The first such distribution found is π(N) ~
N
/
log(N)
, where π(N) is the...
Mathematics, 27.02.2020 05:53 jadentdaniels
The first such distribution found is π(N) ~
N
/
log(N)
, where π(N) is the prime-counting function and log(N) is the natural logarithm of N. This means that for large enough N, the probability that a random integer not greater than N is prime is very close to 1 / log(N). Consequently, a random integer with at most 2n digits (for large enough n) is about half as likely to be prime as a random integer with at most n digits. For example, among the positive integers of at most 1000 digits, about one in 2300 is prime (log(101000) ≈ 2302.6), whereas among positive integers of at most 2000 digits, about one in 4600 is prime (log(102000) ≈ 4605.2). In other words, the average gap between consecutive prime numbers among the first N integers is roughly log(N).[1]
Answers: 2
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