equal to n/2 will have exactly one even semiprime counterpart.
By the prime number counting theorem, the density of primes up to n is Pi(n) ~ n/ln(n)
Accordingly the density of primes up to n/2 is Pi(n/2) ~ (n/2)/ln(n/2))
Then the ratio of primes to even semiprimes is a=Pi(n/2)/ Pi(n)~((n/2)/ln(n/2)))/(n/ln(n)), so a = (log(n))/(2 log(n/2)).
In the limit:Limit[Log[xn]/(2 Log[n/2]), {n -> -Infinity, n -> Infinity}]={1/2, 1/2}
So there are two GIRLs for every BOY.
I have posted some statistics (computed from entries in Sloane) in the comments section. |
