Is the Copeland–Erdős constant a random number? How is it normal?

Question

The Champernowne constant is not random. Is the Copeland–Erdős constant random? Also if Copeland–Erdős number is normal, then shouldnt the number of $5$s and even digits be low because they cannot appear in as the last digit of prime numbers? So how is it normal if certain digits occur less?

Answer 1

For addition to the answer above:

Copeland–Erdős constant is not a Martin-Lof random number. It is a computable real number, which means there exists an algorithm to compute all the digits in the number. Computable number is not random (you can't call something random when you know there is a rule to generate it). Most real numbers are uncomputable, but most real numbers we deal with are computable. It is an open question whether there is any "natural phenomenon" leading to an uncomputable real number (or random number).

Answer 2

I can't tell you it is normal or random (not even sure what random means in this context,) but your argument is not enough to show that it won't be normal, and the Wikipedia article for the number says it is normal in base $10$.

Consider, instead the real number:

$$0.12345678910111213141516171819202122\dots$$

Which is created by concantenating the base $10$ representation of all the natural numbers.

Now, let $C_i(n)$ be defined as number of occurrences of digit $i$ in the first $n$ digits of this number. Since $0$ cannot be the first digit of a natural number, we'd rightly expect that $C_i(n)<\frac{n}{10}$ for most $n$. But that would not prevent $$\lim_{n\to\infty} \frac{C_i(n)}{n} = \frac{1}{10}$$

In your case, while the last digits of primes cannot be $5$ or an even digit, the actual frequency (or density) of last digits is so small that this should have zero effect on the limit.

At heart, any subset of the digits of density $0$ has no effect on normality - we can change them to anything we want and the new number is normal if and only if the old number was normal.