The Copeland-Erdős constant is not strongly normal in base 10

No: the Copeland-Erdős constant is not simply strongly normal in base 10, hence not strongly normal in base 10, and the failure is already visible in the digit 0. The proof compares the zero count at two positions in the expansion. Every prime in [10^(6n), 10^(6n) + 10^(5n)) begins with 1 followed by n zeros, and Ingham's 1937 short-interval theorem with exponent 5/6 supplies about 10^(5n)/(6n log 10) such primes, so the zero excess across that stretch is of order 10^(5n), while simple strong normality would cap the discrepancy at O(sqrt(N log log N)) with N of order 10^(6n), that is at order 10^(3n). Belshaw and Borwein's 2013 Champernowne argument is the model; Ingham replaces the trivial prime count. Strong normality in other bases is not addressed.

No: the Copeland-Erdős constant is not simply strongly normal in base 10, hence not strongly normal in base 10, and the failure is already visible in the digit 0. The proof compares the zero count at two positions in the expansion. Every prime in [10^(6n), 10^(6n) + 10^(5n)) begins with 1 followed by n zeros, and Ingham's 1937 short-interval theorem with exponent 5/6 supplies about 10^(5n)/(6n log 10) such primes, so the zero excess across that stretch is of order 10^(5n), while simple strong normality would cap the discrepancy at O(sqrt(N log log N)) with N of order 10^(6n), that is at order 10^(3n). Belshaw and Borwein's 2013 Champernowne argument is the model; Ingham replaces the trivial prime count. Strong normality in other bases is not addressed.