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A set of numbers is said to satisfy Benford's law if the leading digit d (d ∈ {1, ..., 9}) occurs with probability,

$$ P(d)=\log_{10}(d+1)-\log_{10}d,$$

how do you prove that the prime numbers do not satisfies Benford's law?

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What probability distribution are you using on the prime numbers? –  Chris Eagle Dec 29 '12 at 17:06
    
P(d) is just the frequence of the digit d in the prime numbers, is the limit of the number of primes with leading digit d( looking the first n primes) / n –  user52188 Dec 29 '12 at 17:22
    
Technically, not a probability - there is no uniform probability on any countable set. This term here is "density." –  Thomas Andrews Dec 29 '12 at 17:26
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you have every reason. Now I think that the question is clear. –  user52188 Dec 29 '12 at 19:33

1 Answer 1

up vote 3 down vote accepted

Essentially you face the same problem as the question of whether the natural numbers satisfy Benford's law: if you look at the pattern up to some maximum value $m$ then the distribution depends on what $m$ is, and there is no convergence as $m$ increases because the first digit of $m$ is not stable, though there is a degree of uniformity in the distribution.

From the prime number theorem, the number of primes between $d \times 10^k$ and $(d+1) \times 10^k$ is about $\dfrac{10^k}{\ln((d+\frac12) \times 10^k)}$ while the number between $(d+1) \times 10^k$ and $(d+2) \times 10^k$ is about $\dfrac{10^k}{\ln((d+\frac32) \times 10^k)}$.

The ratio of these tends towards $1$ as $k$ increases, so if you look at the distribution of first digits of primes below $10^k$ then this tends towards uniform as $k$ increases. Nothing like Benford's law.

But if you look at at the distribution of first digits of primes below $3 \times 10^k$, then for large $k$ there will be almost ten times as high a proportion starting with $1$ (or $2$) than with $3$ (or each of the higher digits).

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